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

The solution for y^{''}+4y^'-y=0 is y=c_{1}e^{(-2+sqrt(5))t}+c_{2}e^{(-2-sqrt(5))t}

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

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

y = c_{1} e^{(- 2 + 5) t} + c_{2} e^{(- 2 - 5) t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(- 2 + 5) t} + c_{2} e^{(- 2 - 5) t}

y = c_{1} e^{(- 2 + 5) t} + c_{2} e^{(- 2 - 5) t}

\int_{0}^{2 π} sin^{2} (10 x) d x

y = 7 + 8 x + 2 x e^{x}, (0, 7)

(3.5) (- 1.5)

t^{n}

(x + 2 y) d y - y d x = 0

What is the solution for y^{''}+4y^'-y=0 ?
The solution for y^{''}+4y^'-y=0 is y=c_{1}e^{(-2+sqrt(5))t}+c_{2}e^{(-2-sqrt(5))t}