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

The solution for y^{''}-3y^'-4y=-3,y(0)=0,y^'(0)=0 is y=(3/5-3/4)e^{4t}-3/5 e^{-t}+3/4

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int tan^{2} (2 x se) c^{2} 2 x

y^{^{'}} + y - 2 cos (x) = 0

\int sin (\frac{nπ x}{3}) d x

n = 0 \sum \infty \frac{1}{n + 3}

\int \frac{1}{2 x ^{5}} d x

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