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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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