What is the solution for 4y^{''}+8y^'+3y=0 ?

The solution for 4y^{''}+8y^'+3y=0 is y=c_{1}e^{-t/2}+c_{2}e^{-(3t)/2}

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4y^{''}+8y^'+3y=0

4 y^{^{''}} + 8 y^{^{'}} + 3 y = 0

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

Solution steps

4 y^{''} (t) + 8 y^{'} (t) + 3 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}}

\frac{5 s}{( s ^{2} + 4 ) \cdot ( s + 7 )}

f (x) = (\frac{5 x ^{2} + 6 x + 1}{3 x - 5})^{\frac{3}{2}}

\int - \frac{1}{2} x^{2} d x

\int \frac{6 x + 4}{3 x ^{2} + 4 x + 2} d x

\frac{\partial}{\partial x} (x \cdot ln (x + 2 y))

What is the solution for 4y^{''}+8y^'+3y=0 ?
The solution for 4y^{''}+8y^'+3y=0 is y=c_{1}e^{-t/2}+c_{2}e^{-(3t)/2}