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Popular Calculus >

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

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Graph

Popular Examples

y^'=((e^{x-y}))/(1+e^x)

y^{^{'}} = \frac{( e ^{x - y} )}{1 + e ^{x}}

integral of (24)/(x^2)

\int \frac{24}{x ^{2}} d x

(\frac{x - 1}{x})^{^{'}}

derivative of 2xcos(x-x^2sin(x))

\frac{d}{d x} (2 x cos (x) - x^{2} sin (x))

derivative of log_{6}(5x+1)

\frac{d}{d x} (lo g_{6} (5 x + 1))

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+b^2y=0 ?
The solution for y^{''}+b^2y=0 is y=c_{1}cos(bt)+c_{2}sin(bt)