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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = sin (sin (x)), at x = 2 π

x \to - 1 + lim (arcsin (x))

f (x) = ln (\frac{x + 3}{x - 3})

\frac{d}{d x} ((4 x + 5)^{2})

f (x) = sin (x^{3})

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