What is the solution for y^{''}+8y^'+15y=10,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+8y^'+15y=10,y(0)=2,y^'(0)=1 is y=(5/2+4/3)e^{-3t}-5/2 e^{-5t}+2/3

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y^{''}+8y^'+15y=10,y(0)=2,y^'(0)=1

y^{^{''}} + 8 y^{^{'}} + 15 y = 10, y (0) = 2, y^{^{'}} (0) = 1

y = (\frac{5}{2} + \frac{4}{3}) e^{- 3 t} - \frac{5}{2} e^{- 5 t} + \frac{2}{3}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 15 y = 10

Solve linear ODE:

y = (\frac{5}{2} + \frac{4}{3}) e^{- 3 t} - \frac{5}{2} e^{- 5 t} + \frac{2}{3}

y = (\frac{5}{2} + \frac{4}{3}) e^{- 3 t} - \frac{5}{2} e^{- 5 t} + \frac{2}{3}

What is the solution for y^{''}+8y^'+15y=10,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+8y^'+15y=10,y(0)=2,y^'(0)=1 is y=(5/2+4/3)e^{-3t}-5/2 e^{-5t}+2/3