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Popular >

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

Solution

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

Solution

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

Solution steps

8 y^{''} (t) + 2 y = 0

Solve linear ODE:

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

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

Graph

Popular Examples

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linear y^{''}-6y^'+5y=0 linear

y^{^{''}} - 6 y^{^{'}} + 5 y = 0

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2 y^{^{''}} + 4 y^{^{'}} + 7 y = 0

Frequently Asked Questions (FAQ)

What is the solution for 8y^{''}+2y=0 ?
The solution for 8y^{''}+2y=0 is y=c_{1}cos(t/2)+c_{2}sin(t/2)