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

The solution for y^{''}+3y^'-4y=2,y(0)=0,y^'(0)=0 is y=(-1/10+1/2)e^t+1/10 e^{-4t}-1/2

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

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

y = (- \frac{1}{10} + \frac{1}{2}) e^{t} + \frac{1}{10} e^{- 4 t} - \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{10} + \frac{1}{2}) e^{t} + \frac{1}{10} e^{- 4 t} - \frac{1}{2}

y = (- \frac{1}{10} + \frac{1}{2}) e^{t} + \frac{1}{10} e^{- 4 t} - \frac{1}{2}

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + 4 y = x - sin (3 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = x cos (2 x)

y^{^{''}} - y^{^{'}} - 64 y = 54 cos (3 x)

y^{^{''}} - y^{^{'}} + 14 y = 7 + e^{\frac{x}{2}}

What is the solution for y^{''}+3y^'-4y=2,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+3y^'-4y=2,y(0)=0,y^'(0)=0 is y=(-1/10+1/2)e^t+1/10 e^{-4t}-1/2