1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
2
$\begingroup$

You are both correct.

The value of c depends on the initial conditions.

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

$\endgroup$
0
$\begingroup$

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: $y=x^2+C$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.