# Second order differential with substitution

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).

• Take the derivative ò both sides and use the product rule – Dylan Apr 19 '15 at 19:17
• @Dylan Which part exactly? – MathsIsHard Apr 19 '15 at 19:20
• Take the derivative of $z$ with respect to $x$ – Dylan Apr 19 '15 at 19:23
• @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) – MathsIsHard Apr 19 '15 at 19:25
• You should say that in your question. I'll see if I have an answer. – Dylan Apr 19 '15 at 19:27

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 $$\frac{dy}{dx} = \sqrt{\frac{2}{3}y^{1/2} + \frac{c_1}{y}}$$