Solving First-Order Linear Equation with Reversed Roles

awelex
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Hi,

I have a differential equation that I just can't seem to solve. Now here's the deal: I'm sure there are advanced methods that would easily solve this equation, but the equation is in the chapter on First-Oder Linear Equations, so it shouldn't be anything fancy. There's even a hint: it says that "the roles of the independent and dependent variables may be reversed." What is that supposed to mean?

I tried getting the solution with Mathematica first, and then working backwards, but to no avail. I don't even know where to start.

Here's the equation:

dy/dx = 1/(e^(4y) + 2x)

Any pointers? Thanks!
 
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With first order differential equations, which variable is a function of the other is not really important. That particular equation is given with y a function of x and is very badly non-linear. But if you "flip it over", it becomes
\frac{dx}{dy}= 2x+ e^{4y}
which is a linear equation that is relatively easy to solve for x as a function of y. That's what they mean by "the roles of the independent and dependent variables may be reversed."
 


Just put x=v & y=u & solve for du/dv.
 


@HallsosIvy: Thanks, that makes sense. I solved the resulting DE and get

x = 1/2 * e^(4y) + C*e^(2y)

But how do I find the inverse function of that?
 


Do you have any reason to want to? That is a perfectly good implicit definition of y.
 


x = 1/2 * e^(4y) + C*e^(2y)
But how do I find the inverse function of that?
Let t=e^(2y)
x = (1/2)*t² +C*t
Solve for t
Then y = (1/2)*ln(t)
In fact, ln(abs(t))
 


Your latter "in fact" comment is not strictly necessary, JJ, since t is, by def.>0
 


@Awelex
Which method did you use to solve this equation?
 


sgtkt said:
@Awelex
Which method did you use to solve this equation?
x = (1/2)*t² +C*t
(1/2)t²+Ct-x =0
I suppose that you know how to solve a*X²+b*X+c=0 for X
a=1/2 ; b=C : c=-x and t=X
 
  • #10


I mean which method, linear or separable
 
  • #11


sgtkt said:
I mean which method, linear or separable

Linear method to solve dx/dy -2x = exp(4y) because it is a linear ODE
Then quadratic resoltion to inverse the x(y) function.
 
  • #12


Thankyou
 

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