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 :)