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$$tx'(x'+2)=x$$ First I multiplied it: $$t(x')^2+2tx'=x$$ Then differentiated both sides: $$(x')^2+2tx'x''+2tx''+x'=0$$ substituted $p=x'$ and rewrote it as a multiplication $$(2p't+p)(p+1)=0$$ So either $(2p't+p)=0$ or $p+1=0$

The first one gives $p=\frac{C}{\sqrt{T}}$ The second one gives $p=-1$. My question is how do I take the antidervative of this in order to get the answer for the actual equation?

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EDIT:

$x=t(x')^2+2tx'$

$p=x'$

$x=tp^2+2tp$

We differentiate in respect to $t$:

$p=p^2+t2pp'+2p+2tp' \Rightarrow p'(2tp+2t)=(p-p^2-2p) \Rightarrow p'(p+1)2t=-(p^2+p) \Rightarrow p'(p+1)2t=-p(p+1) \Rightarrow p'(p+1)2t+p(p+1)=0 \Rightarrow (p+1)(2tp'+p)=0 \\ \Rightarrow p+1=0 \text{ or } 2tp'+p=0 \\ \Rightarrow p=-1 \text{ or } p'=-\frac{1}{2t}p \\ \Rightarrow p=-1 \text{ or } \frac{p'}{p}=-\frac{1}{2t} \\ \Rightarrow p=-1 \text{ or } \ln{p}=-\frac{1}{2}\ln{t} +c\\ \Rightarrow p=-1 \text{ or } \ln{p}=\ln{t^{-\frac{1}{2}}} +c\\ \Rightarrow p=-1 \text{ or } p= \pm e^c \frac{1}{\sqrt{t}} \\ \Rightarrow x'=-1 \text{ or } x'= \pm e^c\frac{1}{\sqrt{t}} \\ \Rightarrow x(t)=-t+c_1 \text{ or } x(t)=2 C\sqrt{t}+c_2, \text{ where } C= \pm e^c$.

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  • $x(t)=-t+c_1 \Rightarrow x'=-1$

Replacing this at the initial equation we get:

$x=t(x')^2+2tx' \Rightarrow -t+c_1=t-2t \Rightarrow -t+c_1=-t \Rightarrow c_1=0$

Therefore, $x(t)=-t$.

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  • $x(t)=2 C\sqrt{t}+c_2 \Rightarrow x'(t)=\frac{C}{\sqrt{t}} $

Replacing this at the initial equation we get:

$x=t(x')^2+2tx' \Rightarrow 2 C\sqrt{t}+c_2=t\frac{C^2}{t}+2t\frac{C}{\sqrt{t}} \Rightarrow 2C \sqrt{t}+c_2=C^2+\frac{2Ct}{\sqrt{t}} \\ \Rightarrow 2Ct+c_2\sqrt{t}=C^2\sqrt{t}+2Ct \Rightarrow c_2\sqrt{t}=C^2\sqrt{t} \Rightarrow c_2=C^2$

Therefore, $x(t)=2 C\sqrt{t}+C^2$.

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  • $\begingroup$ $-t+arbitrary constant$ doesnt work here, the C has to be specific and equal to 0. Should I just substitute it into equation and evaluate it? $\endgroup$ – Lugi Aug 20 '14 at 13:50
  • $\begingroup$ @Lugi I edited my answer. $\endgroup$ – Mary Star Aug 20 '14 at 15:41
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Your differential equation is a complicated one. I give some hints on $I=]0,+\infty[$ (the study has to be done also on $]-\infty,0[$).

A) First note that on $I$, your equation is $\displaystyle (x^{\prime}(t)+1)^2=\frac{x(t)+t}{t}$. We see that $x_0(t)=-t$ is a solution. For any solution, we must have $x(t)+t\geq 0$.

Now suppose that on $I$, we have $x(t)+t >0$ for all $t$. Then $x^{\prime}+1\not =0$ for all $t$; by Darboux's theorem, $x^{\prime}+1$ has a constant sign on $I$. With $\varepsilon\in \{\pm 1\}$, we can write $\displaystyle \frac{x^{\prime}+1}{\sqrt{x(t)+t}}=\frac{\varepsilon}{\sqrt{t}}$, and hence $2\sqrt{x(t)+t}=2\varepsilon\sqrt{t}+C$ where $C$ is a constant. Now $x(t)=C^2/4+C\varepsilon \sqrt{t}$. Of course, we have to verify that all these solutions are really solutions(and verify what conditions on $C$ and $\varepsilon$ gives $x(t)+t>0$ for all $t$).

B) To see how the hypothesis $x(t)+t>0$ is important, let $t_0>0$. Put $x(t)=-t$ for $0<t\leq t_0$, and $x(t)=t_0-2\sqrt{t_0t}$ if $t>t_0$. I leave to you the verification that $x(t)$ is a solution, (first show that the derivative at $t_0$ exists and is equal to $-1$). But this solution is not one of the form $C^2/4+C\varepsilon \sqrt{t}$ we have found, as they cannot agree on $]0,t_0[$.

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Group theory gives a pretty direct answer. Use $$ G(t,x)=(\lambda t,\lambda^\beta x)\lambda_o=1 $$Then apply the group to the DEQ: $$ t\bigg(\frac{dx}{dt}\bigg)^2+2t\frac{dx}{dt}=x\rightarrow \lambda t\bigg(\frac{\lambda ^\beta dx}{\lambda dt}\bigg)^2+2\lambda t\frac{\lambda^\beta dx}{\lambda dt}=\lambda^\beta x $$For the group to be invariant, $2\beta-1=\beta$ or $\beta=1$. There are two nontrivial stabilizers for this group: $$ \mu=\frac{x}{t^\beta}=\frac{x}{t} $$and $$ \nu=\frac{x'}{t^{\beta -1}}=x' $$According to Sophus Lie, your DEQ can be written in terms of the stabilizers of an invariant Lie group translation, one that preserves the structure of the smooth manifold. Since our group is an infinite continuous group, we are able to do just that. $$ x'^2+2x'=\frac{x}{t}\rightarrow \nu^2+2\nu=\mu $$This is a quadratic for $\nu$. $$ \nu=-1\pm\sqrt{1+\mu} $$Since $$ t\frac{d\mu}{dt}=\nu-\mu=-1\pm\sqrt{1+\mu}-\mu $$this becomes a separable equation. $$ \frac{d\mu}{-(1+\mu)\pm\sqrt{1+\mu}}=\frac{dx}{x} $$This integrates and simplifies to give $$ 1\mp\sqrt{1+\mu}=\frac{C}{\sqrt{t}} $$and after substituting for $\mu$ and simplifying again, we arrive at the solution: $$ x=C^2-2C\sqrt{t} $$There are other, perhaps easier ways of getting to this solution, but few people know about applied Lie Theory so I like to "beat the drum" whenever possible.

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