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What is a solution to the differential equation #dy/dx=sinx/y#?

Answer 1

Charles Arocha

#y^2=-2cosx+C_2#

#(dy)/(dx)=sinx/y#

separate variables

#intydy=intsinxdx#

#1/2y^2=-cosx+C_1#

#=>y^2=-2cosx+C_2#

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Answer 2

Emilia Aguilar

To solve the differential equation ( \frac{dy}{dx} = \frac{\sin(x)}{y} ), we can rearrange it as ( y,dy = \sin(x),dx ). Integrating both sides yields ( \int y,dy = \int \sin(x),dx ). Integrating ( \int y,dy ) gives ( \frac{1}{2}y^2 + C_1 ), where ( C_1 ) is the constant of integration. Similarly, integrating ( \int \sin(x),dx ) results in ( -\cos(x) + C_2 ), where ( C_2 ) is another constant of integration. Equating these two expressions gives ( \frac{1}{2}y^2 + C_1 = -\cos(x) + C_2 ). Rearranging, we find ( y^2 = -2\cos(x) + C ), where ( C = 2C_2 - C_1 ) is a new constant. Finally, taking the square root of both sides, we obtain the solution ( y = \pm \sqrt{-2\cos(x) + C} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.