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.