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ODE substitution

  • #1

Homework Statement



dy/dx= (4x sec(2y/x) +y) / x IC: y (1) = pi/4

Homework Equations





The Attempt at a Solution



So i can split that up into 4xsec(2y/x)/x +y/x

= 4sec(2y/x) +y/x and let v= y/x dy/dx = xdv/dx +v

4sec(2v) +v = xv' +v
4sec(2v)=xv' which is seperable

1/x dx - 1/4sec(2v) dv =0
1/xdx - cos(2v)/4 dv =0
lnx - sin(2v)/8 = C annnd going back to y/x

lnx-sin(2y/x)/8 = C and with the initial condition

-sin(2pi/4) / 8 = C

C= -1/8
So, lnx-sin(2y/x)/8=-1/8

lnx+1/8=sin(2y/x)/8
8lnx +1 = sin(2y/x)
arcsin(8lnx +1)= 2y/x
(xarcsin(8lnx +1) ) /2 = y Is what i get a solution. However this doesn't make sence because arcsin of 8lnx +1 doesn't really work....

So, im not sure where i went wrong :(
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
12,985
540
On a first sight you didn't go wrong. Arcsin makes sense only in the interval (domain) [-1,1], so that puts a boundary on "x", but the rest is just fine. You can check your final solution y(x) versus the condition y(1)=pi/4. It works.
 
  • #3
hunt_mat
Homework Helper
1,739
18
The inteval of definition of the solution is when 8lmx+1=1 and 8lnx+1=-1, the first condition yields x=1 and the other condition yields x=exp(-0.25), so your solution workk for the range:
[tex]
[e^{-\frac{1}{4}},1]
[/tex]
 

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