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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = \frac{- 6 x}{x ^{2} + 1}, (1, - 3)

y^{^{'}} = y (1 - 0.001 y)

\int_{0}^{1} π (x^{2})^{2} d x

\frac{d}{d t} (6 t)

f (x) = \frac{x}{x + 6}, at x = 1

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