What is the general solution to the differential equation $xy'-2y =x^2$?

[Albert1]

Please find the general solution of :

$xy'-2y =x^2$

 

[MarkFL]

Multiplying through by $x^{-3}$ where $x\ne0$, we obtain:

$$x^{-2}y'-2x^{-3}y=x^{-1}$$

Now we may observe that the left have side is the derivative of a product:

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

Integrate with respect to $x$:

$$\int\frac{d}{dx}\left(x^{-2}y \right)\,dx=\int x^{-1}\,dx$$

$$x^{-2}y=\ln|x|+C$$

Thus, we find the general solution is:

$$y(x)=x^2\left(\ln|x|+C \right)$$

 

[Prove It]

Just because it might not be obvious why we should multiply by [math]\displaystyle \begin{align*} x^{-3} \end{align*}[/math]...

[math]\displaystyle \begin{align*} x\,\frac{dy}{dx} - 2y &= x^2 \\ \frac{dy}{dx} - \frac{2}{x}\,y &= x \end{align*}[/math]

which is now a first order linear DE. The integrating factor is

[math] \displaystyle \begin{align*} e^{ \int{ -\frac{2}{x} \, dx } } = e^{ -2\ln{(x)} } = e^{ \ln{ \left( x^{-2} \right) } } = x^{-2} \end{align*} [/math]

so multiplying both sides of our linear DE by the integrating factor gives

[math]\displaystyle \begin{align*} x^{-2}\,\frac{dy}{dx} - 2x^{-3}\,y &= \frac{1}{x} \end{align*}[/math]

which is the same as multiplying the original equation by [math]\displaystyle \begin{align*} x^{-3} \end{align*}[/math].

 

[Ackbach]

Another method is to recognize that the equation is Cauchy-Euler. Hence, you can substitute $y=x^{r}$ and solve for $r$. You will need reduction of order to get the logarithm function. Alternatively, the substitution $t=\ln(x)$ renders the equation first-order linear with constant coefficients, at which point you employ the usual methods.