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Ordinary DE using integrating factor?

  • #1

Homework Statement


solve the following initial condition problem
[tex]x (d/dx) y(x) + 4xy(x) = -8-y(x) [/tex]
[tex]y(4)=-6 [/tex]

Homework Equations





The Attempt at a Solution


first i rearranged

[tex]xy'+y+4xy=-8 [/tex]
[tex]xy'+y(1+4x)=-8 [/tex]
[tex]y'+y(1/x+4)=-8/x[/tex]

integrating factor:[tex]e^\int(1/x+4)[/tex]
[tex]e^(lnx+4x)[/tex]
[tex]x+e^4x[/tex]

multiplying integrating factor:
[tex]d/dx(x+e^4x) y'+(1/x+4)(x+e^4x)y=-8/x(x+e^4x)[/tex]
[tex]d/dx(x+e^4x])y=-8/x(x+e^4x)[/tex]
[tex]\int d/dx(x+e^4x])y=\int-8/x(x+e^4x)[/tex]
[tex](x+e^4x)y=\int(-8-8e^4x/x)[/tex]

just wondering if im going alright so far... and if i am how do you integrate
[tex] \int (8e^4x/x) [/tex]
 
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Answers and Replies

  • #2
gabbagabbahey
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integrating factor:[tex]e^\int(1/x+4)[/tex]
[tex]e^(lnx+4x)[/tex]
[tex]x+e^4x[/tex]
Careful, [itex]e^{a+b}=e^a e^b \neq e^a+e^b[/itex]:wink:
 
  • #3
so instead of
[tex]x+e^(4x)[/tex] its
[tex]xe^(4x)[/tex]?
 
  • #4
gabbagabbahey
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Yup.:smile:

On a side note, to get exponents with more than one character to display properly in [itex]\LaTeX[/itex], simply enclose them in curly brackets. For example, [itex]xe^{4x}[/itex] is generated using xe^{4x}.
 
  • #5
Yup.:smile:

On a side note, to get exponents with more than one character to display properly in [itex]\LaTeX[/itex], simply enclose them in curly brackets. For example, [itex]xe^{4x}[/itex] is generated using xe^{4x}.
haha cheers for both advices, this should make things easier
 
  • #6
ok after using:
[tex]xe^{4x}[/tex] as an integrating factor
i got
[tex]xe^{4x} y(x)=\int-8e^{4x}[/tex]
[tex]xe^{4x} y(x)= -2e^{4x} +C [/tex]
[tex]y(x)=-2e^{4x}/xe^{4x} + C [/tex]
[tex]y(x)=-2/x +C [/tex]

using initial conditions:
[tex]y(4) = -2/4+C = -6 [/tex]
[tex]c=-11/2[/tex]
[tex]y(x)=-2/x-11/2[/tex]

and it doesnt seem to be the right answer
 
  • #7
gabbagabbahey
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[tex]xe^{4x} y(x)= -2e^{4x} +C [/tex]
[tex]y(x)=-2e^{4x}/xe^{4x} + C [/tex]
Shouldn't you also divide [itex]C[/itex] by [itex]xe^{4x}[/itex] here? :wink:
 
  • #8
Shouldn't you also divide [itex]C[/itex] by [itex]xe^{4x}[/itex] here? :wink:
hmmmm tried doing that before but still didnt get the right answer
 
  • #9
using:

[tex]y(x) = -2/x + C/(xe^{4x})[/tex]

and by substituing initial conditions i got:

[tex]y(4)=-2/4+C/(4e^{16})=-6[/tex]
[tex]-11/2=C/(4e^{16})[/tex]
[tex]C=(-22e^16)/2[/tex]

is this right or did i miss something...
 
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  • #10
gabbagabbahey
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11*4=44, not 22:wink:
 
  • #11
using:

[tex]y(x) = -2/x + C/(xe^{4x})[/tex]

and by substituing initial conditions i got:

[tex]y(4)=-2/4+C/(4e^{16})=-6[/tex]
[tex]-11/2=C/(4e^{16})[/tex]
[tex]C=(-22e^16)/2[/tex]

is this right or did i miss something...
sorry there wasnt supposed to be
[tex]C=(-22e^16)/2[/tex]
divide 2 there, i just simplified and got
[tex]C=(-22e^16)[/tex]
 
  • #12
gabbagabbahey
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Okay, that looks correct then:approve:
 
  • #13
ok i must have done algebra somewhere incorrect then, cant get the right answer out
 
  • #14
gabbagabbahey
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What was your final answer, and what is the answer "supposed" to be according to your assignment sheet/text?
 
  • #15
my answer is [tex] y(x) = -2/x -22e^{-16} [/tex]
as for the solution i dont know since its a computer entered assignment
 
  • #16
gabbagabbahey
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What happened to the [itex]xe^{4x}[/itex] in [itex]y(x) = -2/x + C/(xe^{4x})[/itex]?
 
  • #17
i substituted the initial conditions into the equation and found what the value of C was and entered my answer as that, maybe i'll try keeping the constant and see if thats right
 
  • #18
didnt think ordinary differential equations could be so annoying
haha...
 
  • #19
gabbagabbahey
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i substituted the initial conditions into the equation and found what the value of C was and entered my answer as that, maybe i'll try keeping the constant and see if thats right
You had [itex]y(x) = -2/x + C/(xe^{4x})[/itex] as your general solution, substituting your initial condition gave you [itex]C=-22e^{16}[/itex], so your particular solution should be

[tex]y(x)=-\frac{2}{x}-\frac{22e^{16}}{xe^{4x}}[/tex]

Not what you had in post #15. You need to pay more attention to basic algebra steps like substitution.
 
  • #20
You had [itex]y(x) = -2/x + C/(xe^{4x})[/itex] as your general solution, substituting your initial condition gave you [itex]C=-22e^{16}[/itex], so your particular solution should be

[tex]y(x)=-\frac{2}{x}-\frac{22e^{16}}{xe^{4x}}[/tex]

Not what you had in post #15. You need to pay more attention to basic algebra steps like substitution.
ohhhh... C was over [tex] xe^{4x} [/tex]
keep making stupid mistakes... looks like i need to look at things with more care
thanks a lot :)
 
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