8
$\begingroup$

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.

Thanks!

  • 0
    What tells you that this differential equation has a solution in terms of elementary functions? Where does this problem come from?2012-10-28
  • 0
    @Fabian,It's from my school textbook. Sorry,I am not good at maths.2012-10-28
  • 1
    It doesn't look like a simple homework problem to me.2012-10-28
  • 0
    No, it is not homework.I am studying for a test at school.2012-10-28
  • 0
    @Chris: The solution to this differential equation is... very long.2012-10-28
  • 2
    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.2012-10-28
  • 0
    That's awful. It did appear on a test 3 years back, so I would be indebted if someone gave it a consideration and came up with an answer I can produce in the exam hall.There won't be softwares or calculators allowed.2012-10-28

1 Answers 1

12

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".

  • 0
    You beat me to it. Finding $\alpha, \beta$ is a matter of solving two linear equations in two unknowns.2012-10-28
  • 2
    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 .2012-10-28