What is the solution for y^{''}+2y^'-y=10 ?

The solution for y^{''}+2y^'-y=10 is y=c_{1}e^{(-1+sqrt(2))t}+c_{2}e^{(-1-sqrt(2))t}-10

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y^{''}+2y^'-y=10

y^{^{''}} + 2 y^{^{'}} - y = 10

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} - 10

Solution steps

y^{''} (t) + 2 y^{'} (t) - y = 10

Solve linear ODE:

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} - 10

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} - 10

\frac{d y}{d x} = 1 - 9 y^{2}, y (- 1) = 0

x \to 0 lim (\frac{∣ sin ( x ) ∣}{x})

\frac{d}{d x} (- \frac{600}{x ^{2}} + 4)

f (x) = 4 x - 3 x^{2}, at x = 2

e^{, at} sin (b t)

What is the solution for y^{''}+2y^'-y=10 ?
The solution for y^{''}+2y^'-y=10 is y=c_{1}e^{(-1+sqrt(2))t}+c_{2}e^{(-1-sqrt(2))t}-10