What is the solution for y^{''}+5y^'+4y=24 ?

The solution for y^{''}+5y^'+4y=24 is y=c_{1}e^{-t}+c_{2}e^{-4t}+6

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y^{''}+5y^'+4y=24

y^{^{''}} + 5 y^{^{'}} + 4 y = 24

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + 6

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = 24

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + 6

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + 6

d^{2} (d^{2} + 4) y = 96 x^{2}

\frac{d ^{2} y}{d t ^{2}} + 12 \frac{d y}{d t} + 32 = 32 H (t)

y^{^{''''}} + 2 y^{^{''}} + y = 9 cos (2 x)

y^{^{''}} + y = 2 sec (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = t^{3} e^{2 t}

What is the solution for y^{''}+5y^'+4y=24 ?
The solution for y^{''}+5y^'+4y=24 is y=c_{1}e^{-t}+c_{2}e^{-4t}+6