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
\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 &quot;4 ln(y)&quot;.<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&amp;#039; x^4 with C&amp;#039;= c^4.<br /> <br /> It would have been better to have left the &quot;4&quot; on the right side of the equation:]<br /> \frac{dy}{y}= \frac{4dx}{x}

 

[gomes.]

thanks, I've got it now :)