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Differential Equation -01


Well-known member
Jan 25, 2013
Please find the general solution of :

$xy'-2y =x^2$


Staff member
Feb 24, 2012
Multiplying through by $x^{-3}$ where $x\ne0$, we obtain:

\(\displaystyle x^{-2}y'-2x^{-3}y=x^{-1}\)

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

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

Integrate with respect to $x$:

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

\(\displaystyle x^{-2}y=\ln|x|+C\)

Thus, we find the general solution is:

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
Just because it might not be obvious why we should multiply by \(\displaystyle \displaystyle \begin{align*} x^{-3} \end{align*}\)...

\(\displaystyle \displaystyle \begin{align*} x\,\frac{dy}{dx} - 2y &= x^2 \\ \frac{dy}{dx} - \frac{2}{x}\,y &= x \end{align*}\)

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

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

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

\(\displaystyle \displaystyle \begin{align*} x^{-2}\,\frac{dy}{dx} - 2x^{-3}\,y &= \frac{1}{x} \end{align*}\)

which is the same as multiplying the original equation by \(\displaystyle \displaystyle \begin{align*} x^{-3} \end{align*}\).


Indicium Physicus
Staff member
Jan 26, 2012
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.