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Popular >

y^{''}+7y^'+12y=e^t,y(0)=0,y^'(0)=1

Solution

y^{^{''}} + 7 y^{^{'}} + 12 y = e^{t}, y (0) = 0, y^{^{'}} (0) = 1

Solution

y = (\frac{4}{5} - \frac{1}{20}) e^{- 3 t} - \frac{4}{5} e^{- 4 t} + \frac{1}{20} e^{t}

Solution steps

y^{''} (t) + 7 y^{'} (t) + 12 y = e^{t}

Solve linear ODE:

y = (\frac{4}{5} - \frac{1}{20}) e^{- 3 t} - \frac{4}{5} e^{- 4 t} + \frac{1}{20} e^{t}

y = (\frac{4}{5} - \frac{1}{20}) e^{- 3 t} - \frac{4}{5} e^{- 4 t} + \frac{1}{20} e^{t}

Popular Examples

y^{'''}-2y^{''}-y^'+2y=e^{7t}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{7 t}

y^{''}+3y^'-18y=-66t+36t^2

y^{^{''}} + 3 y^{^{'}} - 18 y = - 66 t + 36 t^{2}

y^{''}+3y^'+2y=4x^2

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 x^{2}

y^{''}+2y^'+5y=3e^{-2t}

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{- 2 t}

(D^2-4D+13)y=24e^{2x}cos(3x)

(D^{2} - 4 D + 13) y = 24 e^{2 x} cos (3 x)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+7y^'+12y=e^t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+7y^'+12y=e^t,y(0)=0,y^'(0)=1 is y=(4/5-1/20)e^{-3t}-4/5 e^{-4t}+1/20 e^t