1. Find the value K

If $\displaystyle siny = xsin(a+y)$ and $\displaystyle \frac{dy}{dx} = K \frac{sin^2y}{x^2}$ , find K

2. Originally Posted by zorro
If $\displaystyle siny = xsin(a+y)$ and $\displaystyle \frac{dy}{dx} = K \frac{sin^2y}{x^2}$ , find K
I'm assuming that a is constant. If we expand your sine relationship and divide by x then $\displaystyle \frac{\sin y}{ x} = \sin a \cos y + \cos a \sin y$ then divide by $\displaystyle \sin y$ so

$\displaystyle \sin a \cot y + \cos a = \frac{1}{x}$ and upon differentiation gives $\displaystyle - \sin a \csc^2 y y' = - \frac{1}{x^2}$ which solving for y ' gives

$\displaystyle y' = \frac{1}{\sin a} \frac{\sin^2 y}{x^2}$

in which we can identify K.

3. Thanks mite....

Originally Posted by Danny
I'm assuming that a is constant. If we expand your sine relationship and divide by x then $\displaystyle \frac{\sin y}{ x} = \sin a \cos y + \cos a \sin y$ then divide by $\displaystyle \sin y$ so

$\displaystyle \sin a \cot y + \cos a = \frac{1}{x}$ and upon differentiation gives $\displaystyle - \sin a \csc^2 y y' = - \frac{1}{x^2}$ which solving for y ' gives

$\displaystyle y' = \frac{1}{\sin a} \frac{\sin^2 y}{x^2}$

in which we can identify K.

thanks mite
cheers