Question

Solve the differential equation = x dy / dx + 2y = x⁴

Answer 1

The given equation can be written as

$\dfrac{dy}{dx} +\dfrac{2y}{x}= x^{3}$ . This is of the form $\dfrac{dy}{dx}+Py=Q$, whose solution is given by

$ye^{\int Pdx} = \int Qe^{\int Pdx} dx + c$.

Here $P = \dfrac{2}{x}, Q= x^{3}$ . Then, $e^{\int Pdx} = e^{\int \frac{2}{x}dx} = e^{2 \log x}= e^{\log x^{2}} = x^{2}.$ Thus the solution is

$\begin{aligned} yx^{2} &= \int x^{3} \cdot x^{2} dx + c\\ yx^{2} &= \int x^{5} dx + c \\ yx^{2} &= \frac {x^{6}}{6} + c \\ \therefore y & = \frac {x^{4}}{6} + cx^{-2} \\ \end{aligned}$