ycos(x)dx-(1+y^2)dy=0

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Solution for ycos(x)dx-(1+y^2)dy=0 equation:


Simplifying
ycos(x) * dx + -1(1 + y2) * dy = 0

Multiply cosy * x
cosxy * dx + -1(1 + y2) * dy = 0

Multiply cosxy * dx
cdosx2y + -1(1 + y2) * dy = 0

Reorder the terms for easier multiplication:
cdosx2y + -1dy(1 + y2) = 0
cdosx2y + (1 * -1dy + y2 * -1dy) = 0
cdosx2y + (-1dy + -1dy3) = 0

Solving
cdosx2y + -1dy + -1dy3 = 0

Solving for variable 'c'.

Move all terms containing c to the left, all other terms to the right.

Add 'dy' to each side of the equation.
cdosx2y + -1dy + dy + -1dy3 = 0 + dy

Combine like terms: -1dy + dy = 0
cdosx2y + 0 + -1dy3 = 0 + dy
cdosx2y + -1dy3 = 0 + dy
Remove the zero:
cdosx2y + -1dy3 = dy

Add 'dy3' to each side of the equation.
cdosx2y + -1dy3 + dy3 = dy + dy3

Combine like terms: -1dy3 + dy3 = 0
cdosx2y + 0 = dy + dy3
cdosx2y = dy + dy3

Divide each side by 'dosx2y'.
c = o-1s-1x-2 + o-1s-1x-2y2

Simplifying
c = o-1s-1x-2 + o-1s-1x-2y2

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