What is the solution for y^{''}-2y^'-3y=4,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}-2y^'-3y=4,y(0)=1,y^'(0)=-1 is y=(-2+7/3)e^{3t}+2e^{-t}-4/3

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

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

y = (- 2 + \frac{7}{3}) e^{3 t} + 2 e^{- t} - \frac{4}{3}

Solution steps

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

Solve linear ODE:

y = (- 2 + \frac{7}{3}) e^{3 t} + 2 e^{- t} - \frac{4}{3}

y = (- 2 + \frac{7}{3}) e^{3 t} + 2 e^{- t} - \frac{4}{3}

y^{^{''}} + 5 y^{^{'}} + 6 y = 20 sin (2 t)

y^{^{''}} - 2 y^{^{'}} + y = 25 sin (2 x)

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 65 x e^{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{t}

y^{^{'''}} - y^{^{''}} - 8 y^{^{'}} + 12 y = 2 e^{t}

What is the solution for y^{''}-2y^'-3y=4,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-2y^'-3y=4,y(0)=1,y^'(0)=-1 is y=(-2+7/3)e^{3t}+2e^{-t}-4/3