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

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

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

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

y = \frac{x ^{3}}{e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{3}}{e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

y = \frac{x ^{3}}{e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

\int (sin (x))^{3} (cos (x))^{4} d x

\int \frac{12 sec ^{2} ( x )}{tan ^{7} ( x )} d x

(2 x^{3} + y) d x - x d y = 0

y^{^{''}} - 2 y^{^{'}} + y = - 2 e^{x}

\int \frac{x}{1 - x} d x

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