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Popular >

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

Solution

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

Solution

y = c_{1} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t)

y = c_{1} cos (2 t)

Graph

Popular Examples

13y^{''}-14y^'+y=0,y(0)=13,y^'(0)= 1/13

13 y^{^{''}} - 14 y^{^{'}} + y = 0, y (0) = 13, y^{^{'}} (0) = \frac{1}{13}

y^{^{''}} - \frac{1}{4} y = 0

y^{''}+a^2*n^2*pi^2*y=0,y^'(0)=0

y^{^{''}} + a^{2} \cdot n^{2} \cdot π^{2} \cdot y = 0, y^{^{'}} (0) = 0

(d^2y)/(dx^2)+k^2*y^'=0

\frac{d ^{2} y}{d x ^{2}} + k^{2} \cdot y^{^{'}} = 0

4y^{''}+4y^'+6y=0

4 y^{^{''}} + 4 y^{^{'}} + 6 y = 0

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_{1}cos(2t)