What is the solution for y^{''}+3y^'-4y=1,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+3y^'-4y=1,y(0)=0,y^'(0)=1 is y=(3/20+1/4)e^t-3/20 e^{-4t}-1/4

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

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

y = (\frac{3}{20} + \frac{1}{4}) e^{t} - \frac{3}{20} e^{- 4 t} - \frac{1}{4}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 4 y = 1

Solve linear ODE:

y = (\frac{3}{20} + \frac{1}{4}) e^{t} - \frac{3}{20} e^{- 4 t} - \frac{1}{4}

y = (\frac{3}{20} + \frac{1}{4}) e^{t} - \frac{3}{20} e^{- 4 t} - \frac{1}{4}

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 e^{- t}, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 7 y^{^{'}} + 10 y = 9 cos (t) + 7 sin (t)

y^{^{''}} - y = cosh (x), y (0) = 2, y^{^{'}} (0) = 12

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{4 t}

What is the solution for y^{''}+3y^'-4y=1,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+3y^'-4y=1,y(0)=0,y^'(0)=1 is y=(3/20+1/4)e^t-3/20 e^{-4t}-1/4