What is the solution for x^{''}-2x-5=0,x(0)=1,x(1)=3 ?

The solution for x^{''}-2x-5=0,x(0)=1,x(1)=3 is

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x^{''}-2x-5=0,x(0)=1,x(1)=3

x^{^{''}} - 2 x - 5 = 0, x (0) = 1, x (1) = 3

x = \frac{11 e ^{2} - 7 e ^{22}}{2 ( e ^{22} - 1 )} + \frac{7}{2} e^{2 t} - \frac{11 e ^{2} - 7 e ^{22}}{2 ( e ^{22} - 1 )} e^{- 2 t} - \frac{5}{2}

Solution steps

x^{''} (t) - 2 x - 5 = 0

Solve linear ODE:

x = \frac{11 e ^{2} - 7 e ^{22}}{2 ( e ^{22} - 1 )} + \frac{7}{2} e^{2 t} - \frac{11 e ^{2} - 7 e ^{22}}{2 ( e ^{22} - 1 )} e^{- 2 t} - \frac{5}{2}

x = \frac{11 e ^{2} - 7 e ^{22}}{2 ( e ^{22} - 1 )} + \frac{7}{2} e^{2 t} - \frac{11 e ^{2} - 7 e ^{22}}{2 ( e ^{22} - 1 )} e^{- 2 t} - \frac{5}{2}

\int \frac{39}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

5 x (1 - \frac{x}{5})^{2}

x \to 2 lim (2 x - 3)

\int 6 z d z

\frac{y ^{2}}{x} \frac{d y}{d x} - (y^{3} + 1) e^{3 x} = 0

What is the solution for x^{''}-2x-5=0,x(0)=1,x(1)=3 ?
The solution for x^{''}-2x-5=0,x(0)=1,x(1)=3 is