What is the solution for y^{''}+4y^'+7=0 ?

The solution for y^{''}+4y^'+7=0 is y=c_{1}+c_{2}e^{-4t}-(7t)/4

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y^{''}+4y^'+7=0

y^{^{''}} + 4 y^{^{'}} + 7 = 0

y = c_{1} + c_{2} e^{- 4 t} - \frac{7 t}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 7 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - \frac{7 t}{4}

y = c_{1} + c_{2} e^{- 4 t} - \frac{7 t}{4}

y^{^{''}} + 4 y^{^{'}} + 2 y = cos (2 x)

y^{^{''}} + 3 y^{^{'}} - 4 y = 14 e^{x}

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 9 t^{2}

y^{^{''}} - 3 y^{^{'}} - 40 y = (t + 1) e^{t}

What is the solution for y^{''}+4y^'+7=0 ?
The solution for y^{''}+4y^'+7=0 is y=c_{1}+c_{2}e^{-4t}-(7t)/4