Answer: The particular solution is

\(x = 25\sin t - 5\cos t + 5e^{-t/3}\).

We solve the differential equation

\(9\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+6\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=50\sin t\)

with the conditions \(x(0)=0\) and \(x'(0)=0\).

First divide through by 9:

\(x''+\frac{2}{3}x'+\frac{1}{9}x=\frac{50}{9}\sin t.\)

We look for a particular solution of the form

\(x_p=A\sin t+B\cos t.\)

Then

\(x_p'=A\cos t-B\sin t,\)

\(x_p''=-A\sin t-B\cos t.\)

Substituting into the differential equation gives

\(9(-A\sin t-B\cos t)+6(A\cos t-B\sin t)+(A\sin t+B\cos t)=50\sin t.\)

Collecting coefficients of \(\sin t\) and \(\cos t\):

\(( -8A-6B )\sin t + ( 6A-8B )\cos t = 50\sin t.\)

So

\(-8A-6B=50,\qquad 6A-8B=0.\)

From \(6A-8B=0\), \(3A=4B\), so \(A=\frac{4}{3}B\).

Substitute into the first equation:

\(-8\left(\frac{4}{3}B\right)-6B=50\)

\(\Rightarrow -\frac{32}{3}B-\frac{18}{3}B=50\)

\(\Rightarrow -\frac{50}{3}B=50\)

\(\Rightarrow B=-3\), hence \(A=-4\).

So a particular solution is \(x_p=-4\sin t-3\cos t\).

The complementary function satisfies

\(9r^2+6r+1=0\),

which gives \((3r+1)^2=0\), so \(r=-\frac13\) is a repeated root. Therefore

\(x_c=(C_1+C_2t)e^{-t/3}.\)

Hence the general solution is

\(x=(C_1+C_2t)e^{-t/3}-4\sin t-3\cos t.\)

Now use the conditions.

At \(t=0\), \(x=0\):

\(C_1-3=0\), so \(C_1=3\).

Differentiate:

\(x' = \left(C_2-\frac13(C_1+C_2t)\right)e^{-t/3}-4\cos t+3\sin t.\)

At \(t=0\), \(x'=0\):

\(C_2-\frac13C_1-4=0.\)

Since \(C_1=3\), this gives \(C_2-1-4=0\), so \(C_2=5\).

Therefore the required solution is

\(x=(3+5t)e^{-t/3}-4\sin t-3\cos t.\)

Answer: (i) \(x=e^{-t}(9\cos3t+2\sin3t)-6\cos3t+\sin3t\).

(ii) For large positive \(t\), \(x\approx-6\cos3t+\sin3t=\sqrt{37}\sin(3t-\phi)\), where \(\tan\phi=6\).

The auxiliary equation for the complementary function is

\(m^2+2m+10=0,\)

so

\(m=-1\pm3i.\)

Hence

\(x_c=e^{-t}(A\cos3t+B\sin3t).\)

For a particular integral, try

\(x_p=p\cos3t+q\sin3t.\)

Then

\(x_p'=-3p\sin3t+3q\cos3t,\qquad x_p''=-9p\cos3t-9q\sin3t.\)

Substitution gives

\((p+6q)\cos3t+(q-6p)\sin3t=37\sin3t.\)

Therefore

\(p+6q=0,\qquad q-6p=37.\)

Solving gives \(p=-6\) and \(q=1\). Thus

\(x=e^{-t}(A\cos3t+B\sin3t)-6\cos3t+\sin3t.\)

Use \(x(0)=3\):

\(A-6=3,\)

so \(A=9\). Differentiating,

\(x'=e^{-t}(-3A\sin3t+3B\cos3t)-e^{-t}(A\cos3t+B\sin3t)+18\sin3t+3\cos3t.\)

Use \(x'(0)=0\):

\(3B-A+3=0.\)

Since \(A=9\), this gives \(B=2\). Therefore

\(x=e^{-t}(9\cos3t+2\sin3t)-6\cos3t+\sin3t.\)

For large positive \(t\), \(e^{-t}\to0\), so

\(x\approx-6\cos3t+\sin3t.\)

Write this as

\(\sqrt{37}\left(\frac1{\sqrt{37}}\sin3t-\frac6{\sqrt{37}}\cos3t\right).\)

If \(\cos\phi=\frac1{\sqrt{37}}\) and \(\sin\phi=\frac6{\sqrt{37}}\), then \(\tan\phi=6\), and

\(x\approx\sqrt{37}\sin(3t-\phi).\)

Answer: \(x=\frac{4}{3}e^{t}-\frac{1}{12}e^{-2t}-t^{2}-\frac{3}{2}t-\frac{5}{4}\).

We solve

\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+\dfrac{\mathrm{d}x}{\mathrm{d}t}-2x=2t^2+t-1\),

with \(x=0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}=0\) when \(t=0\).

First solve the homogeneous equation

\(x''+x'-2x=0\).

The auxiliary equation is

\(m^2+m-2=0\).

Factorising,

\((m-1)(m+2)=0\),

so the roots are \(m=1\) and \(m=-2\).

Hence the complementary function is

\(x_c=Ae^t+Be^{-2t}\).

For a particular integral, since the right-hand side is a quadratic, try

\(x_p=pt^2+qt+r\).

Then

\(x_p'=2pt+q\),

\(x_p''=2p\).

Substitute into the differential equation:

\(2p+(2pt+q)-2(pt^2+qt+r)=2t^2+t-1\).

Expand and collect terms:

\(-2pt^2+(2p-2q)t+(2p+q-2r)=2t^2+t-1\).

Equating coefficients gives

\(-2p=2\), so \(p=-1\),

\(2p-2q=1\),

\(2p+q-2r=-1\).

Using \(p=-1\):

\(-2-2q=1\), so \(q=-\frac{3}{2}\).

Then

\(-2-\frac{3}{2}-2r=-1\),

so \(-\frac{7}{2}-2r=-1\),

\(-2r=\frac{5}{2}\),

\(r=-\frac{5}{4}\).

Therefore

\(x_p=-t^2-\frac{3}{2}t-\frac{5}{4}\).

So the general solution is

\(x=Ae^t+Be^{-2t}-t^2-\frac{3}{2}t-\frac{5}{4}\).

Differentiate:

\(x'=Ae^t-2Be^{-2t}-2t-\frac{3}{2}\).

Now apply the conditions at \(t=0\).

From \(x(0)=0\):

\(A+B-\frac{5}{4}=0\).

From \(x'(0)=0\):

\(A-2B-\frac{3}{2}=0\).

Solve these simultaneously.

From the first, \(A=\frac{5}{4}-B\).

Substitute into the second:

\(\frac{5}{4}-B-2B-\frac{3}{2}=0\).

\(-3B-\frac{1}{4}=0\), so \(B=-\frac{1}{12}\).

Then

\(A=\frac{5}{4}+\frac{1}{12}=\frac{16}{12}=\frac{4}{3}\).

Hence the particular solution satisfying the given conditions is

\(x=\frac{4}{3}e^t-\frac{1}{12}e^{-2t}-t^2-\frac{3}{2}t-\frac{5}{4}\).

Answer: \(x=\mathrm{e}^{-\frac14 t}\left(-\frac19\cos\frac{\sqrt{15}}4 t+\frac{1}{3\sqrt{15}}\sin\frac{\sqrt{15}}4 t\right)+\frac19\mathrm{e}^{-t}.\)

We solve

\(6\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+3\dfrac{\mathrm{d}x}{\mathrm{d}t}+6x=\mathrm{e}^{-t}\), with \(x(0)=0\) and \(x'(0)=0\).

First solve the homogeneous equation:

\(6x''+3x'+6x=0.\)

Its auxiliary equation is

\(6m^2+3m+6=0.\)

So

\(m=\dfrac{-3\pm\sqrt{9-144}}{12}=-\frac14\pm\frac{\sqrt{15}}4 i.\)

Hence the complementary function is

\(x_c=\mathrm{e}^{-\frac14 t}\left(A\cos\frac{\sqrt{15}}4 t+B\sin\frac{\sqrt{15}}4 t\right).\)

For a particular integral, try

\(x_p=k\mathrm{e}^{-t}.\)

Then

\(x_p'=-k\mathrm{e}^{-t},\qquad x_p''=k\mathrm{e}^{-t}.\)

Substituting into the differential equation gives

\(6k\mathrm{e}^{-t}-3k\mathrm{e}^{-t}+6k\mathrm{e}^{-t}=\mathrm{e}^{-t},\)

so \(9k=1\), hence \(k=\frac19\).

Therefore

\(x=\mathrm{e}^{-\frac14 t}\left(A\cos\frac{\sqrt{15}}4 t+B\sin\frac{\sqrt{15}}4 t\right)+\frac19\mathrm{e}^{-t}.\)

Differentiate:

\(x'=\frac{\sqrt{15}}4\mathrm{e}^{-\frac14 t}\left(-A\sin\frac{\sqrt{15}}4 t+B\cos\frac{\sqrt{15}}4 t\right)-\frac14\mathrm{e}^{-\frac14 t}\left(A\cos\frac{\sqrt{15}}4 t+B\sin\frac{\sqrt{15}}4 t\right)-\frac19\mathrm{e}^{-t}.\)

Using \(x(0)=0\):

\(0=A+\frac19\), so \(A=-\frac19\).

Using \(x'(0)=0\):

\(0=\frac{\sqrt{15}}4B-\frac14A-\frac19.\)

Substitute \(A=-\frac19\):

\(0=\frac{\sqrt{15}}4B+\frac1{36}-\frac19=\frac{\sqrt{15}}4B-\frac1{12}.\)

Hence

\(\frac{\sqrt{15}}4B=\frac1{12}\Rightarrow B=\frac{1}{3\sqrt{15}}.\)

So the required particular solution is

\(x=\mathrm{e}^{-\frac14 t}\left(-\frac19\cos\frac{\sqrt{15}}4 t+\frac{1}{3\sqrt{15}}\sin\frac{\sqrt{15}}4 t\right)+\frac19\mathrm{e}^{-t}.\)

Answer: (a) The general solution is

\(x=\mathrm{e}^{-5t}(At+B)+12\sin t-5\cos t\),

where \(A\) and \(B\) are constants.

(b) For large positive values of \(t\), the term \(\mathrm{e}^{-5t}(At+B)\) tends to \(0\), so

\(x\approx 12\sin t-5\cos t=13\sin(t-\phi)\),

with \(R=13\) and

\(\cos\phi=\frac{12}{13},\quad \sin\phi=\frac{5}{13}\),

so \(\phi=\tan^{-1}\!\left(\frac{5}{12}\right)\approx 0.395\text{ rad}\) \((\approx 22.6^\circ)\).

(a) Solve the associated homogeneous equation

\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+10\dfrac{\mathrm{d}x}{\mathrm{d}t}+25x=0\).

The auxiliary equation is

\(m^2+10m+25=0=(m+5)^2\),

so there is a repeated root \(m=-5\).

Hence the complementary function is

\(x_c=\mathrm{e}^{-5t}(At+B)\).

For a particular integral, try

\(x_p=p\sin t+q\cos t\).

Then

\(x_p'=p\cos t-q\sin t\),

\(x_p''=-p\sin t-q\cos t\).

Substituting into the differential equation gives

\((-p\sin t-q\cos t)+10(p\cos t-q\sin t)+25(p\sin t+q\cos t)=338\sin t\).

Collect coefficients:

\((24p-10q)\sin t+(10p+24q)\cos t=338\sin t\).

So

\(24p-10q=338\),

\(10p+24q=0\).

From \(10p+24q=0\), we have \(5p+12q=0\), so \(p=-\dfrac{12}{5}q\).

Substitute into the first equation:

\(24\left(-\dfrac{12}{5}q\right)-10q=338\).

Multiplying by 5:

\(-288q-50q=1690\),

\(-338q=1690\),

so \(q=-5\).

Then \(10p+24(-5)=0\), so \(10p=120\) and \(p=12\).

Thus

\(x_p=12\sin t-5\cos t\).

Therefore the general solution is

\(x=x_c+x_p=\mathrm{e}^{-5t}(At+B)+12\sin t-5\cos t\).

(b) For large positive \(t\), the factor \(\mathrm{e}^{-5t}\to 0\), so for any initial conditions

\(x\approx 12\sin t-5\cos t\).

Write this as \(R\sin(t-\phi)\). Since

\(R\sin(t-\phi)=R\sin t\cos\phi-R\cos t\sin\phi\),

comparing coefficients with \(12\sin t-5\cos t\) gives

\(R\cos\phi=12\),

\(R\sin\phi=5\).

Squaring and adding,

\(R^2=12^2+5^2=169\),

so \(R=13\).

Then

\(\cos\phi=\dfrac{12}{13},\quad \sin\phi=\dfrac{5}{13}\),

hence

\(\phi=\tan^{-1}\!\left(\dfrac{5}{12}\right)\approx 0.395\text{ rad}\).

So for large positive \(t\),

\(x\approx 13\sin(t-0.395)\).

Answer: (a) \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-t+\sqrt{1-t^2}\,\cos^{-1}t\).

(b) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-2\sqrt{1-t^2}-t\cos^{-1}t\).

Given \(x=\sin^{-1}t\) and \(y=t\cos^{-1}t\), where \(0\le t\lt 1\).

(a)

Differentiate with respect to \(t\):

\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{\sqrt{1-t^2}}\).

For \(y=t\cos^{-1}t\), use the product rule:

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=\cos^{-1}t+t\left(-\frac{1}{\sqrt{1-t^2}}\right)=\cos^{-1}t-\frac{t}{\sqrt{1-t^2}}.\)

Hence

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}=\left(\cos^{-1}t-\frac{t}{\sqrt{1-t^2}}\right)\Big/\left(\frac{1}{\sqrt{1-t^2}}\right).\)

So

\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\sqrt{1-t^2}\,\cos^{-1}t-t=-t+\sqrt{1-t^2}\,\cos^{-1}t.\)

(b)

Differentiate \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-t+\sqrt{1-t^2}\,\cos^{-1}t\) with respect to \(t\):

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-1+\frac{\mathrm{d}}{\mathrm{d}t}\left(\sqrt{1-t^2}\,\cos^{-1}t\right).\)

Using the product rule,

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\sqrt{1-t^2}\,\cos^{-1}t\right)=\left(\frac{-t}{\sqrt{1-t^2}}\right)\cos^{-1}t+\sqrt{1-t^2}\left(-\frac{1}{\sqrt{1-t^2}}\right).\)

Therefore

\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-1-\frac{t\cos^{-1}t}{\sqrt{1-t^2}}-1=-2-\frac{t\cos^{-1}t}{\sqrt{1-t^2}}.\)

Now

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\Big/\frac{\mathrm{d}x}{\mathrm{d}t}.\)

Since \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{\sqrt{1-t^2}}\),

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\left(-2-\frac{t\cos^{-1}t}{\sqrt{1-t^2}}\right)\Big/\left(\frac{1}{\sqrt{1-t^2}}\right).\)

Hence

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-2\sqrt{1-t^2}-t\cos^{-1}t.\)

Answer: (a) \(x=A\mathrm{e}^{-t/2}+B\mathrm{e}^{-t/3}+t^2+1\), where \(A\) and \(B\) are constants.

(b) For large positive values of \(t\), \(x\approx t^2+1\).

(a) First solve the homogeneous equation

\(6\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+5\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=0\).

Using \(x=\mathrm{e}^{mt}\), the auxiliary equation is

\(6m^2+5m+1=0\).

This factorises as \((3m+1)(2m+1)=0\), so

\(m=-\dfrac12\) or \(m=-\dfrac13\).

Hence the complementary function is

\(x=A\mathrm{e}^{-t/2}+B\mathrm{e}^{-t/3}\).

For a particular integral, since the right-hand side is quadratic, try

\(x=p+qt+rt^2\).

Then

\(x'=q+2rt\), \(x''=2r\).

Substitute into the differential equation:

\(6(2r)+5(q+2rt)+(p+qt+rt^2)=t^2+10t+13\).

So

\(12r+5q+10rt+p+qt+rt^2=t^2+10t+13\).

Equating coefficients gives

\(r=1\), \(q=0\), \(p=1\).

So the particular integral is \(t^2+1\).

Therefore the general solution is

\(x=A\mathrm{e}^{-t/2}+B\mathrm{e}^{-t/3}+t^2+1\).

(b) For large positive values of \(t\), the exponential terms become negligible, so

\(x\approx t^2+1\).

Answer: \(x=\mathrm{e}^{6t}\left(t-\frac{12}{37}\right)+\frac{35}{37}\sin t+\frac{12}{37}\cos t\).

We solve

\(\displaystyle \frac{\mathrm{d}^2x}{\mathrm{d}t^2}-12\frac{\mathrm{d}x}{\mathrm{d}t}+36x=37\sin t\), with \(x=0\) and \(\frac{\mathrm{d}x}{\mathrm{d}t}=0\) when \(t=0\).

First solve the homogeneous equation:

\(m^2-12m+36=0\)

\((m-6)^2=0\)

So the complementary function is

\(x_c=\mathrm{e}^{6t}(At+B)\).

For a particular integral, try

\(x_p=p\sin t+q\cos t\).

Then

\(x_p'=p\cos t-q\sin t\),

\(x_p''=-p\sin t-q\cos t\).

Substitute into the differential equation:

\((-p\sin t-q\cos t)-12(p\cos t-q\sin t)+36(p\sin t+q\cos t)=37\sin t\).

Collect coefficients:

\((35p+12q)\sin t+(-12p+35q)\cos t=37\sin t\).

Hence

\(35p+12q=37\),

\(-12p+35q=0\).

From \(-12p+35q=0\), we get \(q=\frac{12}{35}p\).

Substitute into the first equation:

\(35p+12\left(\frac{12}{35}p\right)=37\)

\(35p+\frac{144}{35}p=37\)

\(\frac{1369}{35}p=37\)

\(p=\frac{35}{37}\).

Then

\(q=\frac{12}{35}\cdot\frac{35}{37}=\frac{12}{37}\).

So the general solution is

\(x=\mathrm{e}^{6t}(At+B)+\frac{35}{37}\sin t+\frac{12}{37}\cos t\).

Differentiate:

\(x'=A\mathrm{e}^{6t}+6\mathrm{e}^{6t}(At+B)+\frac{35}{37}\cos t-\frac{12}{37}\sin t\).

Use the initial conditions.

When \(t=0\), \(x=0\):

\(B+\frac{12}{37}=0\)

so \(B=-\frac{12}{37}\).

When \(t=0\), \(x'=0\):

\(A+6B+\frac{35}{37}=0\).

Substitute \(B=-\frac{12}{37}\):

\(A-\frac{72}{37}+\frac{35}{37}=0\)

\(A-1=0\)

so \(A=1\).

Therefore, the required solution is

\(x=\mathrm{e}^{6t}\left(t-\frac{12}{37}\right)+\frac{35}{37}\sin t+\frac{12}{37}\cos t\).

Answer: \(y=\mathrm{e}^{-x/3}\left(-2\cos\left(\frac{\sqrt{2}}{3}x\right)+5\sqrt{2}\sin\left(\frac{\sqrt{2}}{3}x\right)\right)+x^2-4x+2\).

We solve \(3\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=x^2\), with \(y(0)=0\) and \(y'(0)=0\).

First solve the homogeneous equation:

\(3y''+2y'+y=0\).

The auxiliary equation is

\(3m^2+2m+1=0\).

So

\(m=\dfrac{-2\pm\sqrt{4-12}}{6}=\dfrac{-2\pm 2\sqrt{2}\,i}{6}=-\dfrac13\pm\dfrac{\sqrt2}{3}i\).

Hence the complementary function is

\(y_c=\mathrm{e}^{-x/3}\left(A\cos\left(\frac{\sqrt2}{3}x\right)+B\sin\left(\frac{\sqrt2}{3}x\right)\right)\).

For a particular integral, try

\(y_p=px^2+qx+r\).

Then

\(y_p'=2px+q,\qquad y_p''=2p\).

Substituting into the differential equation gives

\(3(2p)+2(2px+q)+(px^2+qx+r)=x^2\),

so

\(px^2+(4p+q)x+(6p+2q+r)=x^2\).

Equating coefficients:

\(p=1,\qquad 4p+q=0,\qquad 6p+2q+r=0\).

Thus

\(q=-4,\qquad r=2\).

So

\(y_p=x^2-4x+2\).

The general solution is therefore

\(y=\mathrm{e}^{-x/3}\left(A\cos\left(\frac{\sqrt2}{3}x\right)+B\sin\left(\frac{\sqrt2}{3}x\right)\right)+x^2-4x+2\).

Now use the conditions.

From \(y(0)=0\):

\(0=A+2\), so \(A=-2\).

Differentiate:

\(y'=\mathrm{e}^{-x/3}\left(-\frac{\sqrt2}{3}A\sin\left(\frac{\sqrt2}{3}x\right)+\frac{\sqrt2}{3}B\cos\left(\frac{\sqrt2}{3}x\right)\right)-\frac13\mathrm{e}^{-x/3}\left(A\cos\left(\frac{\sqrt2}{3}x\right)+B\sin\left(\frac{\sqrt2}{3}x\right)\right)+2x-4\).

From \(y'(0)=0\):

\(0=\frac{\sqrt2}{3}B-\frac13A-4\).

Substitute \(A=-2\):

\(0=\frac{\sqrt2}{3}B+\frac23-4=\frac{\sqrt2}{3}B-\frac{10}{3}\).

So

\(\sqrt2 B=10\), hence \(B=5\sqrt2\).

Therefore the required particular solution is

\(y=\mathrm{e}^{-x/3}\left(-2\cos\left(\frac{\sqrt{2}}{3}x\right)+5\sqrt{2}\sin\left(\frac{\sqrt{2}}{3}x\right)\right)+x^2-4x+2\).

Answer: \(x=-38\mathrm{e}^{t/2}+6\mathrm{e}^{t/3}+t^2+11t+44\).

We solve

\(6\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}-5\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=t^2+t+1\), with \(x(0)=12\) and \(x'(0)=-6\).

Complementary function

First solve the homogeneous equation

\(6x''-5x'+x=0\).

Its auxiliary equation is

\(6m^2-5m+1=0\).

Factorising,

\((3m-1)(2m-1)=0\), so the roots are \(m=\frac13\) and \(m=\frac12\).

Hence the complementary function is

\(x_c=A\mathrm{e}^{t/2}+B\mathrm{e}^{t/3}\).

Particular integral

Since the right-hand side is a quadratic, try

\(x_p=pt^2+qt+r\).

Then

\(x_p'=2pt+q\), and \(x_p''=2p\).

Substitute into the differential equation:

\(6(2p)-5(2pt+q)+(pt^2+qt+r)=t^2+t+1\).

Simplifying,

\(pt^2+(-10p+q)t+(12p-5q+r)=t^2+t+1\).

Equating coefficients gives

\(p=1\),

\(-10p+q=1\),

\(12p-5q+r=1\).

So

\(p=1\),

\(q=11\),

\(r=44\).

Therefore

\(x_p=t^2+11t+44\).

So the general solution is

\(x=A\mathrm{e}^{t/2}+B\mathrm{e}^{t/3}+t^2+11t+44\).

Use the initial conditions

Differentiate:

\(x'=\frac12A\mathrm{e}^{t/2}+\frac13B\mathrm{e}^{t/3}+2t+11\).

When \(t=0\),

\(x(0)=A+B+44=12\), so

\(A+B=-32\).

Also

\(x'(0)=\frac12A+\frac13B+11=-6\), so

\(\frac12A+\frac13B=-17\).

Multiply this second equation by 6:

\(3A+2B=-102\).

From \(A+B=-32\), we have \(2A+2B=-64\).

Subtracting,

\(A=-38\).

Then

\(B=6\).

Hence the particular solution satisfying the given conditions is

\(x=-38\mathrm{e}^{t/2}+6\mathrm{e}^{t/3}+t^2+11t+44\).

Answer: \(y=2\sqrt{2}\,\mathrm{e}^{-x}\sin(\sqrt{2}x)+9x^2-12x+2\).

We solve

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+3y=27x^2\).

First solve the associated homogeneous equation

\(y''+2y'+3y=0\).

The auxiliary equation is

\(m^2+2m+3=0\).

So

\(m=\frac{-2\pm\sqrt{4-12}}{2}=-1\pm i\sqrt{2}\).

Hence the complementary function is

\(y_c=\mathrm{e}^{-x}\big(A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x)\big)\).

For a particular integral, try

\(y_p=px^2+qx+r\).

Then

\(y_p'=2px+q\),

\(y_p''=2p\).

Substituting gives

\(2p+2(2px+q)+3(px^2+qx+r)=27x^2\).

So

\(3px^2+(4p+3q)x+(2p+2q+3r)=27x^2\).

Equating coefficients:

\(3p=27\), \(4p+3q=0\), \(2p+2q+3r=0\).

Hence

\(p=9\), \(q=-12\), \(r=2\).

So the general solution is

\(y=\mathrm{e}^{-x}\big(A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x)\big)+9x^2-12x+2\).

Now use the conditions.

When \(x=0\), \(y=2\):

\(2=A+2\), so \(A=0\).

Differentiate:

\(y'=\mathrm{e}^{-x}\big(-\sqrt{2}A\sin(\sqrt{2}x)+\sqrt{2}B\cos(\sqrt{2}x)\big)-\mathrm{e}^{-x}\big(A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x)\big)+18x-12\).

When \(x=0\), \(y'=-8\):

\(-8=\sqrt{2}B-A-12\).

Since \(A=0\),

\(\sqrt{2}B=4\Rightarrow B=2\sqrt{2}\).

Therefore the required particular solution is

\(y=2\sqrt{2}\,\mathrm{e}^{-x}\sin(\sqrt{2}x)+9x^2-12x+2\).

Answer: (a) \(x=\mathrm{e}^{-t/2}\left(A\sin \frac{\sqrt{3}}{2}t+B\cos \frac{\sqrt{3}}{2}t\right)+t^2-2t+1\).

(b) For large positive values of \(t\), \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\approx 2\).

(a) First solve the homogeneous equation

\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+\frac{\mathrm{d}x}{\mathrm{d}t}+x=0\).

The auxiliary equation is

\(m^2+m+1=0\).

So

\(m=\frac{-1\pm\sqrt{1-4}}{2}=-\frac12\pm i\frac{\sqrt3}{2}\).

Hence the complementary function is

\(x_h=\mathrm{e}^{-t/2}\left(A\sin \frac{\sqrt3}{2}t+B\cos \frac{\sqrt3}{2}t\right)\).

For a particular solution, since the right-hand side is \(t^2+1\), try

\(x_p=pt^2+qt+r\).

Then

\(\frac{\mathrm{d}x_p}{\mathrm{d}t}=2pt+q\), \(\frac{\mathrm{d}^2x_p}{\mathrm{d}t^2}=2p\).

Substituting into the differential equation gives

\(2p+(2pt+q)+(pt^2+qt+r)=t^2+1\).

So

\(pt^2+(2p+q)t+(2p+q+r)=t^2+1\).

Equating coefficients:

\(p=1\), \(2p+q=0\Rightarrow q=-2\), and \(2p+q+r=1\Rightarrow r=1\).

Thus

\(x_p=t^2-2t+1\).

Therefore the general solution is

\(x=\mathrm{e}^{-t/2}\left(A\sin \frac{\sqrt3}{2}t+B\cos \frac{\sqrt3}{2}t\right)+t^2-2t+1\).

(b) As \(t\to\infty\), the factor \(\mathrm{e}^{-t/2}\to 0\), so the exponential term and its derivatives become negligible.

Hence for large \(t\),

\(x\approx t^2-2t+1\).

Differentiating twice,

\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\approx 2\).

So for large positive values of \(t\), \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\approx 2\).

Answer: (a) \(y=2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}-3\).

(b) When \(y=0\), \(x=\ln 4\).

(a) Solve the homogeneous equation \(4\dfrac{d^2y}{dx^2}-y=0\).

Using a trial solution \(y=e^{mx}\), the auxiliary equation is

\(4m^2-1=0\),

so \(m=\pm \dfrac12\).

Hence the complementary function is

\(y=Ae^{-\frac{x}{2}}+Be^{\frac{x}{2}}\).

For a particular solution, try a constant \(y=k\). Then \(y''=0\), so substituting into \(4y''-y=3\) gives

\(-k=3\),

hence \(k=-3\).

So the general solution is

\(y=Ae^{-\frac{x}{2}}+Be^{\frac{x}{2}}-3\).

Differentiate:

\(\dfrac{dy}{dx}=-\dfrac12Ae^{-\frac{x}{2}}+\dfrac12Be^{\frac{x}{2}}\).

Use the conditions at \(x=0\):

From \(y(0)=-3\):

\(A+B-3=-3\Rightarrow A+B=0\).

From \(y'(0)=2\):

\(-\dfrac12A+\dfrac12B=2\).

Solving gives \(A=-2\) and \(B=2\).

Therefore

\(y=2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}-3\).

(b) Set \(y=0\):

\(2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}-3=0\).

So

\(2\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)=3\).

Using \(\sinh u=\dfrac{e^u-e^{-u}}{2}\) with \(u=\dfrac{x}{2}\),

\(4\sinh\left(\dfrac{x}{2}\right)=3\).

Hence

\(\dfrac{x}{2}=\sinh^{-1}\left(\dfrac34\right)\).

Now use

\(\sinh^{-1}(t)=\ln\left(t+\sqrt{t^2+1}\right)\).

Then

\(\dfrac{x}{2}=\ln\left(\dfrac34+\sqrt{\dfrac{9}{16}+1}\right)=\ln\left(\dfrac34+\dfrac54\right)=\ln 2\).

Therefore

\(x=2\ln 2=\ln 4\).

Answer: The particular solution satisfying the given conditions is

\(y=\mathrm{e}^{-x/2}\left(-13\cos\frac{x}{2}+13\sin\frac{x}{2}\right)+4x^2-13x+13\).

We solve

\(2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+y=4x^2+3x+3\),

with \(y=0\) and \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\) when \(x=0\).

First solve the homogeneous equation

\(2y''+2y'+y=0\).

Its auxiliary equation is

\(2m^2+2m+1=0\).

So

\(m=\frac{-2\pm\sqrt{4-8}}{4}=-\frac12\pm\frac12 i\).

Hence the complementary function is

\(y_c=\mathrm{e}^{-x/2}\left(A\cos\frac{x}{2}+B\sin\frac{x}{2}\right)\).

For a particular integral, try

\(y_p=px^2+qx+r\).

Then

\(y_p'=2px+q\),

\(y_p''=2p\).

Substituting into the differential equation gives

\(2(2p)+2(2px+q)+(px^2+qx+r)=4x^2+3x+3\),

so

\(px^2+(4p+q)x+(4p+2q+r)=4x^2+3x+3\).

Equating coefficients:

\(p=4\),

\(4p+q=3\),

\(4p+2q+r=3\).

Thus

\(q=-13\), \(r=13\).

So the general solution is

\(y=\mathrm{e}^{-x/2}\left(A\cos\frac{x}{2}+B\sin\frac{x}{2}\right)+4x^2-13x+13\).

Now use the conditions.

From \(y(0)=0\):

\(A+13=0\), so \(A=-13\).

Differentiate:

\(y'=\mathrm{e}^{-x/2}\left(-\frac12A\sin\frac{x}{2}+\frac12B\cos\frac{x}{2}\right)-\frac12\mathrm{e}^{-x/2}\left(A\cos\frac{x}{2}+B\sin\frac{x}{2}\right)+8x-13\).

Using \(y'(0)=0\):

\(\frac12B-\frac12A-13=0\).

Substitute \(A=-13\):

\(\frac12B+\frac{13}{2}-13=0\), so \(B=13\).

Therefore

\(y=\mathrm{e}^{-x/2}\left(-13\cos\frac{x}{2}+13\sin\frac{x}{2}\right)+4x^2-13x+13\).

Answer: (a) The general solution is \(y=A\mathrm{e}^{-x}+B\mathrm{e}^{-2x}+x-1\).

(b) For large positive values of \(x\), an approximate solution is \(y\approx x-1\).

(a) First solve the homogeneous equation

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+3\dfrac{\mathrm{d}y}{\mathrm{d}x}+2y=0\).

Using a trial solution \(y=\mathrm{e}^{mx}\), the auxiliary equation is

\(m^2+3m+2=0\).

So

\((m+1)(m+2)=0\), giving \(m=-1\) and \(m=-2\).

Hence the complementary function is

\(y_c=A\mathrm{e}^{-x}+B\mathrm{e}^{-2x}\).

For a particular integral, since the right-hand side is linear, try

\(y_p=p+qx\).

Then

\(y_p'=q, \quad y_p''=0\).

Substituting into the differential equation gives

\(0+3q+2(p+qx)=2x+1\).

So

\(2qx+(3q+2p)=2x+1\).

Equating coefficients:

\(2q=2\), so \(q=1\),

and \(3q+2p=1\), so \(3+2p=1\), hence \(p=-1\).

Therefore

\(y_p=x-1\).

So the general solution is

\(y=A\mathrm{e}^{-x}+B\mathrm{e}^{-2x}+x-1\).

(b) As \(x\to +\infty\), \(\mathrm{e}^{-x}\to 0\) and \(\mathrm{e}^{-2x}\to 0\).

Therefore, for large positive values of \(x\),

\(y\approx x-1\).

Answer: \(y=\mathrm{e}^x(9x-4)-2\sin x\).

We solve

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}-2\frac{\mathrm{d}y}{\mathrm{d}x}+y=4\cos x\),

with \(y=-4\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=3\) when \(x=0\).

First solve the homogeneous equation

\(y''-2y'+y=0\).

Its auxiliary equation is

\(m^2-2m+1=0\),

so \((m-1)^2=0\).

This gives a repeated root \(m=1\), so the complementary function is

\(y_c=\mathrm{e}^x(Ax+B)\).

For a particular solution, try

\(y_p=p\sin x+q\cos x\).

Then

\(y_p'=p\cos x-q\sin x\),

\(y_p''=-p\sin x-q\cos x\).

Substitute into the differential equation:

\((-p\sin x-q\cos x)-2(p\cos x-q\sin x)+(p\sin x+q\cos x)=4\cos x\).

Collecting coefficients gives

\(2q\sin x-2p\cos x=4\cos x\).

So

\(2q=0\Rightarrow q=0\),

\(-2p=4\Rightarrow p=-2\).

Hence

\(y_p=-2\sin x\).

So the general solution is

\(y=\mathrm{e}^x(Ax+B)-2\sin x\).

Differentiate:

\(y'=A\mathrm{e}^x+\mathrm{e}^x(Ax+B)-2\cos x\).

Now use the initial conditions.

At \(x=0\),

\(y(0)=B=-4\), so \(B=-4\).

Also

\(y'(0)=A+B-2=3\).

Substituting \(B=-4\):

\(A-6=3\Rightarrow A=9\).

Therefore the required particular solution is

\(\boxed{y=\mathrm{e}^x(9x-4)-2\sin x}.\)

\(x=A\mathrm{e}^{-t}+B\mathrm{e}^{9t}-\mathrm{e}^{8t}\).

The auxiliary equation is \(m^2-8m-9=0\), giving \(m=-1\) and \(m=9\). Hence \(x_c=A\mathrm{e}^{-t}+B\mathrm{e}^{9t}\).

Try a particular integral \(x_p=k\mathrm{e}^{8t}\). Then \(\dot{x}_p=8k\mathrm{e}^{8t}\) and \(\ddot{x}_p=64k\mathrm{e}^{8t}\).

Substitution gives \(64k-64k-9k=9\), so \(k=-1\).

Therefore \(x=A\mathrm{e}^{-t}+B\mathrm{e}^{9t}-\mathrm{e}^{8t}\).

\(k=-1\) and \(y=\dfrac12\mathrm{e}^{3x}-x\mathrm{e}^{-x}\).

For \(y=kx\mathrm{e}^{-x}\), \(y'=k\mathrm{e}^{-x}(1-x)\) and \(y''=k\mathrm{e}^{-x}(x-2)\).

Substitution into \(y''-2y'-3y=4\mathrm{e}^{-x}\) gives \(-4k\mathrm{e}^{-x}=4\mathrm{e}^{-x}\), so \(k=-1\).

The auxiliary equation for the homogeneous part is \(m^2-2m-3=0\), giving \(m=-1,3\).

Thus \(y=A\mathrm{e}^{-x}+B\mathrm{e}^{3x}-x\mathrm{e}^{-x}\).

Differentiate: \(y'=-(A-x)\mathrm{e}^{-x}-\mathrm{e}^{-x}+3B\mathrm{e}^{3x}\).

Using \(y(0)=\dfrac12\) and \(y'(0)=\dfrac12\), we get \(A+B=\dfrac12\) and \(-A-1+3B=\dfrac12\). Solving gives \(A=0\), \(B=\dfrac12\).

Therefore \(y=\dfrac12\mathrm{e}^{3x}-x\mathrm{e}^{-x}\).

\(x=6\mathrm{e}^{-5t}-6\mathrm{e}^{-3t}+4\sin3t+\cos3t\).

The auxiliary equation is \(m^2+8m+15=0\), so \(m=-3,-5\). Therefore \(x_c=A\mathrm{e}^{-5t}+B\mathrm{e}^{-3t}\).

For a particular integral, try \(x_p=p\sin3t+q\cos3t\). Then \(\dot{x}_p=3p\cos3t-3q\sin3t\), and \(\ddot{x}_p=-9p\sin3t-9q\cos3t\).

Substitution gives \(6p-24q=0\) and \(24p+6q=102\), i.e. \(p=4\), \(q=1\).

So \(x=A\mathrm{e}^{-5t}+B\mathrm{e}^{-3t}+4\sin3t+\cos3t\).

Now \(\dot{x}=-5A\mathrm{e}^{-5t}-3B\mathrm{e}^{-3t}+12\cos3t-3\sin3t\).

Using \(x(0)=1\) gives \(A+B+1=1\), so \(A+B=0\). Using \(\dot{x}(0)=0\) gives \(-5A-3B+12=0\).

Solving gives \(A=6\), \(B=-6\), hence \(x=6\mathrm{e}^{-5t}-6\mathrm{e}^{-3t}+4\sin3t+\cos3t\).

Answer: (i) The general solution is \(x=(A+Bt)e^{-t}-2\cos t\).

(ii) For large positive \(t\), the exponential terms decay to zero, so an approximate solution is \(x\approx -2\cos t\).

We solve the differential equation \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+2\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=4\sin t\).

First consider the associated homogeneous equation

\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+2\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=0\).

Try \(x=e^{mt}\). Then

\(m^2+2m+1=0\), so \((m+1)^2=0\) and hence \(m=-1\) is a repeated root.

Therefore the complementary function is

\(x_h=(A+Bt)e^{-t}\),

where \(A\) and \(B\) are constants.

Now find a particular integral. Since the forcing term is \(4\sin t\), try

\(x_p=p\sin t+q\cos t\).

Then

\(\dfrac{\mathrm{d}x_p}{\mathrm{d}t}=p\cos t-q\sin t\),

and

\(\dfrac{\mathrm{d}^2x_p}{\mathrm{d}t^2}=-p\sin t-q\cos t\).

Substitute into the differential equation:

\((-p\sin t-q\cos t)+2(p\cos t-q\sin t)+(p\sin t+q\cos t)=4\sin t\).

The \(\sin t\) terms give \(-2q=4\), so \(q=-2\). The \(\cos t\) terms give \(2p=0\), so \(p=0\).

Hence \(x_p=-2\cos t\).

So the general solution is

\(x=(A+Bt)e^{-t}-2\cos t\).

For large positive \(t\), the term \((A+Bt)e^{-t}\) tends to \(0\), so an approximate solution is

\(x\approx -2\cos t\).

Answer: The general solution is \(x=e^{3t}(A\cos 4t+B\sin 4t)+7\sin 2t+4\cos 2t\).

First solve the complementary equation

\(\dfrac{\mathrm d^2x}{\mathrm dt^2}-6\dfrac{\mathrm dx}{\mathrm dt}+25x=0\).

Its auxiliary equation is \(m^2-6m+25=0\), so

\(m=\dfrac{6\pm \sqrt{36-100}}{2}=3\pm 4\mathrm i\).

Hence the complementary function is

\(x_c=e^{3t}(A\cos 4t+B\sin 4t)\).

Now seek a particular integral of the form

\(x_p=p\sin 2t+q\cos 2t\).

Then

\(\dfrac{\mathrm dx_p}{\mathrm dt}=2p\cos 2t-2q\sin 2t\),

\(\dfrac{\mathrm d^2x_p}{\mathrm dt^2}=-4p\sin 2t-4q\cos 2t\).

Substitute into the differential equation:

\(\left(-4p\sin 2t-4q\cos 2t\right)-6\left(2p\cos 2t-2q\sin 2t\right)+25\left(p\sin 2t+q\cos 2t\right)=195\sin 2t\).

Collect coefficients of \(\sin 2t\) and \(\cos 2t\):

\(\left(21p+12q\right)\sin 2t+\left(-12p+21q\right)\cos 2t=195\sin 2t\).

So

\(21p+12q=195\), \(\,-12p+21q=0\).

From \(-12p+21q=0\), we get \(p=\dfrac{7}{4}q\). Substituting into the first equation gives

\(21\cdot \dfrac{7}{4}q+12q=195\), hence \(\dfrac{195}{4}q=195\), so \(q=4\) and \(p=7\).

Therefore

\(x_p=7\sin 2t+4\cos 2t\).

Combining the complementary function and a particular integral gives the general solution:

\(x=e^{3t}(A\cos 4t+B\sin 4t)+7\sin 2t+4\cos 2t\).

Answer: \(x=e^{5t}+2e^{-2t}+0.3\cos t-0.1\sin t\).

We solve the homogeneous equation first:

\(\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2}-3\dfrac{\mathrm{d}x}{\mathrm{d}t}-10x=0\).

Trying \(x=e^{mt}\) gives the auxiliary equation

\(m^2-3m-10=0\), so \((m-5)(m+2)=0\).

Hence the complementary function is

\(x=Ae^{5t}+Be^{-2t}\).

For a particular solution, because the forcing term is \(2\sin t-3\cos t\), let

\(x_p=p\sin t+q\cos t\).

Then

\(\dfrac{\mathrm{d}x_p}{\mathrm{d}t}=p\cos t-q\sin t\),

\(\dfrac{\mathrm{d}^2x_p}{\mathrm{d}t^2}=-p\sin t-q\cos t\).

Substitute into the differential equation:

\(-p\sin t-q\cos t-3(p\cos t-q\sin t)-10(p\sin t+q\cos t)=2\sin t-3\cos t\).

Collecting coefficients of \(\sin t\) and \(\cos t\),

\((-11p+3q)\sin t+(-3p-11q)\cos t=2\sin t-3\cos t\).

So

\(-11p+3q=2\), \(3p+11q=3\).

Solving these gives \(p=-0.1\) and \(q=0.3\).

Therefore

\(x=Ae^{5t}+Be^{-2t}+0.3\cos t-0.1\sin t\).

Now use the initial conditions. When \(t=0\), \(x=3.3\), so

\(A+B+0.3=3.3\), hence \(A+B=3\).

Differentiate:

\(\dfrac{\mathrm{d}x}{\mathrm{d}t}=5Ae^{5t}-2Be^{-2t}-0.3\sin t-0.1\cos t\).

When \(t=0\), \(\dfrac{\mathrm{d}x}{\mathrm{d}t}=0.9\), so

\(5A-2B-0.1=0.9\), hence \(5A-2B=1\).

Solve

\(A+B=3\), \(5A-2B=1\).

From \(B=3-A\),

\(5A-2(3-A)=1 \Rightarrow 7A=7 \Rightarrow A=1\), and then \(B=2\).

So the required solution is

\(x=e^{5t}+2e^{-2t}+0.3\cos t-0.1\sin t\).

Answer: \(y=\left[e^{-x}\left(\frac12\sin2x-\cos2x\right)+3+2x-x^2\right]^{-1}\).

Let \(v=\frac1y\). Then

\(\frac{dv}{dx}=-\frac1{y^2}\frac{dy}{dx}\),

and

\(\frac{d^2v}{dx^2}=-\frac1{y^2}\frac{d^2y}{dx^2}+\frac2{y^3}\left(\frac{dy}{dx}\right)^2\).

Substituting these expressions into the given equation gives

\(\frac{d^2v}{dx^2}+2\frac{dv}{dx}+5v=17+6x-5x^2\), as required.

Now solve the linear differential equation. The complementary equation has auxiliary equation

\(m^2+2m+5=0\),

so \(m=-1\pm2i\). Hence

\(v_c=e^{-x}(A\cos2x+B\sin2x)\).

For a particular integral, try \(v_p=px^2+qx+r\). Then \(v_p'=2px+q\) and \(v_p''=2p\). Substitution gives

\(2p+2(2px+q)+5(px^2+qx+r)=17+6x-5x^2\).

Equating coefficients gives \(5p=-5\), \(4p+5q=6\), and \(2p+2q+5r=17\). Therefore \(p=-1\), \(q=2\), and \(r=3\).

So

\(v=e^{-x}(A\cos2x+B\sin2x)+3+2x-x^2\).

When \(x=0\), \(y=\frac12\), so \(v=2\). Hence \(2=A+3\), giving \(A=-1\).

Also, since \(v=\frac1y\),

\(\frac{dv}{dx}=-\frac1{y^2}\frac{dy}{dx}\).

At \(x=0\), \(y=\frac12\) and \(\frac{dy}{dx}=-1\), so \(\frac{dv}{dx}=4\).

Differentiate \(v\):

\(\frac{dv}{dx}=e^{-x}(-2A\sin2x+2B\cos2x)-e^{-x}(A\cos2x+B\sin2x)+2-2x\).

At \(x=0\), this gives \(4=2B-A+2\). With \(A=-1\), \(4=2B+3\), so \(B=\frac12\).

Thus

\(v=e^{-x}\left(\frac12\sin2x-\cos2x\right)+3+2x-x^2\).

Since \(y=\frac1v\),

\(y=\left[e^{-x}\left(\frac12\sin2x-\cos2x\right)+3+2x-x^2\right]^{-1}\).

Answer: The general solution is \(x=Ae^{-2t}+Be^{-5t}+3\sin 2t-7\cos 2t\).

For large positive values of \(t\), the exponential terms decay to zero, so \(x\approx 3\sin 2t-7\cos 2t\).

First solve the associated homogeneous equation

\(\dfrac{\mathrm d^2x}{\mathrm dt^2}+7\dfrac{\mathrm dx}{\mathrm dt}+10x=0.\)

Try \(x=e^{mt}\). Then \(m^2+7m+10=0\), so

\((m+2)(m+5)=0\), hence \(m=-2\) or \(m=-5\).

So the complementary function is

\(x_h=Ae^{-2t}+Be^{-5t}.\)

Now seek a particular solution in the form

\(x_p=p\sin 2t+q\cos 2t.\)

Then

\(\dfrac{\mathrm dx_p}{\mathrm dt}=2p\cos 2t-2q\sin 2t,\)

and

\(\dfrac{\mathrm d^2x_p}{\mathrm dt^2}=-4p\sin 2t-4q\cos 2t.\)

Substitute into the differential equation:

\(-4p\sin 2t-4q\cos 2t+7(2p\cos 2t-2q\sin 2t)+10(p\sin 2t+q\cos 2t)=116\sin 2t.\)

Collect coefficients of \(\sin 2t\) and \(\cos 2t\):

\((6p-14q)\sin 2t+(14p+6q)\cos 2t=116\sin 2t.\)

Hence

\(6p-14q=116,\qquad 14p+6q=0.\)

From \(14p+6q=0\), divide by 2 to get \(7p+3q=0\), so \(q=-\dfrac{7}{3}p\). Substituting into \(6p-14q=116\):

\(6p-14\left(-\dfrac{7}{3}p\right)=116\)

\(6p+\dfrac{98}{3}p=116\)

\(\dfrac{116}{3}p=116\)

so \(p=3\), and then \(q=-7\).

Therefore \(x_p=3\sin 2t-7\cos 2t\).

So the general solution is

\(x=Ae^{-2t}+Be^{-5t}+3\sin 2t-7\cos 2t.\)

As \(t\to\infty\), the exponential terms vanish, so an approximate solution is

\(x\approx 3\sin 2t-7\cos 2t.\)

Answer: The required solution is \(y=2\ln x+1-\dfrac{1}{x}\bigl(\cos(4\ln x)+\sin(4\ln x)\bigr)\).

Let \(x=e^u\), so \(u=\ln x\). Since \(\frac{du}{dx}=\frac{1}{x}\), the chain rule gives

\(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{x}\frac{dy}{du}\).

Hence

\(x\frac{dy}{dx}=\frac{dy}{du}.\)

For the second derivative,

\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{du}\right).\)

Using the product rule and \(\frac{du}{dx}=\frac{1}{x}\),

\(\frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{du}+\frac{1}{x}\frac{d}{dx}\left(\frac{dy}{du}\right) =-\frac{1}{x^2}\frac{dy}{du}+\frac{1}{x}\frac{d^2y}{du^2}\cdot\frac{du}{dx}\).

So

\(\frac{d^2y}{dx^2}=\frac{1}{x^2}\frac{d^2y}{du^2}-\frac{1}{x^2}\frac{dy}{du}\),

and therefore

\(x^2\frac{d^2y}{dx^2}=\frac{d^2y}{du^2}-\frac{dy}{du}.\)

Now substitute these into

\(x^2\frac{d^2y}{dx^2}+3x\frac{dy}{dx}+17y=34\ln x+21.\)

Since \(\ln x=u\), this becomes

\(\left(\frac{d^2y}{du^2}-\frac{dy}{du}\right)+3\frac{dy}{du}+17y=34u+21,\)

so

\(\frac{d^2y}{du^2}+2\frac{dy}{du}+17y=34u+21.\)

We now solve this ODE in \(u\). The complementary equation is

\(m^2+2m+17=0,\)

so

\(m=-1\pm 4i.\)

Thus

\(y_c=e^{-u}(A\cos 4u+B\sin 4u).\)

For a particular solution, try \(y_p=cu+d\). Then \(y_p'=c\) and \(y_p''=0\), so

\(2c+17(cu+d)=34u+21.\)

Matching coefficients gives \(17c=34\), hence \(c=2\), and then \(2c+17d=21\), so \(4+17d=21\) and \(d=1\). Therefore

\(y=e^{-u}(A\cos 4u+B\sin 4u)+2u+1.\)

Since \(u=\ln x\) and \(e^{-u}=\frac{1}{x}\),

\(y=\frac{1}{x}\big(A\cos(4\ln x)+B\sin(4\ln x)\big)+2\ln x+1.\)

Use the condition \(y=0\) when \(x=1\). Then \(\ln 1=0\), so

\(0=A+1,\)

hence \(A=-1\).

Differentiate with respect to \(x\):

\(\frac{dy}{dx}=-\frac{1}{x^2}\big(A\cos(4\ln x)+B\sin(4\ln x)\big) +\frac{1}{x}\left(-A\sin(4\ln x)\cdot\frac{4}{x}+B\cos(4\ln x)\cdot\frac{4}{x}\right)+\frac{2}{x}.\)

When \(x=1\), this gives

\(-1=-A+4B+2.\)

With \(A=-1\),

\(-1=1+4B+2,\)

so \(4B=-4\) and \(B=-1\).

Therefore

\(y=2\ln x+1-\frac{1}{x}\big(\cos(4\ln x)+\sin(4\ln x)\big)\).

Answer: \(k=2\).

The general solution is \(y=(A+Bx+2x^2)e^{2x}\).

The particular solution is \(y=(3-8x+2x^2)e^{2x}\).

Let us first find a suitable particular integral of the form \(y=kx^2e^{2x}\).

Differentiating,

\(y'=k(2xe^{2x}+2x^2e^{2x})\)

and

\(y''=k(2e^{2x}+8xe^{2x}+4x^2e^{2x})\).

Substitute into \(y''-4y'+4y=4e^{2x}\):

\(k(2e^{2x}+8xe^{2x}+4x^2e^{2x})-4k(2xe^{2x}+2x^2e^{2x})+4kx^2e^{2x}=4e^{2x}\).

The \(xe^{2x}\) and \(x^2e^{2x}\) terms cancel, leaving

\(2ke^{2x}=4e^{2x}\), so \(k=2\).

Hence a particular integral is \(2x^2e^{2x}\).

For the homogeneous equation \(y''-4y'+4y=0\), the auxiliary equation is

\(m^2-4m+4=0\), so \((m-2)^2=0\).

Thus the complementary function is

\(y=(A+Bx)e^{2x}\).

So the general solution is

\(y=(A+Bx+2x^2)e^{2x}\).

Now use the conditions \(y=3\) and \(\frac{dy}{dx}=-2\) when \(x=0\).

At \(x=0\),

\(y(0)=A=3\).

Differentiate \(y=(A+Bx+2x^2)e^{2x}\):

\(y'=(B+4x)e^{2x}+2(A+Bx+2x^2)e^{2x}\).

So at \(x=0\),

\(y'(0)=B+2A\).

Given \(y'(0)=-2\) and \(A=3\),

\(B+6=-2\), hence \(B=-8\).

Therefore the particular solution is

\(y=(3-8x+2x^2)e^{2x}\).

Answer: The general solution is \(x=e^{3t}(A\cos 4t+B\sin 4t)+7\sin 2t+4\cos 2t\).

We solve the complementary function first.

For \(\frac{\mathrm d^2 x}{\mathrm d t^2}-6\frac{\mathrm d x}{\mathrm d t}+25x=0\), the auxiliary equation is \(m^2-6m+25=0\).

So \(m=\frac{6\pm\sqrt{36-100}}{2}=3\pm 4i\).

Hence the complementary function is \(x_c=e^{3t}(A\cos 4t+B\sin 4t)\).

Now choose a particular integral of the form \(x_p=p\sin 2t+q\cos 2t\).

Then

\(\frac{\mathrm d x_p}{\mathrm d t}=2p\cos 2t-2q\sin 2t\),

and

\(\frac{\mathrm d^2 x_p}{\mathrm d t^2}=-4p\sin 2t-4q\cos 2t\).

Substitute into the differential equation:

\((-4p\sin 2t-4q\cos 2t)-6(2p\cos 2t-2q\sin 2t)+25(p\sin 2t+q\cos 2t)=195\sin 2t.\)

Collecting coefficients of \(\sin 2t\) and \(\cos 2t\) gives

\(21p+12q=195,\qquad -12p+21q=0.\)

From \(-12p+21q=0\), we have \(4p=7q\), so \(p=\frac{7}{4}q\). Substituting into the first equation:

\(21\left(\frac{7}{4}q\right)+12q=195\)

\(\frac{147}{4}q+\frac{48}{4}q=195\)

\(\frac{195}{4}q=195\)

so \(q=4\), and therefore \(p=7\).

Thus \(x_p=7\sin 2t+4\cos 2t\).

So the general solution is \(x=e^{3t}(A\cos 4t+B\sin 4t)+7\sin 2t+4\cos 2t\).

Answer: \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For \(\mathbf A\), the eigenvalues corresponding to \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\) and \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\) are \(0\) and \(1\).

A corresponding eigenvector for eigenvalue \(2\) is \(\begin{pmatrix}2\\-1\\0\end{pmatrix}\).

One suitable choice is

\(\mathbf P=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix},\qquad \mathbf D=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix},\)

so that \(\mathbf P^{-1}\mathbf C^2\mathbf P=\mathbf D\).

If \(\mathbf A\mathbf e=\lambda\mathbf e\) and \(\mathbf B\mathbf e=\mu\mathbf e\), then

\(\mathbf{AB}\mathbf e=\mathbf A(\mathbf B\mathbf e)=\mathbf A(\mu\mathbf e)=\mu\mathbf A\mathbf e=\mu\lambda\mathbf e=\lambda\mu\mathbf e.\)

Therefore \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For

\(\mathbf A=\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix},\)

using \(\mathbf v_1=\begin{pmatrix}0\\1\\-1\end{pmatrix}\),

\(\mathbf A\mathbf v_1=\begin{pmatrix}0\\0\\0\end{pmatrix}=0\mathbf v_1,\)

so the corresponding eigenvalue is \(0\).

Using \(\mathbf v_2=\begin{pmatrix}1\\0\\-1\end{pmatrix}\),

\(\mathbf A\mathbf v_2=\begin{pmatrix}1\\0\\-1\end{pmatrix}=1\mathbf v_2,\)

so the corresponding eigenvalue is \(1\).

For eigenvalue \(2\), solve

\((\mathbf A-2\mathbf I)\mathbf v=\mathbf0.\)

One suitable eigenvector is

\(\mathbf v_3=\begin{pmatrix}2\\-1\\0\end{pmatrix},\)

since

\(\mathbf A\mathbf v_3=\begin{pmatrix}4\\-2\\0\end{pmatrix}=2\begin{pmatrix}2\\-1\\0\end{pmatrix}.\)

Now \(\mathbf B\) has the same eigenvectors. Direct multiplication gives the corresponding \(\mathbf B\)-eigenvalues

\(0,\quad -3,\quad -2.\)

Hence \(\mathbf C=\mathbf{AB}\) has corresponding eigenvalues

\(0\cdot0=0,\qquad 1\cdot(-3)=-3,\qquad 2\cdot(-2)=-4.\)

Therefore \(\mathbf C^2\) has eigenvalues

\(0,\quad 9,\quad 16.\)

Taking the common eigenvectors as columns gives

\(\mathbf P=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix}.\)

Thus one suitable diagonal matrix is

\(\mathbf D=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix},\)

and

\(\mathbf P^{-1}\mathbf C^2\mathbf P=\mathbf D.\)

Answer: The general solution is \(x=\mathrm{e}^{-t}(A\cos 2t+B\sin 2t)+2\sin t-\cos t\).

Using \(x(0)=5\) and \(\frac{\mathrm{d}x}{\mathrm{d}t}(0)=2\), the particular solution is \(x=\mathrm{e}^{-t}(6\cos 2t+3\sin 2t)+2\sin t-\cos t\).

For large positive \(t\), \(\mathrm{e}^{-t}\to 0\), so \(x\approx 2\sin t-\cos t\).

First solve the associated homogeneous equation

\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+2\dfrac{\mathrm{d}x}{\mathrm{d}t}+5x=0\).

Try \(x=\mathrm{e}^{mt}\). Then

\(m^2+2m+5=0\).

So

\(m=\dfrac{-2\pm \sqrt{4-20}}{2}=-1\pm 2\mathrm{i}\).

Hence the complementary function is

\(x_c=\mathrm{e}^{-t}(A\cos 2t+B\sin 2t)\).

For a particular solution, since the forcing term is \(10\sin t\), take

\(x_p=p\cos t+q\sin t\).

Then

\(\dot{x}_p=-p\sin t+q\cos t\),

\(\ddot{x}_p=-p\cos t-q\sin t\).

Substitute into the differential equation:

\((-p\cos t-q\sin t)+2(-p\sin t+q\cos t)+5(p\cos t+q\sin t)=10\sin t\).

Collect coefficients of \(\cos t\) and \(\sin t\):

\((4p+2q)\cos t+(-2p+4q)\sin t=0\cdot \cos t+10\sin t\).

So

\(4p+2q=0\), \(-2p+4q=10\).

From \(2p+q=0\), we get \(q=-2p\). Substitute into the second equation:

\(-2p+4(-2p)=10\Rightarrow -10p=10\Rightarrow p=-1\).

Thus \(q=2\), and

\(x_p=2\sin t-\cos t\).

So the general solution is

\(x=\mathrm{e}^{-t}(A\cos 2t+B\sin 2t)+2\sin t-\cos t\).

Now use the initial conditions. When \(t=0\),

\(5=A+0-1\), so \(A=6\).

Differentiate the general solution:

\(\dot{x}=\mathrm{e}^{-t}(-A\cos 2t-B\sin 2t)+\mathrm{e}^{-t}(-2A\sin 2t+2B\cos 2t)+2\cos t+\sin t\).

At \(t=0\), this gives

\(2=-A+2B+2\).

With \(A=6\),

\(2=-6+2B+2\Rightarrow 2B=6\Rightarrow B=3\).

Therefore the particular solution is

\(x=\mathrm{e}^{-t}(6\cos 2t+3\sin 2t)+2\sin t-\cos t\).

As \(t\to \infty\), the exponential term tends to \(0\), so an approximate solution for large positive \(t\) is

\(x\approx 2\sin t-\cos t\).

Answer: The general solution is \(x=Ae^{-2t}+Bte^{-2t}-\frac{1}{8}\cos 2t\).

As \(t\to\infty\), the terms \(e^{-2t}\) and \(te^{-2t}\) both tend to \(0\), so \(x\) approaches \(-\frac{1}{8}\cos 2t\). Hence \(x\) oscillates with constant amplitude \(\frac{1}{8}\) about \(0\).

We first solve the homogeneous equation

\(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+4\frac{\mathrm{d}x}{\mathrm{d}t}+4x=0\).

Its auxiliary equation is \(m^2+4m+4=0\), so \((m+2)^2=0\). Hence the complementary function is

\(x_c=Ae^{-2t}+Bte^{-2t}\).

For a particular solution, since the forcing term is \(\sin 2t\), take

\(x_p=p\sin 2t+q\cos 2t\).

Then

\(\frac{\mathrm{d}x_p}{\mathrm{d}t}=2p\cos 2t-2q\sin 2t\),

\(\frac{\mathrm{d}^2x_p}{\mathrm{d}t^2}=-4p\sin 2t-4q\cos 2t\).

Substitute into the differential equation:

\((-4p\sin 2t-4q\cos 2t)+4(2p\cos 2t-2q\sin 2t)+4(p\sin 2t+q\cos 2t)=\sin 2t.\)

Collecting terms gives

\(-8q\sin 2t+8p\cos 2t=\sin 2t.\)

So

\(-8q=1\) and \(8p=0\),

hence \(q=-\frac{1}{8}\) and \(p=0\). Therefore

\(x_p=-\frac{1}{8}\cos 2t\).

So the general solution is

\(x=Ae^{-2t}+Bte^{-2t}-\frac{1}{8}\cos 2t\).

As \(t\to\infty\), both \(e^{-2t}\to 0\) and \(te^{-2t}\to 0\), so the transient terms decay away. Thus

\(x\sim -\frac{1}{8}\cos 2t\) as \(t\to\infty\).

Therefore \(x\) oscillates indefinitely between \(\pm \frac{1}{8}\), with the exponential terms disappearing over time.

Answer: The general solution is

y = A e^{-x} + B e^{-4x} - 1.40\sin(3x) - 0.20\cos(3x).

For large positive \(x\), the exponential terms tend to \(0\), so

\(y \approx -1.40\sin(3x) - 0.20\cos(3x)\).

This can be written in the form

\(y \approx R\sin(3x+\phi)\), where

\(R = 1.41\) and \(\phi = 3.28\) radians.

First solve the homogeneous equation

\(\dfrac{d^2y}{dx^2}+5\dfrac{dy}{dx}+4y=0\).

Its auxiliary equation is \(m^2+5m+4=0\), so

\((m+1)(m+4)=0\), hence \(m=-1,-4\).

Therefore the complementary function is

\(y_c=Ae^{-x}+Be^{-4x}\).

Now look for a particular integral of the form

\(y_p=P\sin 3x+Q\cos 3x\).

Then

\(y_p' = 3P\cos 3x - 3Q\sin 3x\),

\(y_p'' = -9P\sin 3x - 9Q\cos 3x\).

Substitute into the differential equation:

\((-9P\sin 3x-9Q\cos 3x)+5(3P\cos 3x-3Q\sin 3x)+4(P\sin 3x+Q\cos 3x).\)

Collecting coefficients gives

\((-5P-15Q)\sin 3x + (15P-5Q)\cos 3x.\)

This must equal \(10\sin 3x-20\cos 3x\), so

\(-5P-15Q=10,\qquad 15P-5Q=-20.\)

Dividing the first equation by \(-5\) gives \(P+3Q=-2\). Solving with the second equation:

\(P=-\frac75,\qquad Q=-\frac15.\)

So the general solution is

\(y=Ae^{-x}+Be^{-4x}-\frac75\sin 3x-\frac15\cos 3x.\)

For large positive \(x\), the exponential terms vanish, so

\(y\approx -\frac75\sin 3x-\frac15\cos 3x.\)

Write this as \(R\sin(3x+\phi)\). Since

\(R\sin(3x+\phi)=R(\sin 3x\cos\phi+\cos 3x\sin\phi),\)

we compare coefficients:

\(R\cos\phi=-\frac75,\qquad R\sin\phi=-\frac15.\)

Hence

\(R=\sqrt{\left(\frac75\right)^2+\left(\frac15\right)^2}=\frac{\sqrt{50}}5=\sqrt2\approx 1.41.\)

Also

\(\tan\phi=\frac{(-1/5)}{(-7/5)}=\frac17.\)

Since both \(\sin\phi\) and \(\cos\phi\) are negative, \(\phi\) lies in the third quadrant, so

\(\phi=\pi+\arctan\left(\frac17\right)\approx 3.28.\)

Thus, for large positive \(x\),

\(\boxed{y\approx 1.41\sin(3x+3.28)}\)

with \(R=1.41\) and \(\phi=3.28\) radians.

Answer: (i) The mean value of \(\dfrac{dy}{dx}\) on \(0 \leqslant x \leqslant \dfrac{\pi}{2}\) is

\(\dfrac{1}{\frac{\pi}{2}-0}\int_{0}^{\pi/2} \frac{dy}{dx}\,dx = \dfrac{2}{\pi}\bigl(y(\pi/2)-y(0)\bigr).\)

Since \(y=x^{2}\sin x\), we have \(y(\pi/2)=\left(\dfrac{\pi}{2}\right)^{2}\cdot 1\) and \(y(0)=0\), so

\(\dfrac{2}{\pi}\left(\dfrac{\pi^{2}}{4}\right)=\dfrac{\pi}{2}.\)

So the mean value is \(\dfrac{\pi}{2}\).

(ii) First find \(\dfrac{d^{2}y}{dx^{2}}\).

\(\dfrac{dy}{dx}=2x\sin x+x^{2}\cos x\)

so

\(\dfrac{d^{2}y}{dx^{2}}=2\sin x+4x\cos x-x^{2}\sin x.\)

The mean value on \(0 \leqslant x \leqslant \dfrac{\pi}{2}\) is

\(\dfrac{2}{\pi}\int_{0}^{\pi/2}\left(2\sin x+4x\cos x-x^{2}\sin x\right)dx.\)

Using the fact that this is the change in \(\dfrac{dy}{dx}\) across the interval,

\(\dfrac{2}{\pi}\left[2x\sin x+x^{2}\cos x\right]_{0}^{\pi/2} =\dfrac{2}{\pi}\left(\pi\right)=2.\)

So the mean value of \(\dfrac{d^{2}y}{dx^{2}}\) is \(2\).

For a function \(f(x)\), the mean value on \([a,b]\) is \(\dfrac{1}{b-a}\int_a^b f(x)\,dx\).

(i) Here \(f(x)=\dfrac{dy}{dx}\) and \(a=0\), \(b=\dfrac{\pi}{2}\). Hence

\(\text{MV} = \dfrac{2}{\pi}\int_0^{\pi/2} \dfrac{dy}{dx}\,dx = \dfrac{2}{\pi}\bigl[y\bigr]_0^{\pi/2}.\)

Since \(y=x^2\sin x\),

\(y\left(\dfrac{\pi}{2}\right)=\left(\dfrac{\pi}{2}\right)^2\sin\left(\dfrac{\pi}{2}\right)=\dfrac{\pi^2}{4},\quad y(0)=0.\)

Therefore

\(\text{MV} = \dfrac{2}{\pi}\left(\dfrac{\pi^2}{4}-0\right)=\dfrac{\pi}{2}.\)

(ii) Differentiate twice:

\(\dfrac{dy}{dx}=2x\sin x+x^2\cos x,\)

\(\dfrac{d^2y}{dx^2}=2\sin x+2x\cos x+2x\cos x-x^2\sin x=2\sin x+4x\cos x-x^2\sin x.\)

The mean value of \(\dfrac{d^2y}{dx^2}\) on \([0,\pi/2]\) is

\(\text{MV} = \dfrac{2}{\pi}\int_0^{\pi/2} \dfrac{d^2y}{dx^2}\,dx = \dfrac{2}{\pi}\left[\dfrac{dy}{dx}\right]_0^{\pi/2}.\)

Now

\(\dfrac{dy}{dx}\Big|_{x=\pi/2}=2\cdot\dfrac{\pi}{2}\cdot 1+\left(\dfrac{\pi}{2}\right)^2\cdot 0=\pi,\)

and

\(\dfrac{dy}{dx}\Big|_{x=0}=0.\)

So

\(\text{MV}=\dfrac{2}{\pi}(\pi-0)=2.\)

Therefore the mean values are \(\dfrac{\pi}{2}\) and \(2\).

Answer: a) \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2t-2\sin t}{1+\cos t}\)

b) \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{2t\sin t}{(1+\cos t)^3}\)

c) \(\frac{\mathrm{d}y}{\mathrm{d}x}\) increases with \(x\) over \(-\pi\lt t\lt \pi\).

Differentiate the parametric equations with respect to \(t\):

\(\frac{\mathrm{d}x}{\mathrm{d}t}=1+\cos t\), so \(\frac{\mathrm{d}x}{\mathrm{d}t}\gt 0\) for \(-\pi\lt t\lt \pi\) except at the endpoints where \(\cos t=-1\).

Also \(\frac{\mathrm{d}y}{\mathrm{d}t}=2t-2\sin t\).

Hence

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}=\frac{2t-2\sin t}{1+\cos t}.\)

To find the second derivative, differentiate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(t\), then divide by \(\frac{\mathrm{d}x}{\mathrm{d}t}\):

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2t-2\sin t}{1+\cos t}\right)=\frac{(2-2\cos t)(1+\cos t)-(2t-2\sin t)(-\sin t)}{(1+\cos t)^2}.\)

This simplifies to

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{(2-2\cos t)(1+\cos t)+\sin t(2t-2\sin t)}{(1+\cos t)^2}.\)

Therefore

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\frac{\mathrm{d}}{\mathrm{d}t}(\mathrm{d}y/\mathrm{d}x)}{\mathrm{d}x/\mathrm{d}t}=\frac{(2-2\cos t)(1+\cos t)+\sin t(2t-2\sin t)}{(1+\cos t)^3}.\)

Now simplify the numerator:

\((2-2\cos t)(1+\cos t)=2-2\cos^2 t=2\sin^2 t,\)

so the numerator becomes

\(2\sin^2 t+2t\sin t-2\sin^2 t=2t\sin t.\)

Thus

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{2t\sin t}{(1+\cos t)^3}.\)

For monotonicity, consider the sign of \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\). For \(-\pi\lt t\lt 0\), both \(t\) and \(\sin t\) are negative, so \(t\sin t\gt 0\). For \(0\lt t\lt \pi\), both are positive, so again \(t\sin t\gt 0\). Also \(1+\cos t\gt 0\) throughout the interval \(-\pi\lt t\lt \pi\). Hence \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\gt 0\) for all non-zero \(t\) in the interval.

So \(\frac{\mathrm{d}y}{\mathrm{d}x}\) increases with \(x\) over the given interval.