What is the solution for 2y^{''}+13y^'-7y=0 ?

The solution for 2y^{''}+13y^'-7y=0 is y=c_{1}e^{t/2}+c_{2}e^{-7t}

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2y^{''}+13y^'-7y=0

2 y^{^{''}} + 13 y^{^{'}} - 7 y = 0

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- 7 t}

Solution steps

2 y^{''} (t) + 13 y^{'} (t) - 7 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- 7 t}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- 7 t}

\int \frac{1}{x ^{2} + t ^{2}} d x

\frac{d}{d x} (\frac{1}{3} tan^{3} (x) - tan (x) + x)

f (x) = (x + 9) (x^{2} - 9)

\frac{6}{s ^{4} ( s - 2 ) ^{2}}

\int \frac{( x + 2 )}{x ( x ^{2} + 4 )} d x

What is the solution for 2y^{''}+13y^'-7y=0 ?
The solution for 2y^{''}+13y^'-7y=0 is y=c_{1}e^{t/2}+c_{2}e^{-7t}