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

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

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

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

y = e^{2 t + 2} - e^{1 + t}

Solution steps

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

Solve linear ODE:

y = e^{2 t + 2} - e^{1 + t}

y = e^{2 t + 2} - e^{1 + t}

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

\int e^{3 x} sin (5 x) d x

t^{3} e^{t} t

n \to \infty lim (c^{n})

y = \frac{4 ln ( 3 x )}{4 - 5 x}

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