I had to solve the following differential equation: $(x^2+y)\mathrm{d}x - x \mathrm{d}y=0$. The equation is not exact and so I solved it as a simple linear equation

$$ \frac{\mathrm{d}y}{\mathrm{d}x}-\frac{y}{x}=x$$

The solution I got is $\frac{y}{x}=x+c$. However, the solution in the textbook is $-\frac{y}{x}+x=c$. I tried solve it again and again, and yet could not find where I'm wrong. Or is the solution in the book incorrect?

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    $\begingroup$ Congratulations on getting correct answer ! your answer is equivalent to textbook answer. Just notice that : $-y/x+x = c \implies y/x = x-c$ $\endgroup$ – ganeshie8 Aug 25 '14 at 7:51
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    $\begingroup$ quoting lab bhattacharjee : "Point to be noted, we don't need to match with the answer supplied as long as the approach is right.." $\endgroup$ – Claude Leibovici Aug 25 '14 at 7:57

You are both correct.

The value of c depends on the initial conditions.

Your c will be the negative of the book's c.


Transform your equation with a group $x'=\lambda x$ and $y'=\lambda^\beta y$. Note that $dx'=\lambda dx$ and $dy'=\lambda^\beta dy$.
$$ \frac{\lambda^\beta dy}{\lambda dx}-\frac{\lambda^\beta y}{\lambda x}=\lambda x $$For your ODE to be invariant to the group, $\beta -1=1$ or $\beta =2$. Two stabilizers for this group are $$ \mu=\frac{y}{x^\beta}=\frac{y}{x^2} $$and $$ \nu=\frac{\dot{y}}{x^{\beta -1}}=\frac{\dot{y}}{x} $$where $\dot{y}=\frac{dy}{dx}$. Now your ODE can be written in terms of the group stabilizers: $$ \frac{\dot{y}}{x}-\frac{y}{x^2}=1 \rightarrow \nu-\mu=1 \rightarrow \nu=1+\mu $$Noting that $$ x\frac{d\mu}{dx}=\nu-\beta \mu=1+\mu-2\mu=1-\mu $$we can separate and solve this equation: $$ \frac{d\mu}{1-\mu}=\frac{dx}{x}\rightarrow -ln(1-\mu)=lnx-lnC $$ $$ \frac{C}{x}=1-\mu=1-\frac{y}{x^2}\rightarrow Cx=x^2-y $$Thus your answer is equalivent to the one in the textbook: $$ y=x^2-Cx $$Of course, since C is a constant it can absorb the negative sign and still work as a solution: $y=x^2+C$.

  • $\begingroup$ Well done, but you might have chosen a simpler way of explaining this to a student who has just been introduced to differential eqns. $\endgroup$ – Patrick Shambayati Aug 27 '14 at 1:16
  • $\begingroup$ Ha! I stand corrected, Patrick. You caught me showing off. I've been researching applied Lie Theory for three years, and in the process turned myself into a hammer looking for a nail. Saw this problem and thought, "Easy one!" Didn't think about the needs behind the question. $\endgroup$ – atomteori Aug 27 '14 at 12:48

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