integrating factor or d.e

**yen yen** · June 6th 2010, 10:33 PM

when im doing questions that use integrating factors i always face the same problem.

eg.
dy/dx + P(x)y = Q(x)
and the integrating factor, I, would be e^(integral of P)

however after multiplying the whole equation by I, you would normally get something like

Iy' + IP(x)y = IQ(x)

and the next step would be combining the left hand side into (###)' so that you could equate ### to integral of the right hand side. what i want to ask is how do you get from Iy' + IP(x)y to first derivative of ###?

and example is
dx/dt + 4x/(100-t) = 3/2

im not sure if its correct but i got
P = 4/100-t
=> I = (100-t)^-4 (correct this if im wrong)

then the equation becomes

x'(100-t)^-4 + 4x(100-t)^-5 = 3/2(100-t)^-4

what do i do from here? i always get stuck at this step for all problems using integrating factors.

**TheEmptySet** · June 6th 2010, 10:53 PM

Originally Posted by yen yen

when im doing questions that use integrating factors i always face the same problem.

eg.
dy/dx + P(x)y = Q(x)
and the integrating factor, I, would be e^(integral of P)

however after multiplying the whole equation by I, you would normally get something like

Iy' + IP(x)y = IQ(x)

and the next step would be combining the left hand side into (###)' so that you could equate ### to integral of the right hand side. what i want to ask is how do you get from Iy' + IP(x)y to first derivative of ###?

and example is
dx/dt + 4x/(100-t) = 3/2

im not sure if its correct but i got
P = 4/100-t
=> I = (100-t)^4 (correct this if im wrong)

then the equation becomes

x'(100-t)^4 + 4x(100-t)^3 = 3/2(100-t)^4

what do i do from here? i always get stuck at this step for all problems using integrating factors.

I think you have an error in your integrating factor

$exp(\int Pdt)=exp(\int\frac{4}{100-t}dt)=exp(-4\ln(100-t))=\frac{1}{(100-t)^4}$

**yen yen** · June 6th 2010, 11:08 PM

Originally Posted by TheEmptySet

I think you have an error in your integrating factor

$exp(\int Pdt)=exp(\int\frac{4}{100-t}dt)=exp(-4\ln(100-t))=\frac{1}{(100-t)^4}$

yea i corrected it.. but i still dont know what the next step is supposed to be..

never mind. i figured out how to get it. i only realize now thats what the integrating factor is for.

**TheEmptySet** · June 6th 2010, 11:14 PM

Originally Posted by yen yen

yea i corrected it.. but i still dont know what the next step is supposed to be..

Now multiply the equation by the integrating factor to get

$\frac{1}{(100-t)^4}\frac{dx}{dt}+\frac{4}{(100-t)^5}x=\frac{3}{2}\frac{4}{(100-t)^4}$

Now the left hand side is the derivative of a product and can be written as

$\frac{d}{dt}\left( \frac{1}{(100-t)^4}x\right)=\frac{3}{2}\frac{4}{(100-t)^4}$

Now just integrate both sides with respect to t to solve for x.

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