# homogenous equation

• Feb 21st 2011, 01:30 PM
mathcore
homogenous equation
(Thinking)please give me a hint how to start this@

x(dy/dx) = y + 2xsin^2(y/x)

basically i dont know what the first step is because the way i learnt this, i can't see how to apply that method to this example?
• Feb 21st 2011, 01:31 PM
Ackbeet
You have two equals signs there. Did you mean a plus sign in there somewhere?
• Feb 21st 2011, 01:40 PM
mathcore
yes sorry it is i edited it now
• Feb 21st 2011, 01:47 PM
mathcore
ok i know to divide through x, sub z=y/x. but then how do you find dy/dx? if y=xz? is it chain rule? can someone talk me through this?
• Feb 21st 2011, 04:11 PM
mathcore
i have solved it down to this

-1/(2tan(y/x)) = lnx + c

how can i isolate y please?
• Feb 21st 2011, 04:12 PM
mathcore
isolate y
i have an implicit solution that i need to make plicit

-1/(2tan(y/x)) = lnx + c

how can i isolate y please?
• Feb 21st 2011, 04:28 PM
pickslides
$\displaystyle \frac{-1}{2\tan\frac{y}{x}} = \ln x + c$

$\displaystyle \frac{-1}{\ln x+ c} = 2\tan\frac{y}{x}$

$\displaystyle \frac{-1}{2\ln x+ c} = \tan\frac{y}{x}$

$\displaystyle \tan^{-1}\left(\frac{-1}{2\ln x+ c}\right) = \frac{y}{x}$

$\displaystyle x\tan^{-1}\left(\frac{-1}{2\ln x+ c}\right) = y$
• Feb 21st 2011, 04:46 PM
mathcore
thanks, a lot

should the constant be written as lnc so ln(x + c)?
• Feb 21st 2011, 05:02 PM
skeeter
Quote:

Originally Posted by mathcore
thanks, a lot

should the constant be written as lnc so ln(x + c)?

$\ln{x} + \ln{c} \ne \ln(x+c)$

$\ln{x} + \ln{c} = \ln(cx)$
• Feb 21st 2011, 06:16 PM
mathcore
too right bro. it is, i forgot my log rules but still, i ask the question, is it better to leave the constant as c or stick it in with the ln (x)? i think putting it as ln(cx) seems neater than ln(x) + c? what is convention?
• Feb 21st 2011, 06:55 PM
pickslides
Either way is fine, in general I notice $\displaystyle \ln x + c = \ln x + \ln c = \ln (cx)$ when solving D.E.s

It really depends on the situation and what is required.
• Feb 21st 2011, 06:56 PM
mathcore
i think it is neater
• Feb 22nd 2011, 02:06 AM
Ackbeet
Different people learn this method different ways. I learned it this way:

$\displaystyle x\,\frac{dy}{dx} = y + 2x\sin^{2}\!\left(\dfrac{y}{x}\right)$

$\displaystyle x\,dy-\left(y+2x\sin^{2}\!\left(\dfrac{y}{x}\right)\righ t)dx=0.$

If we follow your substitution of $z=y/x,$ then $y=zx,$ and $dy=z\,dx+x\,dz.$ (This is how I "find" $dy/dx$. If you wish, you can write $dy/dx=z+x(dz/dx)$.) Plugging into the DE yields

$x(z\,dx+x\,dz)-(zx+2x\sin^{2}(z))\,dx=0$

$x^{2}\,dz-2x\sin^{2}(z)\,dx=0$

$x\,dz-2\sin^{2}(z)\,dx=0.$ Separating out yields

$\csc^{2}(z)\,dz-\dfrac{2\,dx}{x}=0$

$-\cot(z)-2\ln|x|=C$

$\cot(z)+2\ln|x|=C$ (absorbing the minus sign into the constant).

$\cot\left(\dfrac{y}{x}\right)=-2\ln|x|+C.$

At this point, as with all solutions to any differential equation, you should verify that this implicitly defined function satisfies the original DE. The reason to differentiate now is that if you do, the constant will go away, which is what you want when you're verifying the solution to a DE. The solution I've got here is, incidentally, the same as yours, I believe. So far, so good. If you do want to isolate $y,$ then take the $\cot^{-1}$ of both sides, and then multiply the resulting equation by $x$ to obtain

$y=x\cot^{-1}\left(-2\ln|x|+C\right).$

On what interval(s) is this solution valid?