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Question here

Hey guys, I was doing this question and am really stuck :/

I got up to taking n as 1 and getting z'=sqrt(y)*y'

Can someone tell me where to go from here?

Edit:

I've done the first part, just not sure how to continue with part (i).

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  • $\begingroup$ Take the derivative ò both sides and use the product rule $\endgroup$ – Dylan Apr 19 '15 at 19:17
  • $\begingroup$ @Dylan Which part exactly? $\endgroup$ – MathsIsHard Apr 19 '15 at 19:20
  • $\begingroup$ Take the derivative of $z$ with respect to $x$ $\endgroup$ – Dylan Apr 19 '15 at 19:23
  • $\begingroup$ @Dylan Oh, I'm sorry I forgot to mention, I can do the first part, just not the part after the first part. As in I don't know how to do part (i) $\endgroup$ – MathsIsHard Apr 19 '15 at 19:25
  • $\begingroup$ You should say that in your question. I'll see if I have an answer. $\endgroup$ – Dylan Apr 19 '15 at 19:27
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Setting $n = 1$, we get $$z = y \left(\frac{dy}{dx}\right)^2$$ and $$ \frac{dz}{dx} = \sqrt{y}\,\frac{dy}{dx} \Rightarrow z = \frac{2}{3}y^{3/2} + c_1 $$

Setting those equal to each other we get $$ \frac{dy}{dx} = \sqrt{\frac{2}{3}y^{1/2} + \frac{c_1}{y}} $$

You now have an IVP to solve.

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  • $\begingroup$ Thank you, I will study this. $\endgroup$ – MathsIsHard Apr 19 '15 at 19:37
  • $\begingroup$ I updated my answer to include a constant $\endgroup$ – Dylan Apr 19 '15 at 19:48

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