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Popular >

y^'=-ysin(2x)

Solution

y^{^{'}} = - y sin (2 x)

Solution

y = e^{\frac{1}{2} c o s (2 x) + c_{1}}

Solution steps

y^{'} (x) = - y sin (2 x)

Solve separable ODE:

y = e^{\frac{1}{2} c o s (2 x) + c_{1}}

y = e^{\frac{1}{2} c o s (2 x) + c_{1}}

Popular Examples

tangent of y=4e^xcos(x),4,\at x=0 tangent of

y = 4 e^{x} cos (x), 4, at x = 0

(dy)/(dx)y=sqrt(4x^2+2)

\frac{d y}{d x} y = 4 x^{2} + 2

inverse oflaplace 5/s inverse laplace

\frac{5}{s}

integral from 1 to 6 of x/2+1/(2x)

\int_{1}^{6} \frac{x}{2} + \frac{1}{2 x} d x

integral of e^{-1/(2x)}

\int e^{- \frac{1}{2 x}} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=-ysin(2x) ?
The solution for y^'=-ysin(2x) is y=e^{1/2 cos(2x)+c_{1}}