What is the solution for (y^{''}+10y^'+149)/(4y)=0 ?

The solution for (y^{''}+10y^'+149)/(4y)=0 is y=c_{1}+c_{2}e^{-10t}-(149t)/(10)

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(y^{''}+10y^'+149)/(4y)=0

\frac{y ^{^{''}} + 10 y ^{^{'}} + 149}{4 y} = 0

y = c_{1} + c_{2} e^{- 10 t} - \frac{149 t}{10}

Solution steps

\frac{y ^{''} ( t ) + 10 y ^{'} ( t ) + 149}{4 y} = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 10 t} - \frac{149 t}{10}

y = c_{1} + c_{2} e^{- 10 t} - \frac{149 t}{10}

x \to 2 lim (+ f (x))

\int \frac{2 x - 3}{x ^{2} + 10 x + 29} d x

\frac{d}{d x} (x + cot (\frac{x}{2}))

\frac{\partial}{\partial y} (10 x y e^{- x^{2} - y^{2}})

\frac{d}{d x} (- 3 cot (1 - x^{2}))

What is the solution for (y^{''}+10y^'+149)/(4y)=0 ?
The solution for (y^{''}+10y^'+149)/(4y)=0 is y=c_{1}+c_{2}e^{-10t}-(149t)/(10)