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

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{4} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{4} e^{3 x}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{4} e^{3 x}

\frac{d}{d x} (r^{2} - 19 x^{2})

x \to 3 lim (x^{2} + 3 x)

x \to 2 lim ((x^{2} - 3) e^{x})

f (x) = x + \frac{3}{x ^{2}}

\frac{d}{d x} (2 x sin (π x))

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