Implicit form question

**bigroo** · December 8th 2009, 04:10 AM

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.

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

Thanks

**galactus** · December 8th 2009, 04:33 AM

Did you try separating variables?.

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

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

**bigroo** · December 8th 2009, 04:35 AM

Originally Posted by galactus

Did you try separating variables?.

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

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

Thank you & I'll give it a go now.

**bigroo** · December 8th 2009, 04:47 AM

Okay so integrating both sides gives me this:

$\frac{y^5/3}{10}=-\frac{1}{2(2-cos x)^2} + c$

Is this the implicit form or do I need to go further?

I have an initial condition $y(0)=1$ which I presume I can solve for c & then substitute back into this equation?

**galactus** · December 8th 2009, 04:54 AM

Yes, use your initial condition

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

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

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

**bigroo** · December 8th 2009, 05:04 AM

Originally Posted by galactus

Yes, use your initial condition

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

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

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

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

**bigroo** · December 8th 2009, 07:05 AM

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 $+C_{1}$ come from?

I'm still not sure about that $+C_{1}$ so I've left it as $+2C$ . Maybe someone will tell me otherwise. Also why did the denominator reverse?

Substituting back into general equation for $y(0)=1$ :

$1=\left(\frac{-1}{(2-cos(0))^{2}}+2C\right)^{3}$

$C=1$

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

Would this satisfy my original question now or have I made it explicit? Thanks

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