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

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

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y^{''}-4y^'+3y=2e^{5x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{1}{4} e^{5 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 2 e^{5 x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{1}{4} e^{5 x}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{1}{4} e^{5 x}

y^{^{''}} - 3 y^{^{'}} - 40 y = 210 e^{2 t}

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = t^{2} - 2 t

y^{^{''}} - 2 y^{^{'}} + 6 y = 18 sin (3 t)

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

y^{^{''''}} + 2 y^{^{''}} + y = 7 + cos (5 t)

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