What is the solution for 4y^{''}-25y=0,y(0)=1,y^'(0)=0 ?

The solution for 4y^{''}-25y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{(5t)/2}+1/2 e^{-(5t)/2}

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4y^{''}-25y=0,y(0)=1,y^'(0)=0

4 y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - \frac{1}{2}) e^{\frac{5 t}{2}} + \frac{1}{2} e^{- \frac{5 t}{2}}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{1}{2}) e^{\frac{5 t}{2}} + \frac{1}{2} e^{- \frac{5 t}{2}}

y = (1 - \frac{1}{2}) e^{\frac{5 t}{2}} + \frac{1}{2} e^{- \frac{5 t}{2}}

y^{^{''}} + a y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 0

y^{^{'''}} + 3 y^{^{''}} - 30 y^{^{'}} - 50 y = 0

y^{^{'''''}} - 4 y^{^{''''}} - 4 y^{^{'''}} + 2 y^{^{''}} = 0

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = - 3

3 y^{^{'''}} + 16 y^{^{''}} + 26 y^{^{'}} + 7 y = 0

What is the solution for 4y^{''}-25y=0,y(0)=1,y^'(0)=0 ?
The solution for 4y^{''}-25y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{(5t)/2}+1/2 e^{-(5t)/2}