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Solve the given initial value problem and determine at least approximately where the solution is valid $(2x - y) dx + (2y- x) dy = 0, y(1) = 3 $.

My Solution: Let $M=2x-y \Rightarrow M_y=-1$ and let $N=2y-x\Rightarrow N_x=-1$, therefore the given DE is exact. Let $\Psi_x=M=2x-y\Rightarrow \Psi = x^2-xy+h(y)\Rightarrow \Psi_y=-x+h'(y)\Rightarrow h'(y)=\Psi_y+x=2y-x+x=2y \Rightarrow \Psi=x^2-xy+2y=C$, initial value condition given is $y(1)=3 \Rightarrow C=(1)^2-(1)(3)+2(3)=4 \Rightarrow x^2-xy+2y=4\Rightarrow y=\frac{4-x^2}{2-x}=2+x$, but the given answer is $y = [x+\sqrt{28-3x^2}]/2, |x|<\sqrt{28/3} $

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up vote 2 down vote accepted

You have made a small error, $h'(y) = 2y$ tells us that $h(y) = y^2+K$, so you should have gotten $$\Psi = x^2-xy+y^2=C.$$ I think you know how to solve the problem from here.

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thanks, silly me :| – Vikram Jul 7 '12 at 16:44

Although nullUeser's answer is enough here, I wanted to note you another approach. Your equation is a homogeneous also, because you can write it as $y'=f(t)=\frac{t-2}{2t-1}$ wherein $t=\frac{y}{x}, x\neq 0$. Now, you can solve the converted equation $$\frac{dt}{f(t)-t}=\frac{dx}{x}$$ instead. Of course, I prefer your own way. :)

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