Reducing First-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. $\frac{\mathrm{d}x}{\mathrm{d}t}=x+t^2$ has a dependent variable of $x$ and an independent variable of $t$.

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

$\frac{\mathrm{d}y}{\mathrm{d}x}=xy\ln y$

where $y=2$ when $x=0$.

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 $y={\mathrm{e}}^z$

$\frac{\mathrm{d}y}{\mathrm{d}z}={\mathrm{e}}^z$

Substitute this into the chain rule

$\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{e}}^z\frac{\mathrm{d}z}{\mathrm{d}x}$

Now substitute this derivative into the original differential equation

${\mathrm{e}}^z\frac{\mathrm{d}z}{\mathrm{d}x}=xy\ln y$

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

Use $y={\mathrm{e}}^z$ and $\ln y=\ln \left({\mathrm{e}}^z\right)=z$ to convert the remaining $y$ terms into $z$ terms

$\begin{array}{rcl}{\mathrm{e}}^z\frac{\mathrm{d}z}{\mathrm{d}x}& =& x\times {\mathrm{e}}^z\times z\\ {\mathrm{e}}^z\frac{\mathrm{d}z}{\mathrm{d}x}& =& xz{\mathrm{e}}^z\\ \frac{\mathrm{d}z}{\mathrm{d}x}& =& xz\end{array}$

Solve this differential equation using separation of variables

• Remember to add a constant of integration

$\begin{array}{rcl}\frac{1}{z}\frac{\mathrm{d}z}{\mathrm{d}x}& =& x\\ \int \frac{1}{z}\mathrm{d}z& =& \int x\mathrm{d}x\\ \ln \left|z\right|& =& \frac{x^2}{2}+c\\ z& =& {\mathrm{e}}^{\frac{x^2}{2}+c}\\ z& =& {\mathrm{e}}^{\frac{x^2}{2}}{\mathrm{e}}^c\\ z& =& A{\mathrm{e}}^{\frac{x^2}{2}}\end{array}$

Transform the general solution back to the variables $\left(x,y\right)$ using $y={\mathrm{e}}^z$ and $\ln y=z$

$\begin{array}{rcl}\ln y& =& A{\mathrm{e}}^{\frac{x^2}{2}}\\ y& =& {\mathrm{e}}^{A{\mathrm{e}}^{\frac{x^2}{2}}}\end{array}$

Now substitute in the boundary conditions $x=0$ and $y=2$ to find $A$

$\begin{array}{rcl}2& =& {\mathrm{e}}^{A{e}^0}\\ 2& =& {\mathrm{e}}^A\\ \ln 2& =& A\end{array}$

Substitute this value of $A$ back into the general solution to get the particular solution

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

This is the answer, but it can also be simplified to $y={\left({\mathrm{e}}^{\ln 2}\right)}^{{\mathrm{e}}^{\frac{x^2}{2}}}=2^{{\mathrm{e}}^{\frac{x^2}{2}}}$

Use the transformation $x=\frac{1}{t}$ where $t>0$ to find the particular solution of the differential equation

$x^2\frac{\mathrm{d}y}{\mathrm{d}x}-xy=-\frac{3}{x}$

where $y=2$ 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 $x=\frac{1}{t}=t^{-1}$ it is easier to use that $\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\begin{array}{rcl}\frac{\mathrm{d}t}{\mathrm{d}x}& =& \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}\\ & =& \frac{1}{-t^{-2}}\\ & =& \frac{1}{\left(-{\displaystyle \frac{1}{t^2}}\right)}\\ & =& 1÷\frac{-1}{t^2}\\ & =& 1\times \frac{t^2}{-1}\\ & =& -t^2\end{array}$

Substitute this into the chain rule

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

Now substitute this derivative into the original differential equation

$x^2\left(-t^2\frac{\mathrm{d}y}{\mathrm{d}t}\right)\mathit{-}xy\mathit{=}\mathit{-}\frac{3}{x}$

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

Use $x=\frac{1}{t}$ to convert the remaining $x$ terms into $t$ terms

$\begin{array}{rcl}{\left(\frac{1}{t}\right)}^2\left(-t^2\right)\frac{\mathrm{d}y}{\mathrm{d}t}\mathit{-}\mathit{\left(}\frac{1}{t}\mathit{\right)}y& \mathit{=}& \mathit{-}\frac{3}{\left({\displaystyle \frac{1}{t}}\right)}\\ \mathit{-}\frac{\mathrm{d}y}{\mathrm{d}t}\mathit{-}\frac{y}{t}& \mathit{=}& \mathit{-}3t\\ \frac{\mathrm{d}y}{\mathrm{d}t}+\frac{y}{t}& \mathit{=}& 3t\end{array}$

This is in the correct form $\frac{\mathrm{d}y}{\mathrm{d}t}+\mathrm{p}\left(t\right)y=\mathrm{q}\left(t\right)$ for the integrating factor

• the integrating factor is ${\mathrm{e}}^{\int \mathrm{p}\left(t\right)\mathrm{d}t}$

• where $\frac{\mathrm{d}}{\mathrm{d}t}\left(y{\mathrm{e}}^{\int \mathrm{p}\left(t\right)\mathrm{d}t}\right)=\mathrm{q}\left(t\right){\mathrm{e}}^{\int \mathrm{p}\left(t\right)\mathrm{d}t}$

Find the integrating factor

${\mathrm{e}}^{\int \frac{1}{t}\mathrm{d}t}={\mathrm{e}}^{\ln t}=t$

Solve the differential equation using the integrating factor

• Remember to add a constant of integration

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}t}\left(yt\right)& =& 3t\times t\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(yt\right)& =& 3t^2\\ yt& =& \int 3t^2\mathrm{d}t\\ yt& =& t^3+c\\ y& =& t^2+\frac{c}{t}\end{array}$

Convert the general solution back into the variables $\left(x,y\right)$ using $x=\frac{1}{t}$ and $t=\frac{1}{x}$

$\begin{array}{rcl}y& =& {\left(\frac{1}{x}\right)}^2+\frac{c}{\left({\displaystyle \frac{1}{x}}\right)}\\ y& =& \frac{1}{x^2}+cx\end{array}$

Find $c$ by substituting in $x=1$ and $y=2$

$\begin{array}{rcl}2& =& \frac{1}{1^2}+c\times 1\\ 1& =& c\end{array}$

Substitute $c=1$ into the general solution to get the particular solution

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

Don't forget to add a constant of integration when using the integrating factor, otherwise entire terms will be missing in the general solution!

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

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

where $y=2$ when $x=1$.

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

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

Substitute this into the original differential equation

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

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

Use $y=xz$ to convert the remaining $y$ terms into $x$ and $z$ terms then simplify

$\begin{array}{rcl}z+x\frac{\mathrm{d}z}{\mathrm{d}x}& =& \frac{\cancel{x}z}{\cancel{x}}+5x^5\\ \cancel{z}+x\frac{\mathrm{d}z}{\mathrm{d}x}& =& \cancel{z}+5x^5\\ x\frac{\mathrm{d}z}{\mathrm{d}x}& =& 5x^5\\ \frac{\mathrm{d}z}{\mathrm{d}x}& =& 5x^4\end{array}$

Solve this differential equation by direct integration of the right-hand side

• Remember to add a constant of integration

$z=x^5+c$

Transform the general solution back to the variables $\left(x,y\right)$ using $y=xz$ so $z=\frac{y}{x}$

$\begin{array}{rcl}\frac{y}{x}& =& x^5+c\\ y& =& x^6+cx\end{array}$

Now substitute in the boundary conditions $x=1$ and $y=2$ to find $c$

$\begin{array}{rcl}2& =& 1^6+c\times 1\\ 1& =& c\end{array}$

Substitute this value of $c$ back into the general solution to get the particular solution

$\begin{array}{rcl}y& =& x^6+x\end{array}$