What is the solution for y^{''}+1y^'+0.25y=0 ?

The solution for y^{''}+1y^'+0.25y=0 is y=c_{1}e^{-t/2}+c_{2}e^{-t/2}t

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y^{''}+1y^'+0.25y=0

y^{^{''}} + 1 y^{^{'}} + 0.25 y = 0

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

Solution steps

y^{''} (t) + 1 y^{'} (t) + 0.25 y = 0

Solve linear ODE:

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

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

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\int_{12}^{13} x x^{2} - 144 d x

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\frac{d}{d x} (\frac{( 6 x ^{2} + 4 )}{( 3 y ^{2} - 4 )})

\int_{- 3}^{0} 1 + 9 - x^{2} d x

What is the solution for y^{''}+1y^'+0.25y=0 ?
The solution for y^{''}+1y^'+0.25y=0 is y=c_{1}e^{-t/2}+c_{2}e^{-t/2}t