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

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

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

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

y = 6 e^{2 t} - 6 e^{t}

Solution steps

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

Solve linear ODE:

y = 6 e^{2 t} - 6 e^{t}

y = 6 e^{2 t} - 6 e^{t}

\frac{d ^{2} y}{d x ^{2}} = 9 \cdot y

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

y^{^{''}} - y^{^{'}} - 120 y = 0

c \cdot y^{^{''}} + y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} + 52 y = 0

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