What is the solution for y^{''}-4y^'-5y=12,5y(0)=0 ?

The solution for y^{''}-4y^'-5y=12,5y(0)=0 is y=c_{1}e^{5t}+c_{2}e^{-t}-12/5

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y^{''}-4y^'-5y=12,5y(0)=0

y^{^{''}} - 4 y^{^{'}} - 5 y = 12, 5 y (0) = 0

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{12}{5}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 12

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{12}{5}

y = c_{1} e^{5 t} + c_{2} e^{- t} - \frac{12}{5}

y^{^{''}} - y^{^{'}} + y = 6 sin (x) - 3 e^{2 x}

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} + x = 10 sin (4 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = cos (2 t)

y^{^{''}} + 10 y^{^{'}} + 24 y = 2 e^{- 4 x} (4 x - 2)

1 y^{^{''}} + 25 y = cos (5 x) csc^{2} (5 x)

What is the solution for y^{''}-4y^'-5y=12,5y(0)=0 ?
The solution for y^{''}-4y^'-5y=12,5y(0)=0 is y=c_{1}e^{5t}+c_{2}e^{-t}-12/5