What is the solution for y^{''}-4y^'+3y=10e^{3x} ?

The solution for y^{''}-4y^'+3y=10e^{3x} is y=c_{1}e^{3x}+c_{2}e^x+5xe^{3x}

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y^{''}-4y^'+3y=10e^{3x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 e^{3 x}

y = c_{1} e^{3 x} + c_{2} e^{x} + 5 x e^{3 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 10 e^{3 x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + 5 x e^{3 x}

y = c_{1} e^{3 x} + c_{2} e^{x} + 5 x e^{3 x}

y^{^{''}} - y^{^{'}} = 1 + 2 x + e^{x} + sin (x) + cos (x)

3 y^{^{''}} - 2 y^{^{'}} - 8 y = (2 x + 3) e^{5 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x ^{2}}

y^{^{''}} - 7 y^{^{'}} = 10

(D^{2} - 4) y = 2 cos (2 x)

What is the solution for y^{''}-4y^'+3y=10e^{3x} ?
The solution for y^{''}-4y^'+3y=10e^{3x} is y=c_{1}e^{3x}+c_{2}e^x+5xe^{3x}