If xsiny=sin(p+y),p∈R,show that sinpdy/dx+sin^2y=0?

Question

Emily Ammerman · Accepted Answer

Kindly refer to a Proof in the Explanation.

Given that, #xsiny=sin(p+y), p in RR#. #:. x=sin(p+y)/siny=(sinpcosy+sinycosp)/siny#, #=(sinpcosy)/siny+(sinycosp)/siny#. # rArr x=sinpcoty+cosp, (p" const.)"# Diff.ing w.r.t. #y#, we get, #dx/dy=-csc^2ysinp+0=-sinp/sin^2y#. #:. dy/dx=1/(dx/dy)=-sin^2y/sinp, or," what is the same as,"# # sinpdy/dx+sin^2y=0#, as desired!

Samantha Adkison · Answer

undefined Given $x \sin(y) = \sin(p+y)$, where $p$ is a constant, differentiate both sides with respect to $x$, using the product rule on the left side and the chain rule on the right side:

$$ \frac{d}{dx} (x \sin(y)) = \frac{d}{dx} (\sin(p+y)) $$

$$ \sin(y) + x \cos(y) \frac{dy}{dx} = \cos(p+y) \frac{dy}{dx} $$

$$ x \cos(y) \frac{dy}{dx} - \cos(p+y) \frac{dy}{dx} = - \sin(y) $$

$$ \frac{dy}{dx} (x \cos(y) - \cos(p+y)) = - \sin(y) $$

$$ \frac{dy}{dx} = \frac{- \sin(y)}{x \cos(y) - \cos(p+y)} $$

Now, we need to express $\sin(p)$ in terms of $y$ and $x$. From the given equation $x \sin(y) = \sin(p+y)$, rearrange to get:

$$ \sin(p) = x \sin(y) - \sin(y) $$

$$ \sin(p) = \sin(y) (x - 1) $$

So, differentiate $\sin(p)$ with respect to $x$:

$$ \frac{d \sin(p)}{dx} = \sin(y) \frac{d}{dx} (x - 1) $$

$$ \frac{d \sin(p)}{dx} = \sin(y) $$

Substitute this into the expression for $\frac{dy}{dx}$:

$$ \frac{dy}{dx} = \frac{- \sin(y)}{x \cos(y) - \cos(p+y)} = \frac{- \sin(y)}{x \cos(y) - \cos(p)} $$

Now, multiply both numerator and denominator by $\sin(y)$:

$$ \frac{dy}{dx} = \frac{- \sin^2(y)}{x \cos(y) \sin(y) - \sin(y) \cos(p)} $$

$$ \frac{dy}{dx} = \frac{- \sin^2(y)}{x \sin(y) \cos(y) - \sin(y) \sin(p)} $$

$$ \frac{dy}{dx} = \frac{- \sin^2(y)}{\sin(y)(x \cos(y) - \sin(p))} $$

$$ \frac{dy}{dx} = \frac{- \sin(y) \sin(y)}{\sin(y)(x \cos(y) - \sin(p))} $$

$$ \frac{dy}{dx} = \frac{- \sin(y)}{x \cos(y) - \sin(p)} $$

So, we've shown that $ \frac{dy}{dx} = \frac{- \sin(y)}{x \cos(y) - \sin(p)} $.

Therefore, $ \sin(p) \frac{dy}{dx} + \sin^2(y) = 0 $.