Find the particular solution of the differential equation

shiri
Messages
85
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top