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Popular >

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

Solution

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

Solution

y = c_{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{2} sin (2 t)

y = c_{2} sin (2 t)

Popular Examples

(d^2q)/(dt^2)+(dq)/(dt)+q=0

\frac{d ^{2} q}{d t ^{2}} + \frac{d q}{d t} + q = 0

y^{''}-y^'-2y=0,y(0)=-8,y^'(0)=-1

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = - 8, y^{^{'}} (0) = - 1

y^{''}+18y=0,y(0)=-3,y^'(0)=-1

y^{^{''}} + 18 y = 0, y (0) = - 3, y^{^{'}} (0) = - 1

(d^2y)/(dx^2)-5(dy)/(dx)-6y=0

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 6 y = 0

y^{''}+8y^'+16y=0,y(-1)=2,y^'(-1)=4

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 4

Frequently Asked Questions (FAQ)

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