What is the solution for 4y^{''}-20y^'+25y=0 ?

The solution for 4y^{''}-20y^'+25y=0 is y=c_{1}e^{(5t)/2}+c_{2}e^{(5t)/2}t

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4y^{''}-20y^'+25y=0

4 y^{^{''}} - 20 y^{^{'}} + 25 y = 0

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{\frac{5 t}{2}} t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{\frac{5 t}{2}} t

y = c_{1} e^{\frac{5 t}{2}} + c_{2} e^{\frac{5 t}{2}} t

What is the solution for 4y^{''}-20y^'+25y=0 ?
The solution for 4y^{''}-20y^'+25y=0 is y=c_{1}e^{(5t)/2}+c_{2}e^{(5t)/2}t