Differential equations reducible by substitutionEdexcel International A Level Further Maths: Revision notes

Some first order equations are neither separable nor linear as they stand. A suitable substitution turns them into one of those types. In this specification the substitution is given in the question. Method:

1. Write the new variable in terms of $x$ and $y$ (or $y$ in terms of $x$ and the new variable).
2. Differentiate to find $\frac{dy}{dx}$ in terms of the new variable and its derivative, using the product or chain rule.
3. Substitute into the original equation and simplify to a separable or linear equation.
4. Solve, then substitute back to give the answer in terms of $x$ and $y$. Check the final answer: if the question asks for $y$ in terms of $x$, do not leave it in terms of $v$ or $z$.

If every term can be written as a function of $\frac yx$, substitute $y=vx$. By the product rule $\frac{dy}{dx}=v+x\frac{dv}{dx}.$ The equation becomes separable in $v$ and $x$. Example: $\frac{dy}{dx}=\frac{x+2y}{x}$. Then $v+x\frac{dv}{dx}=1+2v$, so $x\frac{dv}{dx}=1+v$ and $\int\frac{1}{1+v}\,dv=\int\frac1x\,dx$. Hence $\ln(1+v)=\ln x+c$, $1+v=Ax$ and $y=Ax^2-x$. Check: $y'=2Ax-1$ and $\frac{x+2y}{x}=1+2(Ax-1)=2Ax-1$.

When the equation contains $x+y$ (or similar) as a block, let $z=x+y$. Then $\frac{dz}{dx}=1+\frac{dy}{dx}$ and the equation becomes separable in $z$ and $x$. Example: $\frac{dy}{dx}=(x+y+1)^2$. Let $z=x+y+1$, so $\frac{dz}{dx}=1+z^2$. Then $\int\frac{1}{1+z^2}\,dz=\int dx$ gives $\arctan z=x+C$, so $y=\tan(x+C)-x-1$. With $z=y-x$: $\frac{dz}{dx}=\frac{dy}{dx}-1$.

Some equations become first order linear under a substitution such as $z=\frac1y$ or $z=y^2$. Then use the integrating factor $e^{\int P\,dx}$. Example: $\frac{dy}{dx}+2y=y^2e^x$ with $z=\frac1y$. Dividing by $y^2$: $y^{-2}\frac{dy}{dx}+2y^{-1}=e^x$, so $-\frac{dz}{dx}+2z=e^x$, that is $\frac{dz}{dx}-2z=-e^x$. The integrating factor is $e^{-2x}$, so $\frac{d}{dx}\left(ze^{-2x}\right)=-e^{-x}$ and $ze^{-2x}=e^{-x}+C$. Hence $z=e^x+Ce^{2x}$ and $y=\frac{1}{e^x+Ce^{2x}}$.

Use a given condition after back-substituting (or in terms of the new variable, provided you convert the condition correctly). Example: $xy\frac{dy}{dx}=x^2+y^2$ gives $y^2=2x^2(\ln x+C)$. If $y=2$ when $x=1$, then $C=2$, so $y=x\sqrt{2\ln x+4}$. This is valid when $2\ln x+4>0$, that is $x>e^{-2}$. Always state the domain when the solution contains $\ln$, a square root or a denominator, and choose the root sign that matches the given sign of $y$.