What is the solution for y^{''}-4y^'-5y=2e^{2t} ?

The solution for y^{''}-4y^'-5y=2e^{2t} is y=c_{1}e^{5t}+c_{2}e^{-t}-2/9 e^{2t}

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y^{''}-4y^'-5y=2e^{2t}

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 e^{2 t}

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{2}{9} e^{2 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 2 e^{2 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{2}{9} e^{2 t}

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{2}{9} e^{2 t}

y^{^{''}} + y = csc (x) cot (x)

y^{^{''}} + 4 y = 16 t sin (2 t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 4 e^{- 3 x} - e^{- x}

4 \frac{d ^{2} y}{d x ^{2}} - y = x \cdot e^{\frac{x}{2}}

y^{^{''}} + 9 y = 18 t

What is the solution for y^{''}-4y^'-5y=2e^{2t} ?
The solution for y^{''}-4y^'-5y=2e^{2t} is y=c_{1}e^{5t}+c_{2}e^{-t}-2/9 e^{2t}