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About LinkBacks4) Solve:
I don't even know what I'm supposed to do on this one...
please help...I need this for my exam
Since you are given an equation, I imagine you are supposed to find the general solution!Originally Posted by jangalinn
4) Solve:
I don't even know what I'm supposed to do on this one...
please help...I need this for my exam
There are two ways to do this:
1) Find an integrating factor. That is, find some function u(x) so that multiplying the equation by u(x) makes the left side an "exact differential"- so that. Using the product rule on the right side of that,
so our requirement on u(x) becomes
. That means we must have
or
, a separable equation for u. It "separates" as
which integrates to
. Taking exponentials of both sides,
. We don't need the general solution so take C'= 1,
.
Now= \frac{d(e^xy)}{dx}= xe^x)
integrating the left sides just giveswhile we can integrate the right side using "integration by parts": Let u= x,
so that du= dx and
.
. So
and
2) Separate into "homogeneous" and "non-homogeneous" parts. That is, first find the general solution to the "homogeneous" part,. That is the same as
which separates to
. Integrating both sides, ln(y)= -x+ C so
where
.
Now try to find a "specific", single, solution to. Since we are only trying to find a single solution rather than the general solution we can try "guessing". (educated guessing, of course!) Since the right hand side of the equation is "x", try y= Ax+ B. Then
and the equation becomes A+ Ax+ B= x. Since that must be true for all x, taking x= 0, we get A+ B= 0 or B= -A. Replacing B by -A gives Ax= x and, taking x= 1, A= 1 so B= -1. y(x)= x- 1 satisfies the differential equation
Finally, we add the general solution to the homogenous equation to the specific solution to the non-homogeneous equation to get the general solution to the entire equation:as before.
The second method is easier but requires more theory.
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