Hi, I am able to find the general solution in implicit form for single line equations but I'm really having trouble with this & would appreciate some assistance.
$\displaystyle \frac{dy}{dx}=\frac{6y^\frac{2}{3} sin x}{(2 - cos x)^3}$
Thanks
Hi, I am able to find the general solution in implicit form for single line equations but I'm really having trouble with this & would appreciate some assistance.
$\displaystyle \frac{dy}{dx}=\frac{6y^\frac{2}{3} sin x}{(2 - cos x)^3}$
Thanks
Okay so integrating both sides gives me this:
$\displaystyle \frac{y^5/3}{10}=-\frac{1}{2(2-cos x)^2} + c$
Is this the implicit form or do I need to go further?
I have an initial condition $\displaystyle y(0)=1$ which I presume I can solve for c & then substitute back into this equation?
I'm still not sure about that $\displaystyle +C_{1}$ so I've left it as $\displaystyle +2C$. Maybe someone will tell me otherwise. Also why did the denominator reverse?
Substituting back into general equation for $\displaystyle y(0)=1$:
$\displaystyle 1=\left(\frac{-1}{(2-cos(0))^{2}}+2C\right)^{3}$
$\displaystyle C=1$
Hence $\displaystyle y=\left(\frac{-1}{(cos(x)-2)^{2}}+2\right)^{3}$
Would this satisfy my original question now or have I made it explicit? Thanks