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

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

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y^{''}+y^'-2y=5

y^{^{''}} + y^{^{'}} - 2 y = 5

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 9 y = x e^{2 x} + sin (3 x)

y^{^{''}} - y^{^{'}} - 12 y = 5 - 2 x

2 \frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + y = H (t)

y^{^{''}} - 3 y^{^{'}} - 46 = 2 sin (t)

(d^{2} - 4 d + 3) y = 3 e^{x} \cdot cos (x)

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