# Math Help - integrating factors

1. ## integrating factors

Hi i was wondering if anyone could help me with integrating factors.
y'+1/x(y)=x is the question
Hence the intergrating factor is e^the integral of 1/x. this gives e^lnx +A. which leaves the integratig factor as x. Hence x(y'+1/x(y))=x^2
then xy'+y=x^2. However after this point i am not sure where to go. the example tells me to go to dy/dx(xy)=x^2. But i am not sure why to do that. If anybody could help it would be appreciated and if anybody has any more tips on integrating factors that would be nice! thanks

2. Originally Posted by studentsteve1202
Hi i was wondering if anyone could help me with integrating factors.
y'+1/x(y)=x is the question
Hence the intergrating factor is e^the integral of 1/x. this gives e^lnx +A. which leaves the integratig factor as x. Hence x(y'+1/x(y))=x^2
then xy'+y=x^2. However after this point i am not sure where to go. the example tells me to go to dy/dx(xy)=x^2. But i am not sure why to do that. If anybody could help it would be appreciated and if anybody has any more tips on integrating factors that would be nice! thanks
the whole point of using integrating factors is to turn the left hand side into the result of a product rule derivative. you will notice that xy' + y is the result you get if you differentiate the product xy implicitly. this is the idea. you can therefore go backwards and contract the expression on the right to the original form before it was differentiated and just take the anti-derivative of both sides.

see post #21 here

3. Hello, studentsteve1202!

$y'+\frac{1}{x}\,y\:=\:x$

Hence, the intergrating factor is: $e^{\int\frac{dx}{x}}$
This gives: $e^{\ln x} \:=\:x$

Hence: . $xy' +y\:=\:x^2$

However, after this point i am not sure where to go.

When you multiplied through by the integrating factor,
. . the left side became: . $xy' + y$
which is the derivative of a product . . . namely, the derivative of $xy$
. .
(Check it out for yourself!)

The left side is always the derivative of: . $\text{(Integrating factor)} \times y$

And that's why we can write: . $\frac{d}{dx}(xy) \:=\:x^2\quad\Rightarrow\quad d(xy) \:=\:x^2\,dx$

Now integrate both sides: . $\int d(xy)\;=\;\int x^2\,dx$

. . . . . . . . . .and we get: . . . $xy \;\;\;=\;\;\;\frac{1}{3}x^3 + C$

Therefore: . $y \;=\;\frac{1}{3}x^2 + Cx^{-1}$