What is the solution for 4y^{''}+7y^'-2y=0 ?

The solution for 4y^{''}+7y^'-2y=0 is y=c_{1}e^{t/4}+c_{2}e^{-2t}

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4y^{''}+7y^'-2y=0

4 y^{^{''}} + 7 y^{^{'}} - 2 y = 0

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

Solution steps

4 y^{''} (t) + 7 y^{'} (t) - 2 y = 0

Solve linear ODE:

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

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

\int \frac{( x + 1 )}{x ^{2}} d x

\frac{d}{d y} (ln (y) - \frac{y ^{2}}{8})

\int cos (4 x) d x

\int \frac{x + 2}{x ^{2} - 6 x + 5} d x

\frac{2}{s ^{2} + 3 s - 4}

What is the solution for 4y^{''}+7y^'-2y=0 ?
The solution for 4y^{''}+7y^'-2y=0 is y=c_{1}e^{t/4}+c_{2}e^{-2t}