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

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

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

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

y = (1 - \frac{4}{3}) e^{2 t} + \frac{4}{3} e^{t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{4}{3}) e^{2 t} + \frac{4}{3} e^{t}

y = (1 - \frac{4}{3}) e^{2 t} + \frac{4}{3} e^{t}

\frac{d}{d x} (\frac{ln ( x )}{x ^{2}})

\frac{\partial}{\partial x} (\frac{( 2 x + y )}{x + 2 y})

\int \frac{2 x}{x ^{2} - x + 1} d x

\int_{0}^{π} π (sin^{2} (x))^{2} d x

\int + x ln (x) d x

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