What is the solution for y^{''}-2y^'+y=8e^x ?

The solution for y^{''}-2y^'+y=8e^x is y=c_{1}e^x+c_{2}xe^x+4x^2e^x

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y^{''}-2y^'+y=8e^x

y^{^{''}} - 2 y^{^{'}} + y = 8 e^{x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 8 e^{x}

Solve linear ODE:

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

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

x \to 0 + lim ((ln (x))^{2})

sec^{2} (x) + tan^{2} (x)

f (x) = 15 x - 1.86 x^{2}, at x = a

\frac{d}{d x} (7^{x^{2}})

\int - 3 x^{2} - 20 x + 2000 d x

What is the solution for y^{''}-2y^'+y=8e^x ?
The solution for y^{''}-2y^'+y=8e^x is y=c_{1}e^x+c_{2}xe^x+4x^2e^x