# Homework Help: Ordinary DE using integrating factor?

1. Apr 2, 2010

### cheddacheeze

1. The problem statement, all variables and given/known data
solve the following initial condition problem
$$x (d/dx) y(x) + 4xy(x) = -8-y(x)$$
$$y(4)=-6$$

2. Relevant equations

3. 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
$$\int (8e^4x/x)$$

Last edited: Apr 2, 2010
2. Apr 2, 2010

### gabbagabbahey

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

3. Apr 2, 2010

### cheddacheeze

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

4. Apr 2, 2010

### 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}.

5. Apr 2, 2010

### cheddacheeze

haha cheers for both advices, this should make things easier

6. Apr 2, 2010

### 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 doesnt seem to be the right answer

7. Apr 2, 2010

### gabbagabbahey

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

8. Apr 2, 2010

### cheddacheeze

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

9. Apr 2, 2010

### 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...

Last edited: Apr 2, 2010
10. Apr 2, 2010

### gabbagabbahey

11*4=44, not 22

11. Apr 2, 2010

### cheddacheeze

sorry there wasnt supposed to be
$$C=(-22e^16)/2$$
divide 2 there, i just simplified and got
$$C=(-22e^16)$$

12. Apr 2, 2010

### gabbagabbahey

Okay, that looks correct then

13. Apr 3, 2010

### cheddacheeze

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

14. Apr 3, 2010

### gabbagabbahey

15. Apr 3, 2010

### cheddacheeze

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

16. Apr 3, 2010

### gabbagabbahey

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

17. Apr 3, 2010

### 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 thats right

18. Apr 3, 2010

### cheddacheeze

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

19. Apr 3, 2010

### gabbagabbahey

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.

20. Apr 3, 2010

### cheddacheeze

ohhhh... C was over $$xe^{4x}$$
keep making stupid mistakes... looks like i need to look at things with more care
thanks a lot :)

Last edited: Apr 3, 2010