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

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

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

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

y = \frac{3}{2} e^{2 t} - \frac{1}{2}

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = 1

Solve linear ODE:

y = \frac{3}{2} e^{2 t} - \frac{1}{2}

y = \frac{3}{2} e^{2 t} - \frac{1}{2}

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 6 y = x e^{x}

(D^{2} - 4 D + 4) y = e^{- 4 x} + 5 cos (3 x)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 10 y = 50 x

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

y^{^{''}} + 8 y^{^{'}} + 16 y = 4 e^{- 4 t}

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