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How do we solve the differential equation $$\frac{dy}{dx}=\frac{3x+4y+7}{x-2y-11}$$?

I tried substituting $v=yx$ but I do not seem to be getting anywhere.Putting $u=x-2y$ yielded nothing better.


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What tells you that this differential equation has a solution in terms of elementary functions? Where does this problem come from? – Fabian Oct 28 '12 at 10:39
@Fabian,It's from my school textbook. Sorry,I am not good at maths. – user43081 Oct 28 '12 at 10:40
It doesn't look like a simple homework problem to me. – Fabian Oct 28 '12 at 10:41
No, it is not homework.I am studying for a test at school. – user43081 Oct 28 '12 at 10:42
Yes, the answer is very long. $$10\sqrt{15}\arctan\left(\frac{\sqrt{\frac{5}{3}} (-1+3 x+2y)}{-11+x-2y}\right)=3\left(4C+10\ln(-3+x)+5\ln\left(\frac{23+3 (-2+x) x+y(7+3 x+2 y)}{5 (-3+x)^2}\right)\right)$$. This is given by Mathematica. – user46090 Oct 28 '12 at 10:53
up vote 12 down vote accepted

A hint: Introduce new variables $X$, $Y$ via $$x:=X+\alpha, \quad y:=Y+\beta$$ and choose the constants $\alpha$, $\beta$ such that the $7$ and the $-11$ on the right side of your equation disappear. In terms of the new variables your equation now has the form $$Y'={3X+4Y\over X-2Y} ={3+4{Y\over X}\over 1-2{Y\over X}}\ .$$ This is a standard type of ODE, sometimes called "homogeneous".

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You beat me to it. Finding $\alpha, \beta$ is a matter of solving two linear equations in two unknowns. – GEdgar Oct 28 '12 at 12:22
It did the trick.Thank you.I followed your hint, obtained the values of $\alpha$ and $\beta$, put $\frac{Y}{X}=v$ and the rest followed through . – user43081 Oct 28 '12 at 12:24

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