Answer: Either: Let \(y=z^3\). Then \(\dfrac{dy}{dx}=3z^2\dfrac{dz}{dx}\) and \(\dfrac{d^2y}{dx^2}=6z\left(\dfrac{dz}{dx}\right)^2+3z^2\dfrac{d^2z}{dx^2}\). Hence the given differential equation becomes \(\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}+5y=5x+2\).

So \(y''+2y'+5y=5x+2\). A particular solution is \(y=x\). The complementary function is \(y=e^{-x}(A\cos 2x+B\sin 2x)\). Therefore \(y=e^{-x}(A\cos 2x+B\sin 2x)+x\).

Using \(x=0\), \(z=1\), we get \(y=1\). Also \(y'=3z^2z'\), so when \(x=0\), \(y' = 3(1)^2\left(-\frac23\right)=-2\).

Now \(y(0)=A=1\). Also \(y'=e^{-x}[-A\cos 2x-B\sin 2x-2A\sin 2x+2B\cos 2x]+1\), so at \(x=0\), \(y'(0)=-A+2B+1=-2\). With \(A=1\), this gives \(2B=-2\), so \(B=-1\).

Thus \(y=e^{-x}(\cos 2x-\sin 2x)+x\), and since \(y=z^3\),

\(z=\left[e^{-x}(\cos 2x-\sin 2x)+x\right]^{1/3}.\)

For large positive \(x\), the exponential term tends to \(0\), so \(y\approx x\). Hence \(z^3\approx x\), and therefore \(z\approx x^{1/3}\).

Or: For \(y=\dfrac{x(x+1)}{(x-1)^2}\), the vertical asymptote is \(x=1\). Also, as \(|x|\to\infty\), the highest powers give \(y\to 1\), so the horizontal asymptote is \(y=1\).

To find where the curve meets its asymptotes, set \(y=1\): \(\dfrac{x(x+1)}{(x-1)^2}=1\). Then \(x(x+1)=(x-1)^2\), so \(x^2+x=x^2-2x+1\), giving \(3x=1\) and \(x=\frac13\). Hence the intersection point is \(\left(\frac13,1\right)\).

Differentiate using the quotient rule:

\(\dfrac{dy}{dx}=\dfrac{(2x+1)(x-1)^2-2x(x+1)(x-1)}{(x-1)^4}=\dfrac{-(3x+1)}{(x-1)^3}.\)

Stationary points satisfy \(\dfrac{dy}{dx}=0\), so \(3x+1=0\), giving \(x=-\frac13\). Then

\(y=\dfrac{(-1/3)(2/3)}{(-4/3)^2}=\dfrac{-2/9}{16/9}=-\frac18.\)

So the stationary point is \(\left(-\frac13,-\frac18\right)\).

The gradient is negative when \(\dfrac{-(3x+1)}{(x-1)^3}\lt 0\), which gives \(x\lt -\frac13\) or \(x\gt 1\).

Either

Let \(y=z^3\). Then by the chain rule, \(\dfrac{dy}{dx}=3z^2\dfrac{dz}{dx}\), and differentiating again gives \(\dfrac{d^2y}{dx^2}=6z\left(\dfrac{dz}{dx}\right)^2+3z^2\dfrac{d^2z}{dx^2}\).

Substitute these into the given differential equation:

\(3z^2z''+6z^2z'+6z(z')^2+5z^3=5x+2.\)

The first and third terms combine to give \(y''\), the second term is \(2y'\), and \(5z^3=5y\). Hence

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

Now solve the linear differential equation. For the complementary function, assume \(y=e^{rx}\), so

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

Thus \(r=-1\pm 2i\), and

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

For a particular solution, since the right-hand side is linear in \(x\), try \(y_p=ax+b\). Then \(y_p'=a\), \(y_p''=0\), so

\(2a+5(ax+b)=5x+2.\)

Comparing coefficients gives \(5a=5\), so \(a=1\), and \(2a+5b=2\), so \(2+5b=2\), hence \(b=0\). Therefore \(y_p=x\).

So the general solution is

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

We are given that when \(x=0\), \(z=1\) and \(\dfrac{dz}{dx}=-\dfrac23\). Since \(y=z^3\), we have \(y(0)=1\). Also

\(y'=3z^2z',\)

so at \(x=0\), \(y'(0)=3(1)^2\left(-\dfrac23\right)=-2.\)

Now use the general solution. At \(x=0\),

\(y(0)=A=1.\)

Differentiate \(y\):

\(y'=e^{-x}\big(-A\cos 2x-B\sin 2x-2A\sin 2x+2B\cos 2x\big)+1.\)

So at \(x=0\),

\(y'(0)=-A+2B+1.\)

Since \(y'(0)=-2\) and \(A=1\),

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

hence \(2B=-2\) and \(B=-1\).

Therefore

\(y=e^{-x}(\cos 2x-\sin 2x)+x.\)

Since \(y=z^3\),

\(z=\left[e^{-x}(\cos 2x-\sin 2x)+x\right]^{1/3}.\)

For large positive \(x\), the term \(e^{-x}(\cos 2x-\sin 2x)\) tends to \(0\), so \(y\approx x\). Hence \(z^3\approx x\), and therefore

\(z\approx x^{1/3}.\)

Or

The curve is \(y=\dfrac{x(x+1)}{(x-1)^2}\).

(i) The vertical asymptote occurs when the denominator is zero: \(x=1\). For large \(|x|\), the highest powers dominate, so

\(y\sim \dfrac{x^2}{x^2}=1,\)

giving the horizontal asymptote \(y=1\).

(ii) A point of intersection with the asymptotes must lie on \(y=1\). So solve

\(\dfrac{x(x+1)}{(x-1)^2}=1.\)

Then \(x(x+1)=(x-1)^2\), so

\(x^2+x=x^2-2x+1.\)

Thus \(3x=1\), so \(x=\dfrac13\). Substituting into \(y=1\), the point is \(\left(\dfrac13,1\right)\).

(iii) Using the quotient rule,

\(\dfrac{dy}{dx}=\dfrac{(2x+1)(x-1)^2-2x(x+1)(x-1)}{(x-1)^4}.\)

Factorising gives

\(\dfrac{dy}{dx}=\dfrac{-(3x+1)}{(x-1)^3}.\)

(a) Stationary points satisfy \(\dfrac{dy}{dx}=0\), so \(3x+1=0\), hence \(x=-\dfrac13\). Then

\(y=\dfrac{(-1/3)(2/3)}{(-4/3)^2}=-\dfrac18.\)

So the stationary point is \(\left(-\dfrac13,-\dfrac18\right)\).

(b) The gradient is negative when

\(\dfrac{-(3x+1)}{(x-1)^3}\lt 0.\)

Checking the sign of numerator and denominator gives

\(x\lt -\dfrac13\) or \(x\gt 1\).

(iv) For the sketch: draw a vertical asymptote \(x=1\) and a horizontal asymptote \(y=1\). The curve passes through \((0,0)\) and \((-1,0)\), meets the asymptote \(y=1\) at \(\left(\dfrac13,1\right)\), and has a minimum at \(\left(-\dfrac13,-\dfrac18\right)\). The left-hand branch approaches \(y=1\) as \(x\to -\infty\), rises through the x-axis, dips to the minimum, then rises to meet \(y=1\). The right-hand branch tends to \(+\infty\) as \(x\to 1^+\) and approaches \(y=1\) as \(x\to +\infty\).