Question
Solve the following differential equation:
$\frac{dy}{dx} -\frac{y}{x} = 2x^{2}$
$\frac{dy}{dx} -\frac{y}{x} = 2x^{2}$
$= e^{-\int\frac{- logx}{}} = \frac{1}{x} $
The solution of different equation is
$ y^\frac{1} {x} = {\int\frac{2x^{2}}{x}} dx + c$
$y^{{}{}} \frac{1}{x} = x^{2} + c$ or
$y = x^{3} + cx$
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| Differential equation | Function |
| $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$ | $\text{y}=\pm\sqrt{\text{a}^2-\text{x}^2}$ |