What is the solution for y^{''}-2y^'+y=24x^2e^x ?

The solution for y^{''}-2y^'+y=24x^2e^x is y=c_{1}e^x+c_{2}xe^x+2e^xx^4

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y^{''}-2y^'+y=24x^2e^x

y^{^{''}} - 2 y^{^{'}} + y = 24 x^{2} e^{x}

y = c_{1} e^{x} + c_{2} x e^{x} + 2 e^{x} x^{4}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 24 x^{2} e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 2 e^{x} x^{4}

y = c_{1} e^{x} + c_{2} x e^{x} + 2 e^{x} x^{4}

e^{c o t (x)}

(sin (θ))^{^{'}}

\int \frac{x}{t} d t

\frac{d}{d x} (x^{3} + 3 x^{2})

\frac{d}{d x} (24 cos (3 x))

What is the solution for y^{''}-2y^'+y=24x^2e^x ?
The solution for y^{''}-2y^'+y=24x^2e^x is y=c_{1}e^x+c_{2}xe^x+2e^xx^4