Reducing Second-Order Differential Equations (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Be careful in modelling questions, as the letters can change, e.g. $\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}+x=t^2$ has a dependent variable of $x$ and an independent variable of $t$.

Use the transformation $y=\frac{1}{z}$ to find the general solution of the differential equation

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}-\frac{2}{y}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2-5\frac{\mathrm{d}y}{\mathrm{d}x}+6y=-{\mathrm{e}}^x{y}^2$

Answer:

Identify the variables being transformed

$\left(x,y\right)\to \left(x,z\right)$

This is a transformation of the dependent variable, $y$

Write $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\frac{\mathrm{d}z}{\mathrm{d}x}$ using the chain rule

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}z}\times \frac{\mathrm{d}z}{\mathrm{d}x}$

Find $\frac{\mathrm{d}y}{\mathrm{d}z}$ from $y=\frac{1}{z}=z^{-1}$

$\frac{\mathrm{d}y}{\mathrm{d}z}=-{\mathrm{z}}^{-2}=-\frac{1}{{\mathrm{z}}^2}$

Substitute this into the chain rule

$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}$

Now differentiate both sides of this derivative by $\frac{\mathrm{d}}{\mathrm{d}x}$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}\right)$

Simplify the left-hand side and use the product rule on the right-hand side

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(-\frac{1}{z^2}\right)\frac{\mathrm{d}z}{\mathrm{d}x}+\left(-\frac{1}{z^2}\right)\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(-z^{-2}\right)\frac{\mathrm{d}z}{\mathrm{d}x}-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}\end{array}$

Use implicit differentiation on the part $\begin{array}{rcl}& & \frac{\mathrm{d}}{\mathrm{d}x}\left(-z^{-2}\right)\end{array}$

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{\mathrm{d}}{\mathrm{d}z}\left(-z^{-2}\right)\times \frac{\mathrm{d}z}{\mathrm{d}x}\times \frac{\mathrm{d}z}{\mathrm{d}x}-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{\mathrm{d}}{\mathrm{d}z}\left(-z^{-2}\right){\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& 2z^{-3}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{2}{z^3}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}\end{array}$

Now substitute $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$ into the original differential equation

$\left(\begin{array}{rcl}& & \frac{2}{z^3}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}\end{array}\right)-\frac{2}{y}{\left(-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-5\left(-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}\right)+6y=-{\mathrm{e}}^x{y}^2$

This is not yet in the form $\left(x,z\right)$, as there are still $y$ terms on both sides of the equation

Use $y=\frac{1}{z}$ to convert the remaining $y$ terms into $z$ terms, then simplify

$\begin{array}{rcl}\frac{2}{z^3}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}-2z{\left(-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-5\left(-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}\right)+\frac{6}{z}& =& -\frac{{\mathrm{e}}^x}{{\mathrm{z}}^2}\\ \frac{2}{z^3}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}-2z\times \frac{1}{z^4}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2+\frac{5}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}+\frac{6}{z}& =& -\frac{{\mathrm{e}}^x}{{\mathrm{z}}^2}\\ \cancel{\frac{2}{z^3}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2}-\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}\cancel{-\frac{2}{z^3}{\left(\frac{\mathrm{d}z}{\mathrm{d}x}\right)}^2}+\frac{5}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}+\frac{6}{z}& =& -\frac{{\mathrm{e}}^x}{{\mathrm{z}}^2}\\ -\frac{1}{z^2}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}+\frac{5}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}+\frac{6}{z}& =& -\frac{{\mathrm{e}}^x}{z^2}\end{array}$

Multiply both sides by $-z^2$

$\begin{array}{rcl}\frac{{\mathrm{d}}^2\mathrm{z}}{\mathrm{d}{x}^2}-5\frac{\mathrm{d}z}{\mathrm{d}x}-6z& =& {\mathrm{e}}^x\end{array}$

Solve the auxiliary equation

$\begin{array}{rcl}{m}^2-5m-6& =& 0\\ \left(m-6\right)\left(m+1\right)& =& 0\\ m& =& 6\mathrm{or}m=-1\end{array}$

Write out the complementary function

$A{\mathrm{e}}^{6x}+B{\mathrm{e}}^{-x}$

Find the particular integral, of the form $z=\lambda {\mathrm{e}}^x$

$\begin{array}{rcl}\lambda {\mathrm{e}}^x-5{\mathrm{\lambda e}}^x-6{\mathrm{\lambda e}}^x& \equiv & {\mathrm{e}}^x\\ -10\mathrm{\lambda }& =& 1\\ \mathrm{\lambda }& =& -\frac{1}{10}\end{array}$

Find the general solution for $z$ in terms of $x$

$z=A{\mathrm{e}}^{6x}+B{\mathrm{e}}^{-x}-\frac{1}{10}{\mathrm{e}}^x$

Use the transformation $y=\frac{1}{z}$ to find the general solution for $y$ in terms of $x$

$y=\frac{1}{A{\mathrm{e}}^{6x}+B{\mathrm{e}}^{-x}-\frac{1}{10}{\mathrm{e}}^x}$

This is the general solution, but it can also be rearranged (with new arbitrary constants) to $y=\frac{10}{C{\mathrm{e}}^{6x}+D{\mathrm{e}}^{-x}-{\mathrm{e}}^x}$

Use the transformation $x={\mathrm{e}}^t$ to find the particular solution of the differential equation

$x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+5x\frac{\mathrm{d}y}{\mathrm{d}x}+4y=\ln x$

where $y=\frac{3}{4}$ and $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{3}{4}$when $x=1$.

Answer:

Identify the variables being transformed

$\left(x,y\right)\to \left(t,y\right)$

This is a transformation of the independent variable, $x$

Write $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\frac{\mathrm{d}y}{\mathrm{d}t}$ using the chain rule

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x}$

To find $\frac{\mathrm{d}t}{\mathrm{d}x}$ from $x={\mathrm{e}}^t$ it is easier to use that $\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\begin{array}{rcl}\frac{\mathrm{d}t}{\mathrm{d}x}& =& \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}\\ & =& \frac{1}{{\mathrm{e}}^t}\end{array}$

Substitute this into the chain rule

$\begin{array}{rcl}\frac{\mathrm{d}y}{\mathrm{d}x}& =& \frac{\mathrm{d}y}{\mathrm{d}t}\times \left(\frac{1}{{\mathrm{e}}^t}\right)\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =& {\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}\end{array}$

Now differentiate both sides of this derivative by $\frac{\mathrm{d}}{\mathrm{d}x}$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\begin{array}{rcl}& & {\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}\end{array}\right)$

Both terms on the right-hand side are in terms of $t$ so you can use implicit differentiation before the product rule

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(\begin{array}{rcl}& & {\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}\end{array}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}$

Simplify the left-hand side and use the product rule on the right-hand side

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \left(-{\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}+{\mathrm{e}}^{-t}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\right)\frac{\mathrm{d}t}{\mathrm{d}x}\end{array}$

To find $\frac{\mathrm{d}t}{\mathrm{d}x}$ from $x={\mathrm{e}}^t$ use that $\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \left(-{\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}+{\mathrm{e}}^{-t}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\right)\frac{1}{{\mathrm{e}}^t}\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& -{\mathrm{e}}^{-2t}\frac{\mathrm{d}y}{\mathrm{d}t}+{\mathrm{e}}^{-2t}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\end{array}$

Now substitute $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$ into the original differential equation

$x^2\left(\begin{array}{rcl}& & -{\mathrm{e}}^{-2t}\frac{\mathrm{d}y}{\mathrm{d}t}+{\mathrm{e}}^{-2t}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\end{array}\right)+5x\left(\begin{array}{rcl}& & {\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}\end{array}\right)+4y=\ln x$

This is not yet in the form $\left(t,y\right)$, as there are still $x$ terms remaining

Use $x={\mathrm{e}}^t$ to convert the remaining $x$ terms into $t$ terms, then simplify

$\begin{array}{rcl}{\left({\mathrm{e}}^t\right)}^2\left(-{\mathrm{e}}^{-2t}\frac{\mathrm{d}y}{\mathrm{d}t}+{\mathrm{e}}^{-2t}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\right)+5\left({\mathrm{e}}^t\right)\left({\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}\right)+4y& =& \ln \left({\mathrm{e}}^t\right)\\ {\mathrm{e}}^{2t}\left(-{\mathrm{e}}^{-2t}\frac{\mathrm{d}y}{\mathrm{d}t}+{\mathrm{e}}^{-2t}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\right)+5{\mathrm{e}}^t\left({\mathrm{e}}^{-t}\frac{\mathrm{d}y}{\mathrm{d}t}\right)+4y& =& t\\ -\frac{\mathrm{d}y}{\mathrm{d}t}+\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}+5\frac{\mathrm{d}y}{\mathrm{d}t}+4y& =& t\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}+4\frac{\mathrm{d}y}{\mathrm{d}t}+4y& =& t\end{array}$

Solve the auxiliary equation

$\begin{array}{rcl}{m}^2+4m+4& =& 0\\ {\left(m+2\right)}^2& =& 0\\ m& =& -2\mathrm{repeated}\end{array}$

Write out the complementary function

$\left(A+Bt\right){\mathrm{e}}^{-2t}$

Find the particular integral, of the form $y=\lambda +\mu t$

$\begin{array}{rcl}4\mu +4\lambda +4\mu t& \equiv & t\\ 4\mu & =& 1\Rightarrow \mu =\frac{1}{4}\\ \mu +\lambda & =& 0\Rightarrow \lambda =-\frac{1}{4}\end{array}$

Find the general solution for $y$ in terms of $t$

$y=\left(A+Bt\right){\mathrm{e}}^{-2t}-\frac{1}{4}+\frac{1}{4}t$

Use the transformation $x={\mathrm{e}}^t$ (i.e. $t=\ln x$) to find the general solution for $y$ in terms of $t$

$\begin{array}{l}y=\left(A+B\ln x\right){\mathrm{e}}^{-2\ln x}-\frac{1}{4}+\frac{1}{4}\ln x\\ y=\left(A+B\ln x\right){\mathrm{e}}^{\ln \left(x^{-2}\right)}-\frac{1}{4}+\frac{1}{4}\ln x\\ y=\left(A+B\ln x\right)x^{-2}-\frac{1}{4}+\frac{1}{4}\ln x\\ y=A{x}^{-2}+B{x}^{-2}\ln x-\frac{1}{4}+\frac{1}{4}\ln x\end{array}$

This is the general solution so to find the particular solution first use that $y=\frac{3}{4}$ when $x=1$

$\begin{array}{rcl}\frac{3}{4}& =& A+B\times 0-\frac{1}{4}+\frac{1}{4}\times 0\\ 1& =& A\end{array}$

Then use that $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{3}{4}$when $x=1$, which requires differentiating the general solution (with $A=1$) first

$\begin{array}{rcl}y& =& x^{-2}+B{x}^{-2}\ln x-\frac{1}{4}+\frac{1}{4}\ln x\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =& -2x^{-3}+B\left(-2x^{-3}\ln x+x^{-2}\times \frac{1}{x}\right)-0+\frac{1}{4}\times \frac{1}{x}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =& -2x^{-3}+B\left(-2x^{-3}\ln x+x^{-3}\right)+\frac{1}{4x}\end{array}$

Substitute in $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{3}{4}$ when $x=1$

$\begin{array}{rcl}-\frac{3}{4}& =& -2+B\left(-2\times 0+1\right)+\frac{1}{4}\\ -1& =& -2+B\\ 1& =& B\end{array}$

Substitute $A=1$ and $B=1$ into the general solution to get the particular solution

$y=x^{-2}+x^{-2}\ln x-\frac{1}{4}+\frac{1}{4}\ln x$

$y=\frac{1}{x^2}+\frac{\ln x}{x^2}-\frac{1}{4}+\frac{1}{4}\ln x$

Use the transformation $y=xz$ to find the general solution of the differential equation

$x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+2\left(x^2-x\right)\frac{\mathrm{d}y}{\mathrm{d}x}+\left(5x^2-2x+2\right)y=x^3\sin x$

Answer:

Identify the variables being transformed

$\left(x,y\right)\to \left(x,z\right)$

This is a transformation of the dependent variable, $y$

Write $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\frac{\mathrm{d}z}{\mathrm{d}x}$ using the product rule

$\begin{array}{rcl}\frac{\mathrm{d}y}{\mathrm{d}x}& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(xz\right)\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =& 1\times z+x\times \frac{\mathrm{d}z}{\mathrm{d}x}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =& z+x\frac{\mathrm{d}z}{\mathrm{d}x}\end{array}$

Now differentiate both sides of this derivative by $\frac{\mathrm{d}}{\mathrm{d}x}$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\begin{array}{rcl}& & z+x\frac{\mathrm{d}z}{\mathrm{d}x}\end{array}\right)$

Simplify the left-hand side and find the derivative in $x$ of the two terms on the right-hand side

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{\mathrm{d}z}{\mathrm{d}x}+\frac{\mathrm{d}}{\mathrm{d}x}\left(x\frac{\mathrm{d}z}{\mathrm{d}x}\right)\end{array}$

Use the product rule for the last term, then simplify

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{\mathrm{d}z}{\mathrm{d}x}+\frac{\mathrm{d}z}{\mathrm{d}x}+x\frac{{\mathrm{d}}^{2}z}{\mathrm{d}{x}^2}\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& 2\frac{\mathrm{d}z}{\mathrm{d}x}+x\frac{{\mathrm{d}}^{2}z}{\mathrm{d}{x}^2}\end{array}$

Now substitute $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$ into the original differential equation

$x^2\left(\begin{array}{rcl}& & 2\frac{\mathrm{d}z}{\mathrm{d}x}+x\frac{{\mathrm{d}}^{2}z}{\mathrm{d}{x}^2}\end{array}\right)+2\left(x^2-x\right)\left(\begin{array}{rcl}& & z+x\frac{\mathrm{d}z}{\mathrm{d}x}\end{array}\right)+\left(5x^2-2x+2\right)y=x^3\sin x$

This is not yet in the form $\left(x,z\right)$, as there is still a $y$ term on the left-hand side

Use $y=xz$ to convert the remaining $y$ term int a $z$ term, then simplify

$\begin{array}{rcl}{x}^2\left(2\frac{\mathrm{d}z}{\mathrm{d}x}+x\frac{{\mathrm{d}}^{2}z}{\mathrm{d}{x}^2}\right)+2\left(x^2-x\right)\left(z+x\frac{\mathrm{d}z}{\mathrm{d}x}\right)+\left(5x^2-2x+2\right)xz& =& x^3\sin x\\ \cancel{2x^2\frac{\mathrm{d}z}{\mathrm{d}x}}+x^3\frac{{\mathrm{d}}^{2}z}{\mathrm{d}{x}^2}+\cancel{2x^{2}z}-\bcancel{2xz}+2x^3\frac{\mathrm{d}z}{\mathrm{d}x}-\cancel{2x^2\frac{\mathrm{d}z}{\mathrm{d}x}}+5x^{3}z\cancel{-2x^{2}z}+\bcancel{2xz}& =& x^3\sin x\\ x^3\frac{{\mathrm{d}}^{2}z}{\mathrm{d}{x}^2}+2x^3\frac{\mathrm{d}z}{\mathrm{d}x}+5x^{3}z& =& x^3\sin x\end{array}$

Divide both sides by $x^3$

$\begin{array}{rcl}\frac{{\mathrm{d}}^{2}z}{\mathrm{d}{x}^2}+2\frac{\mathrm{d}z}{\mathrm{d}x}+5z& =& \sin x\end{array}$

Solve the auxiliary equation

$\begin{array}{rcl}{m}^2+2m+5& =& 0\\ m& =& \frac{-2\pm \sqrt{2^2-4\times 1\times 5}}{2}\\ m& =& \frac{-2\pm \sqrt{-16}}{2}\\ m& =& \frac{-2\pm 4\mathrm{i}}{2}\\ m& =& -1\pm 2\mathrm{i}\end{array}$

Write out the complementary function

${\mathrm{e}}^{-x}\left(A\cos 2x+B\sin 2x\right)$

Find the particular integral, of the form $z=\lambda \cos x+\mu \sin x$

$\begin{array}{l}\left(-\lambda \cos x-\mu \sin x\right)+2\left(-\lambda \sin x+\mu \cos x\right)+5\left(\lambda \cos x+\mu \sin x\right)\equiv \sin x\\ -\lambda +2\mu +5\lambda =0\Rightarrow \lambda =-\frac{1}{2}\mu \\ -\mu -2\lambda +5\mu =1\Rightarrow \mu =\frac{1}{5}\mathrm{and}\lambda =-\frac{1}{10}\end{array}$

Find the general solution for $z$ in terms of $x$

$z={\mathrm{e}}^{-x}\left(A\cos 2x+B\sin 2x\right)-\frac{1}{10}\cos x+\frac{1}{5}\sin x$

Use the transformation $y=xz$, i.e. $z=\frac{y}{x}$, to find the general solution for $y$ in terms of $x$

$y=x\left({\mathrm{e}}^{-x}\left(A\cos 2x+B\sin 2x\right)-\frac{1}{10}\cos x+\frac{1}{5}\sin x\right)$

$y=x{\mathrm{e}}^{-x}\left(A\cos 2x+B\sin 2x\right)-\frac{1}{10}x\cos x+\frac{1}{5}x\sin x$