Ordinary DE using integrating factor?

[cheddacheeze]

Homework Statement

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

Homework Equations

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

 

[gabbagabbahey]

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

Careful, $e^{a+b}=e^a e^b \neq e^a+e^b$

 

[cheddacheeze]

so instead of
$$x+e^(4x)$$ its
$$xe^(4x)$$?

 

[gabbagabbahey]

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}.

 

[cheddacheeze]

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

 

[cheddacheeze]

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 doesn't seem to be the right answer

 

[gabbagabbahey]

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

Shouldn't you also divide $C$ by $xe^{4x}$ here?

 

[cheddacheeze]

Shouldn't you also divide $C$ by $xe^{4x}$ here?

hmmmm tried doing that before but still didnt get the right answer

 

[cheddacheeze]

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...

 

[gabbagabbahey]

11*4=44, not 22

 

[cheddacheeze]

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]

Okay, that looks correct then

 

[cheddacheeze]

ok i must have done algebra somewhere incorrect then, can't get the right answer out

 

[gabbagabbahey]

What was your final answer, and what is the answer "supposed" to be according to your assignment sheet/text?

 

[cheddacheeze]

my answer is $$y(x) = -2/x -22e^{-16}$$
as for the solution i don't know since its a computer entered assignment

 

[gabbagabbahey]

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

 

[cheddacheeze]

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

 

[cheddacheeze]

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

 

[gabbagabbahey]

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

 

[cheddacheeze]

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 :)