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

The solution for y^{''}-3y^'+2y=0,y(0)= 3/5 ,y^'(0)=-1/5 is y=(3/5-7/5)e^{2t}+7/5 e^t

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

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

y = (\frac{3}{5} - \frac{7}{5}) e^{2 t} + \frac{7}{5} e^{t}

Solution steps

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

Solve linear ODE:

y = (\frac{3}{5} - \frac{7}{5}) e^{2 t} + \frac{7}{5} e^{t}

y = (\frac{3}{5} - \frac{7}{5}) e^{2 t} + \frac{7}{5} e^{t}

\frac{d ^{4} y}{d x ^{4}} + 8 \frac{d ^{2} y}{d x ^{2}} + 6 y = 0

(D^{4} - D^{2}) y = 0

\frac{1}{2} x^{^{''}} + 3 x^{^{'}} + 4 x = 0, x (0) = 3, x^{^{'}} (0) = 0

y^{^{'''}} - 7 y^{^{''}} + 10 y = 0, y (0) = 7

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

What is the solution for y^{''}-3y^'+2y=0,y(0)= 3/5 ,y^'(0)=-1/5 ?
The solution for y^{''}-3y^'+2y=0,y(0)= 3/5 ,y^'(0)=-1/5 is y=(3/5-7/5)e^{2t}+7/5 e^t