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

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

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

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

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 1

Solve linear ODE:

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

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

a^{2} \cdot y + y^{^{''}} = cos (b \cdot t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} = 1 - x^{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{x}

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x}

(d^{3} - 3 d^{2} - 10 d + 24) y = 2 x^{2} + 5

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