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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 5 e^{3 x}

\frac{( 10 s - 3 )}{s ^{2} + 4 s + 29}

\int \frac{x}{( 1 + 4 x ^{2} ) ^{2}} d x

f (x) = 4 x + 10

\frac{d}{d x} ((6 x + 7)^{2})

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