What is the solution for y^{''}-3y^'+2y=0,y(0)=-1,y^'(0)=-9/4 ?

The solution for y^{''}-3y^'+2y=0,y(0)=-1,y^'(0)=-9/4 is y=(-1-1/4)e^{2t}+1/4 e^t

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y^{''}-3y^'+2y=0,y(0)=-1,y^'(0)=-9/4

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = - \frac{9}{4}

y = (- 1 - \frac{1}{4}) e^{2 t} + \frac{1}{4} e^{t}

Solution steps

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

Solve linear ODE:

y = (- 1 - \frac{1}{4}) e^{2 t} + \frac{1}{4} e^{t}

y = (- 1 - \frac{1}{4}) e^{2 t} + \frac{1}{4} e^{t}

y^{^{''}} + 25 y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 0

10 y^{^{''}} + 6 y^{^{'}} + 10.9 y = 0

y^{^{''}} (s) - 3 y^{^{'}} (s) + y (s) 3 = 0 es

y^{^{''''}} - k^{2} y^{^{''}} = 0

y^{^{''}} + \frac{2}{5} y^{^{'}} + \frac{1}{25} y = 0, y (0) = 0, y^{^{'}} (0) = 5

What is the solution for y^{''}-3y^'+2y=0,y(0)=-1,y^'(0)=-9/4 ?
The solution for y^{''}-3y^'+2y=0,y(0)=-1,y^'(0)=-9/4 is y=(-1-1/4)e^{2t}+1/4 e^t