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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

D^{2} (D^{2} - 4 D + 5)^{3} y = 0

20 y^{^{''}} + 27 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 8

3 y^{^{''}} + 2 y^{^{'}} + \frac{y}{4} = 0

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (1) = 1, y^{^{'}} (1) = 1

y^{^{''}} + y = 0, y (0) = 19, y^{^{'}} (0) = 36

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