What is the solution for 4y^{''}+4y^'+17=0 ?

The solution for 4y^{''}+4y^'+17=0 is y=c_{1}+c_{2}e^{-t}-(17t)/4

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4y^{''}+4y^'+17=0

4 y^{^{''}} + 4 y^{^{'}} + 17 = 0

y = c_{1} + c_{2} e^{- t} - \frac{17 t}{4}

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + 17 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} - \frac{17 t}{4}

y = c_{1} + c_{2} e^{- t} - \frac{17 t}{4}

\frac{d}{d x} (\frac{x + 7}{4 x ^{2} e ^{x}})

t an g e n t (x^{3} + 2)^{5}, at x = - 1

\frac{\partial}{\partial x} (56 y^{4} x^{6} - 42 y^{3} x^{5})

d er i v a t i v e (x - 1)^{2} (x - 2)^{3}

n = 1 \sum \infty 8 ((\frac{1}{5})^{n})

What is the solution for 4y^{''}+4y^'+17=0 ?
The solution for 4y^{''}+4y^'+17=0 is y=c_{1}+c_{2}e^{-t}-(17t)/4