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

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

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

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

y = c_{2} e^{2 t} sin (4 t)

Solution steps

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

Solve linear ODE:

y = c_{2} e^{2 t} sin (4 t)

y = c_{2} e^{2 t} sin (4 t)

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = - 2, y^{^{'}} (0) = 0

2 y^{^{''}} + 7 y^{^{'}} = 0

y^{^{''}} - 83 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 73

x^{^{''}} - 9 x = 0

\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 4 x = 0, x (0) = 1

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