Here's the question you clicked on:
Dave321
Solve dy/dx=(-5x+y)^2 -4 I got 1/6ln((u-3)/(u+3))=x using a substitution of u=(-5x+y). Is that right?
the problem isnt stated properly... it looks wrong though. some clarification would help.
\[dy/dx=(-5x+y)^2-4\] Is that better?
That u-substitution works for this differential equation. The substitution reduces it to the variables-separable equation \[du/(u^{2}-9) + dx = 0.\]
But then you still have to integrate it, right?
Did you get the same answer I got?
(1/6)(ln|u - 3| - ln|u + 3| = -x + ln(C)/6 as an intermediate solution. Once that is taken back to original variables, we're finished.
My final solution is 5x - y + 3 = C e^(6x)(5x - y - 3) in implicit form.
Its explicit form is \[y = 5 x + 3(1 + C e^{6x})/(-1 + C e^{6x}).\]
For increasingly large values of x, all solution curves more closely approximate the linear function 5x + 3, so the line y = 5x + 3 is the asymtote of all solution curves.
That's a good problem, Dave :^)
I'll tell my professor;-) Thanks!
Major ten-four, good buddy :^)