What is the solution for y^{''}-3y^'+2y=0,y(0)=-1/5 ,y^'(0)=-8/5 ?

The solution for y^{''}-3y^'+2y=0,y(0)=-1/5 ,y^'(0)=-8/5 is y=(-1/5-6/5)e^{2t}+6/5 e^t

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y^{''}-3y^'+2y=0,y(0)=-1/5 ,y^'(0)=-8/5

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - \frac{1}{5}, y^{^{'}} (0) = - \frac{8}{5}

y = (- \frac{1}{5} - \frac{6}{5}) e^{2 t} + \frac{6}{5} e^{t}

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = (- \frac{1}{5} - \frac{6}{5}) e^{2 t} + \frac{6}{5} e^{t}

y = (- \frac{1}{5} - \frac{6}{5}) e^{2 t} + \frac{6}{5} e^{t}

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What is the solution for y^{''}-3y^'+2y=0,y(0)=-1/5 ,y^'(0)=-8/5 ?
The solution for y^{''}-3y^'+2y=0,y(0)=-1/5 ,y^'(0)=-8/5 is y=(-1/5-6/5)e^{2t}+6/5 e^t