• Support PF! Buy your school textbooks, materials and every day products Here!

Dif. eq. problem - dont know why it is wrong

  • #1
1,197
1
dif. eq. problem -- dont know why it is wrong

Solve for y: [tex]x \frac{dy}{dx} + 6y=5x[/tex]

with y(1)=2

[tex]\frac{dy}{dx} =5 - 6\frac{y}{x} [/tex]
[tex]\frac{6y}{x} \frac{dy}{dx} =5[/tex]
[tex] 3y^2=\frac{5}{2}x^2 +C[/tex]
[tex] y= \sqrt{\frac{5x^2}{6}+\frac{C}{3}}[/tex]

solve for when y(1)=2

[tex]4=\frac{5}{6}+\frac{C}{3}[/tex]
[tex]C=\frac{19}{2}[/tex]

so..

[tex]y= \sqrt{\frac{5x^2}{6}+\frac{19}{6}}[/tex]

what did i do wrong? because this is not the answer
 
Last edited:

Answers and Replies

  • #2
1,074
1
y/x does not equal (dy)/(dx), You need to divide by x and then find an integrating factor.
 
  • #3
1,197
1
sorry, that was typo! could you check it now?
 
  • #4
1,074
1
It looks even worse, you somehow make addition into multiplaication. I already told you what you need to do. You need to divide everything by x to get the y' by itself and then find an integrating factor, you should know how to do that if you're being given this kind of problem because it certainly isn't separable,
 
  • #5
1,197
1
oh thanks. by the way, when I am given a dif. eq. and they ask me "What are the constant solutions of this equation? "

what exactly are do they want me to find?
 
  • #6
ek
181
0
As said before, you're taking the wrong approach.

You need to divide it by x to get it into standard form or whatever it's called, and then get an integrating factor.

I get an answer of y = 5x/7 +c/x6
 
Last edited:
  • #7
1,074
1
UrbanXrisis said:
oh thanks. by the way, when I am given a dif. eq. and they ask me "What are the constant solutions of this equation? "

what exactly are do they want me to find?
Umm... I think they might mean for you to find the solutions that are just a constant function ie. y=c c is just some constant.
 
  • #8
1,074
1
ek said:
As said before, you're taking the wrong approach.

You need to divide it by x to get it into standard form or whatever it's called, and then get an integrating factor.

I get an answer of y = 5x/7 +c
You forgot to divide c by x6
 
  • #9
ek
181
0
d_leet said:
You forgot to divide c by x6
Ya. Actually I forgot to put in period and added it in haphazardly without thinking. I'm quite absent minded some times. I'll edit my post. I edited my last post.
 
Last edited:
  • #10
69
0
hmmm

[tex] x\frac{dy}{dx}+6y=5x[/tex]


[tex] \frac{dy}{dx}+\frac{6}{x}y=\frac{5}[/tex]

then your integrating factor is

[tex]e^{6\int\frac{dx}{x}}=x^6[/tex]

multiply through by integrating factor

[tex] x^6\frac{dy}{dx}+6x^5y=5x^6[/tex]

then integrate both sides

[tex]x^6y=\frac{5x^7}{7}+C [/tex]

Divide through by x^6

[tex]y=\frac{5x}{7}+Cx^{-6}[/tex]

do you see where your mistake is?

now solve for C

[tex] y(1)=2 [/tex]

[tex] 2=\frac{5}{7}1+C [/tex]

[tex] 2-\frac{5}{7}=C [/tex]

[tex] \frac{9}{7}=C [/tex]

so the specific solution is

[tex]y=\frac{5x}{7}+\frac{9}{7}x^{-6}[/tex]
 
Last edited:
  • #11
1,074
1
Valhalla said:
hmmm

[tex] x\frac{dy}{dx}+6y=5[/tex]
The term on the right hand side should be 5x not 5 so your answer is wrong as well.
 
  • #12
69
0
d_leet said:
The term on the right hand side should be 5x not 5 so your answer is wrong as well.
shizer your right

fixed
 
Last edited:

Related Threads on Dif. eq. problem - dont know why it is wrong

Replies
2
Views
1K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
1
Views
827
Replies
5
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
7
Views
589
Replies
3
Views
2K
Replies
8
Views
957
Replies
0
Views
2K
Top