What is the solution for y^{''}+6y^'+5y=10 ?

The solution for y^{''}+6y^'+5y=10 is y=c_{1}e^{-t}+c_{2}e^{-5t}+2

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y^{''}+6y^'+5y=10

y^{^{''}} + 6 y^{^{'}} + 5 y = 10

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

Solution steps

y^{''} (t) + 6 y^{'} (t) + 5 y = 10

Solve linear ODE:

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

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

\int \frac{1}{h} d h

\frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 5

\int x^{2} \cdot e^{- x^{3}} d x

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

\frac{x}{x ^{2} + 64}

What is the solution for y^{''}+6y^'+5y=10 ?
The solution for y^{''}+6y^'+5y=10 is y=c_{1}e^{-t}+c_{2}e^{-5t}+2