Ordinary DE using integrating factor?

In summary, the student attempted to solve an initial condition problem but made mistakes. When substituting the initial condition into the equation, they found that C=-22e^{16}.
  • #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 I am going alright so far... and if i am how do you integrate
[tex] \int (8e^4x/x) [/tex]
 
Last edited:
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  • #2
cheddacheeze said:
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
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
gabbagabbahey said:
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 doesn't seem to be the right answer
 
  • #7
cheddacheeze said:
[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
gabbagabbahey said:
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
11*4=44, not 22:wink:
 
  • #11
cheddacheeze said:
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
Okay, that looks correct then:approve:
 
  • #13
ok i must have done algebra somewhere incorrect then, can't get the right answer out
 
  • #14
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 don't know since its a computer entered assignment
 
  • #16
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 that's right
 
  • #18
didnt think ordinary differential equations could be so annoying
haha...
 
  • #19
cheddacheeze said:
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 that's 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
gabbagabbahey said:
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 :)
 
Last edited:

What is Ordinary DE using integrating factor?

Ordinary differential equations (ODEs) are equations that involve derivatives of a function with respect to a single independent variable. Integrating factor is a technique used to solve certain types of ODEs.

What are the steps involved in solving an ODE using integrating factor?

The steps involved in solving an ODE using integrating factor are as follows:
1. Identify the ODE as a first order linear equation.
2. Write the equation in the standard form: y' + p(x)y = g(x).
3. Find the integrating factor, which is the exponential of the integral of p(x).
4. Multiply both sides of the equation by the integrating factor.
5. Use the product rule to simplify the left side of the equation.
6. Integrate both sides of the equation and solve for y.

What are the advantages of using integrating factor to solve ODEs?

Integrating factor allows for the solving of certain types of ODEs that cannot be solved using other techniques. It also provides a systematic approach to solving linear ODEs.

Can all ODEs be solved using integrating factor?

No, integrating factor can only be used to solve first order linear ODEs. It cannot be used to solve higher order ODEs or nonlinear ODEs.

How can I check my solution when using integrating factor to solve an ODE?

To check your solution, plug it back into the original equation and see if it satisfies the equation. Additionally, you can take the derivative of your solution and see if it matches the right side of the equation.

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