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anonymous
 4 years ago
Solve dy/dx=(5x+y)^2 4
I got 1/6ln((u3)/(u+3))=x using a substitution of u=(5x+y). Is that right?
anonymous
 4 years ago
Solve dy/dx=(5x+y)^2 4 I got 1/6ln((u3)/(u+3))=x using a substitution of u=(5x+y). Is that right?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the problem isnt stated properly... it looks wrong though. some clarification would help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[dy/dx=(5x+y)^24\] Is that better?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That usubstitution works for this differential equation. The substitution reduces it to the variablesseparable equation \[du/(u^{2}9) + dx = 0.\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But then you still have to integrate it, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did you get the same answer I got?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(1/6)(lnu  3  lnu + 3 = x + ln(C)/6 as an intermediate solution. Once that is taken back to original variables, we're finished.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My final solution is 5x  y + 3 = C e^(6x)(5x  y  3) in implicit form.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its explicit form is \[y = 5 x + 3(1 + C e^{6x})/(1 + C e^{6x}).\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's a good problem, Dave :^)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll tell my professor;) Thanks!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Major tenfour, good buddy :^)
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