Question

# Solve the differential equations (1) xy'-2y=x^3e^x (2) (2ydx+dy)e^2x=0

Second order linear equations
Solve the differential equations
(1) $$\displaystyle{x}{y}'-{2}{y}={x}^{{3}}{e}^{{x}}$$
(2) $$\displaystyle{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{2}}{x}={0}$$

2021-02-06
(1)Here the differential equation is given by $$\displaystyle{x}{y}'-{2}{y}={x}^{{3}}{e}^{{x}}$$.
This implies that $$\displaystyle{y}'-\frac{{2}}{{x}},{y}={x}^{{2}}{e}^{{x}}$$.The integrating factor is
$$\displaystyle{e}^{{-\int\frac{{2}}{{x}}{\left.{d}{x}\right.}}}={e}^{{-{2}{\log{{x}}}}}=\frac{{1}}{{x}^{{2}}}$$.Multiplying both side by $$\displaystyle\frac{{1}}{{x}^{{2}}}$$ we get
$$\displaystyle{y}\frac{'}{{x}^{{2}}}-\frac{{2}}{{x}^{{3}}}{y}={e}^{{x}}$$.
Taking integral we get
$$\displaystyle\frac{{y}}{{x}^{{2}}}=\int{e}^{{x}}{\left.{d}{x}\right.}+{c}$$
$$\displaystyle\Rightarrow\frac{{y}}{{x}^{{2}}}={e}^{{x}}{\left.{d}{x}\right.}+{c}$$
$$\displaystyle\Rightarrow{y}={x}^{{2}}{e}^{{x}}{\left.{d}{x}\right.}+{c}{x}^{{2}}$$,
where c is an arbitrary constant.
(2)Here the differential equation is given by $$\displaystyle{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{{2}{x}}}={0}$$.Now
$$\displaystyle{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{{2}{x}}}={0}$$
$$\displaystyle\Rightarrow{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{{2}{x}}}={0}$$
$$\displaystyle\Rightarrow\frac{{\left.{d}{y}\right.}}{{y}}=-{2}{\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow\int\frac{{{\left.{d}{y}\right.}}}{{y}}=-\int{2}{\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow{\log{{y}}}=-{2}{x}+{c}$$
$$\displaystyle\Rightarrow{y}={e}^{{-{2}{x}+{c}}}$$,
where c is an arbitrary constant.