Question

Solve the equation xfrac{dy}{dx}-ay=x+1 where a is a constant

Answer 1

First we find the general solution of $xy'-ay=0$.

The chatacteristic equation of $xy'-ay=0$ is $\lambda-a=0$, which has solution $\lambda_0=a$.

We obtain that the general solution of $xy'-ay=0$ is $y=Cx^{\lambda_0}=Cx^a$

Next, we find the partial solution of $xy'-ay=x+1$ of the form $y=Ax+B$

We have $xy'-ay=x(Ax+B)'-a(Ax+B)=A(1-a)x-aB$.

So we obtain that if $a\neq 0$ and $a\neq 1$ , then $A(1-a)x-aB=x+1$, that is $A(1-a)=1$ and $-aB=1$. Then the general solution of $xy'-ay=x+1$ is $Cx^a+Ax+B=Cx^a+\frac{1}{1-a}x-\frac{1}{a}$

Consider the case $a=0$. In this case we have the equation $xy'=x+1$, that is $y'=1+\frac{1}{x}$, so $y=\int\left(1+\frac{1}{x}\right)dx=x+\ln|x|+C$.

Consider case $a=1$. In this case we have the equation $x\frac{dy}{dx}-y=x+1$, that is $xdy-ydx=(x+1)dx$. Divide this equation by $x^2$ and obtain $\frac{xdy-ydx}{x^2}=\left(\frac{1}{x}+\frac{1}{x^2}\right)dx$

$\frac{xdy-ydx}{x^2}=d\left(\frac{y}{x}\right)$ and $\left(\frac{1}{x}+\frac{1}{x^2}\right)dx=d\left(\ln|x|-\frac{1}{x}\right)$ , so we obtain the solution $\frac{y}{x}=\ln|x|-\frac{1}{x}+C$, that is $y=x\ln|x|-1+Cx$

Answer:$y=\begin{cases} Cx^a+\frac{1}{1-a}x-\frac{1}{a},&\text{if $a\not\in\{0,1\}$}\\ x+ln|x|+C,&\text{if $a=0$}\\ xln|x|-1+Cx,&\text{if $a=1$} \end{cases}$