What is the solution for y^{''}-4y=e^{2t} ?

The solution for y^{''}-4y=e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-2t}+(e^{2t}t)/4

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y^{''}-4y=e^{2t}

y^{^{''}} - 4 y = e^{2 t}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{2 t} t}{4}

Solution steps

y^{''} (t) - 4 y = e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{2 t} t}{4}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{2 t} t}{4}

3 x^{2} ln (7 x)

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

\int \frac{1}{tan ^{2} ( θ )} sec^{2} (θ) d θ

25 - x^{2}, x^{2} - 25, - 5, 5

\int sec^{2} (\frac{x}{a}) d x

What is the solution for y^{''}-4y=e^{2t} ?
The solution for y^{''}-4y=e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-2t}+(e^{2t}t)/4