# Ordinary DE using integrating factor?

## Homework Statement

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

## The Attempt at a Solution

first i rearranged

$$xy'+y+4xy=-8$$
$$xy'+y(1+4x)=-8$$
$$y'+y(1/x+4)=-8/x$$

integrating factor:$$e^\int(1/x+4)$$
$$e^(lnx+4x)$$
$$x+e^4x$$

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

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

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gabbagabbahey
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integrating factor:$$e^\int(1/x+4)$$
$$e^(lnx+4x)$$
$$x+e^4x$$
Careful, $e^{a+b}=e^a e^b \neq e^a+e^b$

$$x+e^(4x)$$ its
$$xe^(4x)$$?

gabbagabbahey
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Yup.

On a side note, to get exponents with more than one character to display properly in $\LaTeX$, simply enclose them in curly brackets. For example, $xe^{4x}$ is generated using xe^{4x}.

Yup.

On a side note, to get exponents with more than one character to display properly in $\LaTeX$, simply enclose them in curly brackets. For example, $xe^{4x}$ is generated using xe^{4x}.
haha cheers for both advices, this should make things easier

ok after using:
$$xe^{4x}$$ as an integrating factor
i got
$$xe^{4x} y(x)=\int-8e^{4x}$$
$$xe^{4x} y(x)= -2e^{4x} +C$$
$$y(x)=-2e^{4x}/xe^{4x} + C$$
$$y(x)=-2/x +C$$

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

and it doesnt seem to be the right answer

gabbagabbahey
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$$xe^{4x} y(x)= -2e^{4x} +C$$
$$y(x)=-2e^{4x}/xe^{4x} + C$$
Shouldn't you also divide $C$ by $xe^{4x}$ here?

Shouldn't you also divide $C$ by $xe^{4x}$ here?
hmmmm tried doing that before but still didnt get the right answer

using:

$$y(x) = -2/x + C/(xe^{4x})$$

and by substituing initial conditions i got:

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

is this right or did i miss something...

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gabbagabbahey
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11*4=44, not 22

using:

$$y(x) = -2/x + C/(xe^{4x})$$

and by substituing initial conditions i got:

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

is this right or did i miss something...
sorry there wasnt supposed to be
$$C=(-22e^16)/2$$
divide 2 there, i just simplified and got
$$C=(-22e^16)$$

gabbagabbahey
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Okay, that looks correct then

ok i must have done algebra somewhere incorrect then, cant get the right answer out

gabbagabbahey
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my answer is $$y(x) = -2/x -22e^{-16}$$
as for the solution i dont know since its a computer entered assignment

gabbagabbahey
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What happened to the $xe^{4x}$ in $y(x) = -2/x + C/(xe^{4x})$?

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

didnt think ordinary differential equations could be so annoying
haha...

gabbagabbahey
Homework Helper
Gold Member
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 $y(x) = -2/x + C/(xe^{4x})$ as your general solution, substituting your initial condition gave you $C=-22e^{16}$, so your particular solution should be

$$y(x)=-\frac{2}{x}-\frac{22e^{16}}{xe^{4x}}$$

Not what you had in post #15. You need to pay more attention to basic algebra steps like substitution.

You had $y(x) = -2/x + C/(xe^{4x})$ as your general solution, substituting your initial condition gave you $C=-22e^{16}$, so your particular solution should be

$$y(x)=-\frac{2}{x}-\frac{22e^{16}}{xe^{4x}}$$

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

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