 # Solve the differential equations (1) xy'-2y=x^3e^x (2) (2ydx+dy)e^2x=0 Aneeka Hunt 2021-02-05 Answered
Solve the differential equations
(1) $x{y}^{\prime }-2y={x}^{3}{e}^{x}$
(2) $\left(2ydx+dy\right){e}^{2}x=0$
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(1)Here the differential equation is given by $x{y}^{\prime }-2y={x}^{3}{e}^{x}$.
This implies that ${y}^{\prime }-\frac{2}{x},y={x}^{2}{e}^{x}$.The integrating factor is
${e}^{-\int \frac{2}{x}dx}={e}^{-2\mathrm{log}x}=\frac{1}{{x}^{2}}$.Multiplying both side by $\frac{1}{{x}^{2}}$ we get
$y\frac{{}^{\prime }}{{x}^{2}}-\frac{2}{{x}^{3}}y={e}^{x}$.
Taking integral we get
$\frac{y}{{x}^{2}}=\int {e}^{x}dx+c$
$⇒\frac{y}{{x}^{2}}={e}^{x}dx+c$
$⇒y={x}^{2}{e}^{x}dx+c{x}^{2}$,
where c is an arbitrary constant.
(2)Here the differential equation is given by $\left(2ydx+dy\right){e}^{2x}=0$.Now
$\left(2ydx+dy\right){e}^{2x}=0$
$⇒\left(2ydx+dy\right){e}^{2x}=0$
$⇒\frac{dy}{y}=-2dx$
$⇒\int \frac{dy}{y}=-\int 2dx$
$⇒\mathrm{log}y=-2x+c$
$⇒y={e}^{-2x+c}$,
where c is an arbitrary constant.

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