# Find the general solution of the given differential equation. x^2y'+x(x+7)y=e^x y=

Find the general solution of the given differential equation.
${x}^{2}{y}^{\prime }+x\left(x+7\right)y={e}^{x}$
$y=$
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Consider the differential equation
${x}^{2}{y}^{\prime }+x\left(x+7\right)y={e}^{x}$
On rewriting it, we get
${y}^{\prime }+\left(1+\frac{2}{x}\right)y=\frac{{e}^{x}}{{x}^{2}}...$(1)
Standard form of the linear differential equation $\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)$
Clearly equation (1) is in the standard form
Compare the differential equation (1) with standard orm, and identity
$P\left(x\right)=\frac{2+2}{x}$ and $f\left(x\right)=\frac{{e}^{x}}{{x}^{2}}$
The dunctions $P\left(x\right)=\frac{2+2}{x}$ and $f\left(x\right)=\frac{{e}^{x}}{{x}^{2}}$ are continuous on $\left(0,\mathrm{\infty }\right)$
The integrating facrote is
${e}^{\int P\left(x\right)dx}={e}^{\int \left(1+\frac{2}{x}\right)dx}$
$={e}^{x+2\mathrm{ln}x}$
${e}^{x}{e}^{{\mathrm{ln}x}^{2}}$
$={x}^{2}{e}^{x}$
Multiply the standart form, with the integrating factor
${x}^{2}{e}^{x}\left({y}^{\prime }+\left(1+\frac{2}{x}\right)y\right)={x}^{2}{e}^{x}\cdot \frac{{e}^{x}}{{x}^{2}}$
${x}^{2}{e}^{x}{y}^{\prime }+{x}^{2}{e}^{x}y+2x{e}^{x}y={e}^{2x}$
${x}^{2}{e}^{x}\frac{dy}{dx}+{x}^{2}y\frac{d}{dx}\left({e}^{x}\right)+{e}^{x}y\frac{d}{dx}\left({x}^{2}\right)={e}^{2x}$
$\frac{d}{dx}\left({x}^{2}{e}^{x}y\right)={e}^{2x}$
${x}^{2}{e}^{x}y=\int {e}^{2x}dx$
${x}^{2}{e}^{x}y=\frac{1}{2}{e}^{2x}+c$
${x}^{2}{e}^{x}y=\frac{1}{2{x}^{2}}{e}^{x}+c{x}^{-2}{e}^{-x}$
Hence the general solution of the given differential equation is

###### Not exactly what you’re looking for?
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answer Part 1 $\frac{1}{2{x}^{2}}{e}^{x}+C{x}^{-2}{e}^{-x}$
Part 2 $\left(0,\mathrm{\infty }\right)$
Part 3 $C{x}^{-2}{e}^{-x}$
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