What is the solution for y^{''}-8y^'+16=0,y(0)=4,y(1)=5 ?

The solution for y^{''}-8y^'+16=0,y(0)=4,y(1)=5 is y=(4-(4e^8-3)/(e^8-1))e^{8t}+(4e^8-3)/(e^8-1)+2t

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y^{''}-8y^'+16=0,y(0)=4,y(1)=5

y^{^{''}} - 8 y^{^{'}} + 16 = 0, y (0) = 4, y (1) = 5

y = (4 - \frac{4 e ^{8} - 3}{e ^{8} - 1}) e^{8 t} + \frac{4 e ^{8} - 3}{e ^{8} - 1} + 2 t

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 = 0

Solve linear ODE:

y = (4 - \frac{4 e ^{8} - 3}{e ^{8} - 1}) e^{8 t} + \frac{4 e ^{8} - 3}{e ^{8} - 1} + 2 t

y = (4 - \frac{4 e ^{8} - 3}{e ^{8} - 1}) e^{8 t} + \frac{4 e ^{8} - 3}{e ^{8} - 1} + 2 t

4 y^{^{''}} + 6 y^{^{'}} + 25 y = 4 cos (2 t)

y^{^{''}} + 8 y^{^{'}} + 6 y = 18 x^{2}

y^{^{''}} + 4 y + 5 y = 35 e^{- 4 x}, y (0) = - 6, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + y = 6 sin (x)

x y^{^{''}} + x y^{^{'}} = x

What is the solution for y^{''}-8y^'+16=0,y(0)=4,y(1)=5 ?
The solution for y^{''}-8y^'+16=0,y(0)=4,y(1)=5 is y=(4-(4e^8-3)/(e^8-1))e^{8t}+(4e^8-3)/(e^8-1)+2t