It is given that \(t \neq 0\) and
\(t \frac{\mathrm{~d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{~d} x}{\mathrm{~d} t}+9 t x=3 t^{2}+1\)
(i) Show that if \(y=t x\) then
\(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+9 y=3 t^{2}+1\)

(ii) Find \(x\) in terms of \(t\), given that \(x=\frac{1}{9} \pi\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{2}{3}\) when \(t=\frac{1}{3} \pi\).

Use the substitution \(z=x+y\) to find the solution of the differential equation
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+3 x+3 y}{3 x+3 y-1}\)
for which \(y=0\) when \(x=1\). Give your answer in the form \(a \ln (x+y)+b(x-y)+c=0\), where \(a\), \(b\) and \(c\) are constants to be determined.

It is given that \(v=y^{4}\) and
\(y^{3} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}+3 y^{2}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{2}+y^{3} \frac{\mathrm{~d} y}{\mathrm{~d} x}+y^{4}=\mathrm{e}^{-2 x} .\)
(a) Show that
\(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}+\frac{\mathrm{d} v}{\mathrm{~d} x}+4 v=4 \mathrm{e}^{-2 x}\)
(b) Find \(y\) in terms of \(x\), given that, when \(x=0, y=1\) and \(\frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{3}{8}\).

Use the substitution \(y=v x\) to find the solution of the differential equation
\(x \frac{\mathrm{~d} y}{\mathrm{~d} x}=y+\sqrt{9 x^{2}+y^{2}}\)
for which \(y=0\) when \(x=1\). Give your answer in the form \(y=\mathrm{f}(x)\), where \(\mathrm{f}(x)\) is a polynomial in \(x\).

(a) Use the substitution \(u=1-(\theta-1)^{2}\) to find
\(\int \frac{\theta-1}{\sqrt{1-(\theta-1)^{2}}} \mathrm{~d} \theta .\)
(b) Find the solution of the differential equation
\(\theta \frac{\mathrm{d} y}{\mathrm{~d} \theta}-y=\theta^{2} \sin ^{-1}(\theta-1),\)
where \(0\lt \theta\lt 2\), given that \(y=1\) when \(\theta=1\). Give your answer in the form \(y=\mathrm{f}(\theta)\).

It is given that \(x=t^{3} y\) and
\[t^{3} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} t^{2}}+\left(4 t^{3}+6 t^{2}\right) \frac{\mathrm{d} y}{\mathrm{~d} t}+\left(13 t^{3}+12 t^{2}+6 t\right) y=61 \mathrm{e}^{\frac{1}{2} t}\]
(a) Show that
\[\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4 \frac{\mathrm{~d} x}{\mathrm{~d} t}+13 x=61 \mathrm{e}^{\frac{1}{2} t}\]
(b) Find the general solution for \(y\) in terms of \(t\).

It is given that \(y=x^{2} w\) and
\[x^{2} \frac{\mathrm{~d}^{2} w}{\mathrm{~d} x^{2}}+4 x(x+1) \frac{\mathrm{d} w}{\mathrm{~d} x}+\left(5 x^{2}+8 x+2\right) w=5 x^{2}+4 x+2\]
(a) Show that
\[\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+4 \frac{\mathrm{~d} y}{\mathrm{~d} x}+5 y=5 x^{2}+4 x+2\]
(b) Find the general solution for \(w\) in terms of \(x\).

The variables \(x\) and \(y\) are related by the differential equation
\(y^{2} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} x^{2}}+2 y^{2} \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^{2}-5 y^{3}=8 \mathrm{e}^{-x} .\)

Given that \(v=y^{3}\), show that
\(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}+2 \frac{\mathrm{~d} v}{\mathrm{~d} x}-15 v=24 \mathrm{e}^{-x}\)

Hence find the general solution for \(y\) in terms of \(x\).