Answer: (i) \(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\cot y\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2\).

(ii) \(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{4}{3\sqrt{3}}=-\dfrac{4\sqrt{3}}{9}\).

(i) From \(\cos y=x\), differentiate implicitly with respect to \(x\):

\(-\sin y\,\dfrac{\mathrm dy}{\mathrm dx}=1\), so \(\dfrac{\mathrm dy}{\mathrm dx}=-(\sin y)^{-1}\).

Differentiate again:

\(\dfrac{\mathrm d^2y}{\mathrm dx^2}=(\sin y)^{-2}\cos y\,\dfrac{\mathrm dy}{\mathrm dx}\).

Since \(\dfrac{\mathrm dy}{\mathrm dx}=-(\sin y)^{-1}\), this becomes

\(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{\cos y}{\sin^3y}=-\cot y\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2\),

as required.

(ii) At \(y=\dfrac{\pi}{3}\),

\(\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{1}{\sin(\pi/3)}=-\dfrac{2}{\sqrt3}\), and \(\cot(\pi/3)=\dfrac{1}{\sqrt3}\).

Therefore

\(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{1}{\sqrt3}\left(-\dfrac{2}{\sqrt3}\right)^2=-\dfrac{4}{3\sqrt3}=-\dfrac{4\sqrt3}{9}\).

Answer: Alternative 1

(i) \(\dfrac{\mathrm d^2w}{\mathrm dx^2}+2\dfrac{\mathrm dw}{\mathrm dx}+w=-\mathrm e^{-2x}\).

(ii) \(y=\cos^{-1}\!\left(\left(\dfrac32-x\right)\mathrm e^{-x}-\mathrm e^{-2x}\right)\), with the branch satisfying \(y(0)=\dfrac{\pi}{3}\).

OR

Alternative 2

(i) \(\alpha=\ln(1+\sqrt2)\) and \(r=4\sqrt2\).

(ii) The sketch should show \(C_1\) starting at \(r=4\) on the initial line, \(C_2\) starting at the pole, and the curves meeting at \(P\).

(iii) The area is \(5\ln(1+\sqrt2)+(1+\sqrt2)^2-(1+\sqrt2)^{-2}-\dfrac18\left((1+\sqrt2)^4-(1+\sqrt2)^{-4}\right)\), which is \(5.82\) square units to 3 significant figures.

Alternative 1

(i) Since \(w=\cos y\),

\(\dfrac{\mathrm dw}{\mathrm dx}=-\sin y\,\dfrac{\mathrm dy}{\mathrm dx}\),

and

\(\dfrac{\mathrm d^2w}{\mathrm dx^2}=-\sin y\,\dfrac{\mathrm d^2y}{\mathrm dx^2}-\cos y\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2\).

Now multiply the given differential equation by \(\cos y\). Since \(\tan y\cos y=\sin y\) and \(\sec y\cos y=1\), we get

\(\sin y\,\dfrac{\mathrm d^2y}{\mathrm dx^2}+\cos y\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2\sin y\,\dfrac{\mathrm dy}{\mathrm dx}=\cos y+\mathrm e^{-2x}\).

Using the expressions for \(w'\) and \(w''\),

\(w''+2w'+w=-\left(\sin y\,y''+\cos y\,(y')^2+2\sin y\,y'\right)+\cos y\).

Substitute the equation above:

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

Hence

\(\dfrac{\mathrm d^2w}{\mathrm dx^2}+2\dfrac{\mathrm dw}{\mathrm dx}+w=-\mathrm e^{-2x}\).

(ii) Solve

\(w''+2w'+w=-\mathrm e^{-2x}\).

The auxiliary equation is

\(m^2+2m+1=0\), so \((m+1)^2=0\). Therefore the complementary function is

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

For a particular integral, try \(w_p=k\mathrm e^{-2x}\). Then

\(w_p''+2w_p'+w_p=k\mathrm e^{-2x}\), so \(k=-1\).

Thus

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

When \(x=0\), \(y=\dfrac{\pi}{3}\), so \(w=\cos\dfrac{\pi}{3}=\dfrac12\). Hence

\(B-1=\dfrac12\), so \(B=\dfrac32\).

Also

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

At \(x=0\),

\(w'=-\sin\left(\dfrac{\pi}{3}\right)\dfrac{1}{\sqrt3}=-\dfrac12\).

Therefore

\(A-B+2=-\dfrac12\).

Using \(B=\dfrac32\), we get \(A=-1\).

So

\(w=\left(\dfrac32-x\right)\mathrm e^{-x}-\mathrm e^{-2x}\).

Since \(w=\cos y\), the particular solution is

\(y=\cos^{-1}\!\left(\left(\dfrac32-x\right)\mathrm e^{-x}-\mathrm e^{-2x}\right)\),

with the branch chosen so that \(y(0)=\dfrac{\pi}{3}\).

Alternative 2

(i) At the intersection point \(P\), the two polar radii are equal:

\(\mathrm e^{2\alpha}-\mathrm e^{-2\alpha}=2(\mathrm e^\alpha+\mathrm e^{-\alpha})\).

Factor the left-hand side:

\((\mathrm e^\alpha-\mathrm e^{-\alpha})(\mathrm e^\alpha+\mathrm e^{-\alpha})=2(\mathrm e^\alpha+\mathrm e^{-\alpha})\).

Since \(\mathrm e^\alpha+\mathrm e^{-\alpha}>0\), divide by this factor to get

\(\mathrm e^\alpha-\mathrm e^{-\alpha}=2\).

Multiplying by \(\mathrm e^\alpha\) gives

\(\mathrm e^{2\alpha}-2\mathrm e^\alpha-1=0\).

Let \(u=\mathrm e^\alpha\). Then

\(u^2-2u-1=0\), so \(u=1+\sqrt2\), since \(u>0\). Hence

\(\alpha=\ln(1+\sqrt2)\).

At \(P\),

\(r=2(\mathrm e^\alpha+\mathrm e^{-\alpha})=2\left((1+\sqrt2)+(\sqrt2-1)\right)=4\sqrt2\).

(ii) The sketch should show \(C_1\) starting on the initial line at \(r=4\), while \(C_2\) starts at the pole. The curves meet at \(P\), where \(\theta=\alpha\), and the relative positions should match this single intersection before \(\theta=\dfrac{\pi}{2}\).

(iii) For \(0\leq\theta\leq\alpha\), the required area is

\(A=\dfrac12\int_0^\alpha\left(r_1^2-r_2^2\right)\,\mathrm d\theta\).

Therefore

\(A=2\int_0^\alpha(\mathrm e^\theta+\mathrm e^{-\theta})^2\,\mathrm d\theta-\dfrac12\int_0^\alpha(\mathrm e^{2\theta}-\mathrm e^{-2\theta})^2\,\mathrm d\theta\).

Expanding the integrand gives

\(A=\int_0^\alpha\left(5+2\mathrm e^{2\theta}+2\mathrm e^{-2\theta}-\dfrac12\mathrm e^{4\theta}-\dfrac12\mathrm e^{-4\theta}\right)\,\mathrm d\theta\).

So

\(A=\left[5\theta+\mathrm e^{2\theta}-\mathrm e^{-2\theta}-\dfrac18\mathrm e^{4\theta}+\dfrac18\mathrm e^{-4\theta}\right]_0^\alpha\).

Using \(\alpha=\ln(1+\sqrt2)\),

\(A=5\ln(1+\sqrt2)+(1+\sqrt2)^2-(1+\sqrt2)^{-2}-\dfrac18\left((1+\sqrt2)^4-(1+\sqrt2)^{-4}\right)\).

Numerically,

\(A=5.821\ldots\),

so the area is \(5.82\) square units to 3 significant figures.

Answer: The particular solution is \(y=-(49+56x)e^{-x/7}+49(x+1)\).

The auxiliary equation is

\(49m^2+14m+1=0\),

so

\(\left(7m+1\right)^2=0\),

and therefore \(m=-\frac{1}{7}\) is a repeated root. Hence the complementary function is

\(y_c=(A+Bx)e^{-x/7}.\)

For a particular integral, try \(y_p=px+q\). Then \(y_p'=p\) and \(y_p''=0\). Substituting in

\(49y''+14y'+y=49x+735\)

gives

\(14p+px+q=49x+735.\)

Comparing coefficients gives \(p=49\) and \(14p+q=735\), so \(q=49\). Thus

\(y=(A+Bx)e^{-x/7}+49x+49.\)

Using \(y(0)=0\),

\(A+49=0\), so \(A=-49\).

Now

\(y'=\left(B-\frac{A+Bx}{7}\right)e^{-x/7}+49.\)

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

\(B-\frac{A}{7}+49=0.\)

With \(A=-49\), this gives \(B+7+49=0\), so \(B=-56\).

Therefore

\(\boxed{y=-(49+56x)e^{-x/7}+49(x+1)}.\)

Answer: (i) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1\).

(ii) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac{4}{3}\).

Let \(p=\dfrac{\mathrm{d}y}{\mathrm{d}x}\). At \(x=0\), \(y=-1\), so the equation gives

\((0+p)^3=(-1)^2+0\),

hence \(p^3=1\) and therefore

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1\).

Now differentiate

\(\left(x+\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^3=y^2+x\)

with respect to \(x\):

\(3\left(x+\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\left(1+\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)=2y\dfrac{\mathrm{d}y}{\mathrm{d}x}+1.\)

Substitute \(x=0\), \(y=-1\), and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1\):

\(3(1)^2\left(1+\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)=2(-1)(1)+1=-1.\)

So

\(1+\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac13\),

and therefore

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac43\).