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

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

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

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

y = 3 e^{2 t} - 4 e^{t}

Solution steps

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

Solve linear ODE:

y = 3 e^{2 t} - 4 e^{t}

y = 3 e^{2 t} - 4 e^{t}

\frac{d ^{2} x}{d t ^{2}} - 6 \frac{d x}{d t} + 9 x = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y^{^{'}} (- 1) = 9, y (- 1) = 3

y^{^{'''}} - 2 y^{^{''}} - 9 y^{^{'}} + 4 y = 0

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

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