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Popular Calculus >

3y^{''}+27y=0

Solution

3 y^{^{''}} + 27 y = 0

Solution

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

3 y^{''} (t) + 27 y = 0

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Graph

Popular Examples

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d er i v a t i v e \frac{32}{x}

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(\partial)/(\partial x)(600-4x+10x^2y)

\frac{\partial}{\partial x} (600 - 4 x + 10 x^{2} y)

derivative of f(x)=(x^2-5x+4)/(x^2+5x+4)

d er i v a t i v e f (x) = \frac{x ^{2} - 5 x + 4}{x ^{2} + 5 x + 4}

tangent of y=sqrt(x+5),(4,3)

t an g e n t y = x + 5, (4, 3)

Frequently Asked Questions (FAQ)

What is the solution for 3y^{''}+27y=0 ?
The solution for 3y^{''}+27y=0 is y=c_{1}cos(3t)+c_{2}sin(3t)