What is the solution for x^{''}-2x^'+x=24t^2e^t ?

The solution for x^{''}-2x^'+x=24t^2e^t is x=c_{1}e^t+c_{2}te^t+2e^tt^4

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x^{''}-2x^'+x=24t^2e^t

x^{^{''}} - 2 x^{^{'}} + x = 24 t^{2} e^{t}

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

Solution steps

x^{''} (t) - 2 x^{'} (t) + x = 24 t^{2} e^{t}

Solve linear ODE:

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

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

f (x) = x^{8 x}

\int \frac{sec ^{2} ( 8 x )}{tan ( 8 x ) - 6} d x

\frac{d y}{d x} = x e^{x^{2} - 4}

n = 1 \sum \infty (- 8)^{- n}

\frac{\partial}{\partial z} (e^{x y} sin (z))

What is the solution for x^{''}-2x^'+x=24t^2e^t ?
The solution for x^{''}-2x^'+x=24t^2e^t is x=c_{1}e^t+c_{2}te^t+2e^tt^4