What is the solution for 25y^{''}-20y^'+4y=0,y(5)=0,y^'(0)=-e^2 ?

The solution for 25y^{''}-20y^'+4y=0,y(5)=0,y^'(0)=-e^2 is y=-5e^{2+(2t)/5}+e^{2+(2t)/5}t

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25y^{''}-20y^'+4y=0,y(5)=0,y^'(0)=-e^2

25 y^{^{''}} - 20 y^{^{'}} + 4 y = 0, y (5) = 0, y^{^{'}} (0) = - e^{2}

y = - 5 e^{2 + \frac{2 t}{5}} + e^{2 + \frac{2 t}{5}} t

Solution steps

25 y^{''} (t) - 20 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = - 5 e^{2 + \frac{2 t}{5}} + e^{2 + \frac{2 t}{5}} t

y = - 5 e^{2 + \frac{2 t}{5}} + e^{2 + \frac{2 t}{5}} t

x = y^{2} + 1, x = y + 3

\frac{\partial}{\partial x} (6 x - 6 y + 5 x y)

\frac{d}{d x} (tan (sin (5 x)))

\int 6 x^{6} - x d x

\frac{\partial}{\partial x} (y^{2} + cos^{2} (x))

What is the solution for 25y^{''}-20y^'+4y=0,y(5)=0,y^'(0)=-e^2 ?
The solution for 25y^{''}-20y^'+4y=0,y(5)=0,y^'(0)=-e^2 is y=-5e^{2+(2t)/5}+e^{2+(2t)/5}t