# Gerechte23's question

• HallsofIvy
In summary, Gerechte23 is trying to solve a nonlinear equation and has made a mistake. He needs help from someone who is better at solving these equations.f

#### HallsofIvy

Homework Helper
gerechte23 added this to a thread on a different problem. I am starting a new thread for it

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Hello, I am calculating a very important thing. But to find it, I should be able to resolve this equation. Y^2xY’’=C (C: Real number). Please help me solve it. Thanks in advance!

(YY)xY''=C (C:Real nnumber)

Why do you think you ought to be able to resolve it? That looks to me like a very nasty non-linear equation.

Hold on, I was assuming that the "x" in your equation was the independent variable. Did you mean it simply as "times"? (unlikely- by the time you are doing differential equations you should know better!

IF that was what you meant, then since the independent variable does not appear explicitely, we can use "quadrature". Let v= Y'. Then Y"= V'= dV/dx= (dV/dY)(dY/dx)= V(dV/dY) so the equation becomes the separable first order equation Y^2V dV/dY= C which can be written V dV= CdY/Y^2 and integrates to $(1/2)V^2= -C/Y+ D$ where D is the constant of integration. Now, since V= Y', we can write that as the first order equation (1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y, again a separable equation.

$$\frac{YdY}{DY- C}= 2dx$$
To integrate the left side, let u= DY- C so that du= Ddy and we have
$$\frac{1}{D}\frac{du}{u}= 2dx$$

That is easy to integrate.

Of course, that only works if you meant Y2Y"= C. If your really meant xY2Y"= C with x the independent variable, I'm afraid it cannot be integrated in terms of elementary functions.

friend, i was working on the result and then i realized this :(1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y . The problem in this is Y'=[2(DY- C)/Y]^1/2. So, a little error came in. But I've tried solving it, but no. I am not good at all in differential equations. Please give me a hand again in this.
Gerechte23

$$\int\sqrt{Y}$$/$$\int\sqrt{DY-C}$$ = $$\int\sqrt{2}dt$$ . When i resolve this equation using variable change i found the first part like: -$$\sqrt{[(DY-C)/D]}$$ which seems not correct when i derivate it, it doesn't give me the result i was expecting. But when i use trigonometry variable changes, it give me something with logarithm and square root. I don't know how to calculate Y in such a situation. I will post the result tomorrow.

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Excuse me, but can you help me in this. I just can't locate where to click so that i start a new thread in here. Please help

Excuse me, but can you help me in this. I just can't locate where to click so that i start a new thread in here. Please help

go to forums, the sub-forums then make new topic by clicking on that particular button. This is standard procedure for all forums.....

friend, i was working on the result and then i realized this :(1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y . The problem in this is Y'=[2(DY- C)/Y]^1/2. So, a little error came in. But I've tried solving it, but no. I am not good at all in differential equations. Please give me a hand again in this.
Gerechte23

$$\sqrt{\frac{Y}{DY- C}}dY = \sqrt{2}dx$$

Integrate

$$\sqrt{Dy}\sqrt{Dy-C}+C\log(\sqrt{Dy}+\sqrt{Dy-C}) = t\sqrt{2D^3}+ E$$ :yuck:

http://integrals.wolfram.com/index.jsp?expr=Sqrt[x/(d*x-c)]&random=false"

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Thanks for all guys. Now my only problem is to calculate y as a function of t [Y(t)]. It seems quite complicated for me though.
Thanks again

Yes, it is! Are you required to do that?

hi again, yes. I want to calculate it, but it's not a homeworh though. I've proposed myself to find it. Cause with it, i'd be able to calculate a certain acceleration depending on time. Don't you know a way that could help me solve it. If yes, please give a hand