MCQ
The general solution of the differential equation $\frac{{dy}}{{dx}} + \frac{2}{x}y = {x^2}$ is
- A$y = c{x^{ - 3}} - \frac{{{x^2}}}{4}$
- B$y = c{x^3} - \frac{{{x^2}}}{4}$
- C$y = c{x^2} + \frac{{{x^3}}}{5}$
- ✓$y = c{x^{-2}} + \frac{{{x^3}}}{5}$
$\frac{d y}{d x}+\frac{2}{x}, y=x^{2}$
This is of thelinear form
$\therefore P=\frac{2}{x}, Q=x^{2}$
I.F $=e^{\int \frac{2}{x} d x}=e^{\log x^{2}}=x^{2}$
Solution is
$y \cdot x^{2}=\int x^{2} x^{2} d x+c=\frac{x^{5}}{5}+c$
$y=\frac{x^{3}}{5}+c x^{-2}$
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