Find the particular solution of the differential equation

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SUMMARY

The discussion focuses on solving the differential equation dy/dx = (3x + 42y)/7x with the initial condition y(1) = 5. The user initially derived the solution y(x) = 3x/49 + 242/(49x^6), which was incorrect. The correct particular solution is y(x) = -3/35 x + 178/(35x^6). The error stemmed from a miscalculation in the integration step involving the integrating factor.

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shiri
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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.
 
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shiri said:
u(x) = -6 ∫(1/x)dx = -6ln|x|
e^(u(x)) = -x^6
e^u=e^{-6\ln|x|}=\left(e^{\ln|x|}\right)^{-6}=|x|^{-6}=x^{-6}\neq -x^6

: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?
 
gabbagabbahey said:
e^u=e^{-6\ln|x|}=\left(e^{\ln|x|}\right)^{-6}=|x|^{-6}=x^{-6}\neq -x^6

: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

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

right?
 
Last edited:
shiri said:
Just to be sure, the answer is

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

right?

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

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