What is the solution for 4y^{''}-21y^'+20y=0,y(0)=0 ?

The solution for 4y^{''}-21y^'+20y=0,y(0)=0 is y=-c_{2}e^{4t}+c_{2}e^{(5t)/4}

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4y^{''}-21y^'+20y=0,y(0)=0

4 y^{^{''}} - 21 y^{^{'}} + 20 y = 0, y (0) = 0

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

Solution steps

4 y^{''} (t) - 21 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 5 y = 0

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

8 y^{^{''}} + 80 y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

u^{^{''}} + H (t) = 0, H (0) = 0, u^{^{'}} (0) = 2

2 y^{^{''}} + 2 y^{^{'}} + 8 y = 0

What is the solution for 4y^{''}-21y^'+20y=0,y(0)=0 ?
The solution for 4y^{''}-21y^'+20y=0,y(0)=0 is y=-c_{2}e^{4t}+c_{2}e^{(5t)/4}