What is the solution for y^{''}+3y^'-10y=0,y(0)=4,y^'(0)=-2 ?

The solution for y^{''}+3y^'-10y=0,y(0)=4,y^'(0)=-2 is y=(4-10/7)e^{2t}+10/7 e^{-5t}

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y^{''}+3y^'-10y=0,y(0)=4,y^'(0)=-2

y^{^{''}} + 3 y^{^{'}} - 10 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

y = (4 - \frac{10}{7}) e^{2 t} + \frac{10}{7} e^{- 5 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 10 y = 0

Solve linear ODE:

y = (4 - \frac{10}{7}) e^{2 t} + \frac{10}{7} e^{- 5 t}

y = (4 - \frac{10}{7}) e^{2 t} + \frac{10}{7} e^{- 5 t}

3 y^{^{''}} = 5 y^{^{'}}

y^{^{''}} - 14 y^{^{'}} + 49 y = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} - 6 y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = 0

What is the solution for y^{''}+3y^'-10y=0,y(0)=4,y^'(0)=-2 ?
The solution for y^{''}+3y^'-10y=0,y(0)=4,y^'(0)=-2 is y=(4-10/7)e^{2t}+10/7 e^{-5t}