What is the solution for y^{''}+2y^'+0.75y=0 ?

The solution for y^{''}+2y^'+0.75y=0 is y=c_{1}e^{-t/2}+c_{2}e^{-(3t)/2}

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y^{''}+2y^'+0.75y=0

y^{^{''}} + 2 y^{^{'}} + 0.75 y = 0

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

Solution steps

y^{''} (t) + 2 y^{'} (t) + 0.75 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

y = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

y^{^{''}} - 24 y^{^{'}} + 12 y = 0

\frac{d ^{2} x}{d t ^{2}} - 4 x = 0

x^{^{''}} - (a - 2) x^{^{'}} - (a - 1) x = 0

3 y^{^{''}} - 6 y^{^{'}} + 9 y = 0

y^{^{''}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

What is the solution for y^{''}+2y^'+0.75y=0 ?
The solution for y^{''}+2y^'+0.75y=0 is y=c_{1}e^{-t/2}+c_{2}e^{-(3t)/2}