What is the solution for y^{''}-6y^'+5y=0,y(0)=1,y(1)=3 ?

The solution for y^{''}-6y^'+5y=0,y(0)=1,y(1)=3 is y=(1-(e^5-3)/(e^5-e))e^{5t}+(e^5-3)/(e^5-e)e^t

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y^{''}-6y^'+5y=0,y(0)=1,y(1)=3

y^{^{''}} - 6 y^{^{'}} + 5 y = 0, y (0) = 1, y (1) = 3

y = (1 - \frac{e ^{5} - 3}{e ^{5} - e}) e^{5 t} + \frac{e ^{5} - 3}{e ^{5} - e} e^{t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = (1 - \frac{e ^{5} - 3}{e ^{5} - e}) e^{5 t} + \frac{e ^{5} - 3}{e ^{5} - e} e^{t}

y = (1 - \frac{e ^{5} - 3}{e ^{5} - e}) e^{5 t} + \frac{e ^{5} - 3}{e ^{5} - e} e^{t}

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

- x^{^{''}} = b x

9 y^{^{''}} - 30 y^{^{'}} - 24 y = 0, y (0) = 1, y^{^{'}} (0) = 0

f^{^{''}} (x) = - \frac{f ^{^{'}} ( x )}{2 ( n + 2 )}

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

What is the solution for y^{''}-6y^'+5y=0,y(0)=1,y(1)=3 ?
The solution for y^{''}-6y^'+5y=0,y(0)=1,y(1)=3 is y=(1-(e^5-3)/(e^5-e))e^{5t}+(e^5-3)/(e^5-e)e^t