1. ## Implicit form question

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

2. Did you try separating variables?.

$\displaystyle \frac{1}{6}\int y^{\frac{-2}{3}}dy=\int\frac{sin(x)}{(2-cos(x))^{3}}dx$

Integrate both sides and don't forget the constant on the right.

3. Originally Posted by galactus
Did you try separating variables?.

$\displaystyle \frac{1}{6}\int y^{\frac{-2}{3}}dy=\int\frac{sin(x)}{(2-cos(x))^{3}}dx$

Integrate both sides and don't forget the constant on the right.
Thank you & I'll give it a go now.

4. 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?

5. Yes, use your initial condition

$\displaystyle \frac{1}{2}y^{\frac{1}{3}}=\frac{-1}{2(cos(x)-2)^{2}}+C$

$\displaystyle y^{\frac{1}{3}}=\frac{-1}{(cos(x)-2)^{2}}+2C$

$\displaystyle y=\left(\frac{-1}{(cos(x)-2)^{2}}+C_{1}\right)^{3}$

6. Originally Posted by galactus

$\displaystyle \frac{1}{2}y^{\frac{1}{3}}=\frac{-1}{2(cos(x)-2)^{2}}+C$

$\displaystyle y^{\frac{1}{3}}=\frac{-1}{(cos(x)-2)^{2}}+2C$

$\displaystyle y=\left(\frac{-1}{(cos(x)-2)^{2}}+C_{1}\right)^{3}$
Right I can see how you've done that (including the initial fraction for y which I left undone!) but where does the $\displaystyle +C_{1}$ come from?

7. Originally Posted by bigroo
Right I can see how you've done that (including the initial fraction for y which I left undone!) but where does the $\displaystyle +C_{1}$ come from?
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