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

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

Question
Differential equations
asked 2021-02-05
Solve the differential equations
(1) \(\displaystyle{x}{y}'-{2}{y}={x}^{{3}}{e}^{{x}}\)
(2) \(\displaystyle{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{2}}{x}={0}\)

Answers (1)

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.
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