What is the solution for y^{''}-3y^'+2y=0,y(0)=2,y^'(0)=-6 ?

The solution for y^{''}-3y^'+2y=0,y(0)=2,y^'(0)=-6 is y=-8e^{2t}+10e^t

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y^{''}-3y^'+2y=0,y(0)=2,y^'(0)=-6

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = - 6

y = - 8 e^{2 t} + 10 e^{t}

Solution steps

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

Solve linear ODE:

y = - 8 e^{2 t} + 10 e^{t}

y = - 8 e^{2 t} + 10 e^{t}

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1

y^{^{''}} + 15 y^{^{'}} + 44 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

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

\frac{d ^{2} y}{d x ^{2}} - 2 \cdot \frac{d y}{d x} = 0

What is the solution for y^{''}-3y^'+2y=0,y(0)=2,y^'(0)=-6 ?
The solution for y^{''}-3y^'+2y=0,y(0)=2,y^'(0)=-6 is y=-8e^{2t}+10e^t