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Find the particular solution of the differential equation

  • Thread starter shiri
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  • #1
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Find the particular solution of the differential equation

dy/dx = (3x+42y)/7x

satisfying the initial condition y(1) = 5.

Attempt:
dy/dx = 3/7 + 6y/x

dy/dx - 6y/x = 3/7

p(x) = -6/x
q(x) = 3/7

u(x) = -6 ∫(1/x)dx = -6ln|x|
e^(u(x)) = -x^6


1/e^(u(x)) ∫e^(u(x))q(x)dx

= 1/(-x^6) ∫(-x^6)(3/7)dx

= 3/(7(-x^6)) [(-x^7/7)+c]

=3x/49 - 3c/(7x^6)


y(1) = 5 = 3(1)/49 - 3c/(7(1)^6)

5 = 3/49 - 3c/7

242/49 = -3c/7

c = -242/21


So, what I got is

y(x) = 3x/49 + 242/(49x^6)

however, my professor said it's wrong. So, can anybody tell me what I did wrong? I'll be very appreciated. thanks.
 

Answers and Replies

  • #2
gabbagabbahey
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u(x) = -6 ∫(1/x)dx = -6ln|x|
e^(u(x)) = -x^6

[tex]e^u=e^{-6\ln|x|}=\left(e^{\ln|x|}\right)^{-6}=|x|^{-6}=x^{-6}\neq -x^6[/tex]

:wink:

So, what I got is

y(x) = 3x/49 + 242/(49x^6)

however, my professor said it's wrong. So, can anybody tell me what I did wrong? I'll be very appreciated. thanks.
It's always a good idea to check your answers before submitting them. Does this solution satisfy your original ODE?
 
  • #3
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[tex]e^u=e^{-6\ln|x|}=\left(e^{\ln|x|}\right)^{-6}=|x|^{-6}=x^{-6}\neq -x^6[/tex]

:wink:



It's always a good idea to check your answers before submitting them. Does this solution satisfy your original ODE?
Just to be sure, the answer is

[tex]y(x) = -3x/35 + 178/(35x^{6})[/tex]

right?
 
Last edited:
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Just to be sure, the answer is

[tex]y(x) = -3x/35 + 178/(35x^{6})[/tex]

right?
You mean [tex]y(x) = -\frac{3}{35}x + \frac{178}{35}x^6[/tex]...right? If so, then yes.
 

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