First Order ODE Help: Troubleshooting Tips for Solving Differential Equations

[gomes.]

Having a bit of trouble, what do i do next? Thanks.

 

[micromass]

Uh, you've already solved the problem. There's nothing left for you to do here.
I do see that you didn't find an explicit version of y (i.e. nothing of the form y=...), but in many cases of ODE's, this is not even possible or desirable. The solution you provided is implicit, but it's a valid solution nonetheless. Except when your instructor told you to find an explicit solution, then you're not done yet...

 

[HallsofIvy]

You have the equation xy'- 4y= 0 and have separated it as
[tex]\frac{dy}{4y}= \frac{dx}{x}[/itex]<br /> You then integrate to get <br /> $$4 ln(y)= ln(x)+ C$$<br /> <br /> That is incorrect: <br /> $$\int \frac{dy}{4y}= \frac{1}{4}\int\frac{dy}{y}= \frac{1}{4}ln(y)$$<br /> NOT "4 ln(y)".<br /> <br /> You should then have $ln(y^{1/4})= ln(cx)$ where C= ln(c).<br /> <br /> That will then give you $y^{1/4}= cx$ or $y= c^4x^4$ which you could also write as $y= C' x^4$ with $C'= c^4$.<br /> <br /> It would have been better to have left the "4" on the right side of the equation:]<br /> $$\frac{dy}{y}= \frac{4dx}{x}$$[/tex]

 

[gomes.]

thanks, I've got it now :)