What is the solution for x^4y^'+3x^3y=x^2e^x ?

The solution for x^4y^'+3x^3y=x^2e^x is y=(e^x)/(x^2)-(e^x)/(x^3)+(c_{1})/(x^3)

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x^4y^'+3x^3y=x^2e^x

x^{4} y^{^{'}} + 3 x^{3} y = x^{2} e^{x}

y = \frac{e ^{x}}{x ^{2}} - \frac{e ^{x}}{x ^{3}} + \frac{c _{1}}{x ^{3}}

Solution steps

x^{4} y^{'} (x) + 3 x^{3} y = x^{2} e^{x}

Solve linear ODE:

y = \frac{e ^{x}}{x ^{2}} - \frac{e ^{x}}{x ^{3}} + \frac{c _{1}}{x ^{3}}

y = \frac{e ^{x}}{x ^{2}} - \frac{e ^{x}}{x ^{3}} + \frac{c _{1}}{x ^{3}}

\frac{d}{d x} (\frac{3 x - 9}{2 x + 4})

\int \frac{1}{16 x ^{2} + 25} d x

\int 15 - 2 x - x^{2} d x

\frac{d y}{d t} + k (340 - y) = 0

e

What is the solution for x^4y^'+3x^3y=x^2e^x ?
The solution for x^4y^'+3x^3y=x^2e^x is y=(e^x)/(x^2)-(e^x)/(x^3)+(c_{1})/(x^3)