What is the solution for y^{''}+3y^'-4y=2,y(0)=0 ?

The solution for y^{''}+3y^'-4y=2,y(0)=0 is y=(-c_{2}+1/2)e^t+c_{2}e^{-4t}-1/2

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y^{''}+3y^'-4y=2,y(0)=0

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y (0) = 0

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

Solution steps

y^{''} (t) + 3 y^{'} (t) - 4 y = 2

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + y = sin (t)

y^{^{''}} - 2 a y^{^{'}} + a^{2} y = ∣ t ∣

y^{^{''}} - 10 y^{^{'}} + 25 y = 2 e^{5} x

6 y^{^{''}} + y^{^{'}} - y = - 25 e^{2 x} + 7 e^{- 2 x}

(d^{3} - 3 d - 2) y = x^{2}

What is the solution for y^{''}+3y^'-4y=2,y(0)=0 ?
The solution for y^{''}+3y^'-4y=2,y(0)=0 is y=(-c_{2}+1/2)e^t+c_{2}e^{-4t}-1/2