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Popular >

4y^{''}+16y=0

Solution

4 y^{^{''}} + 16 y = 0

Solution

y = c_{1} cos (2 t) + c_{2} sin (2 t)

Solution steps

4 y^{''} (t) + 16 y = 0

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t)

Popular Examples

y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=4

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 4

y^{''}-4y=0,y(0)=2,y^'(0)=1

y^{^{''}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{''}+13y^'+40y=0

y^{^{''}} + 13 y^{^{'}} + 40 y = 0

y^{'''}+6y^{''}+13y^'=0

y^{^{'''}} + 6 y^{^{''}} + 13 y^{^{'}} = 0

0.2y^{''}+5y^'+1/(0.002)y=0

0.2 y^{^{''}} + 5 y^{^{'}} + \frac{1}{0.002} y = 0

Frequently Asked Questions (FAQ)

What is the solution for 4y^{''}+16y=0 ?
The solution for 4y^{''}+16y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)