SOLUTION: Find the family of solutions of the following equation using integrating factors found by inspection. 1. 3x^2 y + (y^5 - x^3) y' = 0

Algebra ->  Finance -> SOLUTION: Find the family of solutions of the following equation using integrating factors found by inspection. 1. 3x^2 y + (y^5 - x^3) y' = 0      Log On


   



Question 1188306: Find the family of solutions of the following equation using integrating factors found by inspection.
1. 3x^2 y + (y^5 - x^3) y' = 0

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2y+%2B+%28y%5E5+-+x%5E3%29%2A%22y%27%22+=+0

3x%5E2y+%2B+%28y%5E5+-+x%5E3%29%2Aexpr%28dy%2Fdx%29+=+0

3x%5E2y+%2B+y%5E5%2Aexpr%28dy%2Fdx%29+-+x%5E3%2Aexpr%28dy%2Fdx%29+=+0

3x%5E2y%2Adx+%2B+y%5E5%2Ady+-+x%5E3%2Ady+=+0

The trick here is to get the left side into the expression of either the product expression of 
the differential of uv, which is:

u%2Adv+%2B+v%2Adu

or the quotient expression for the differential of u%2Fv, which is:

%28v%2Adu-u%2Adv%29%2Fv%5E2

We notice by inspection that the third term has x3 and the first term has 3x2dx, which is the 
differential of x3.  So we will get those two terms together and the other term
on the other side:

3x%5E2y%2Adx+-+x%5E3%2Ady+=+y%5E5%2Ady

The product rule isn't going to work because there's a MINUS sign between the terms, not a 
PLUS sign.  So, we think of trying to make the left side into the form 

%28v%2Adu-u%2Adv%29%2Fv%5E2, the differential form for u%2Fv

The du can be the 3x2dx (with u as x3), and the v can be the y. So we'll write the y first 
in the first term:  

y%2A3x%5E2%2Adx+-+x%5E3%2Ady+=+y%5E5%2Ady

Now on the left we have the numerator of the differential form for x%5E3%2Fy.

Now all we need to make the left side into the quotient differential form of x%5E3%2Fy%5E%22%22 is to 
divide both sides of the equation through by y2:

%28y%2A3x%5E2%2Adx+-+x%5E3%2Ady%29%2Fy%5E2+=+expr%28y%5E5%2Fy%5E2%29%2Ady

So the integrating factor used here is 1%2Fy%5E2.

%28y%2A3x%5E2%2Adx+-+x%5E3%2Ady%29%2Fy%5E2%22%22=%22%22y%5E3%2Ady

Now we can integrate both sides of the equation:

int%28%28y%2A3x%5E2%2Adx+-+x%5E3%2Ady%29%2Fy%5E2%29%22%22=%22%22int%28y%5E3%2Ady%29

The whole left side integrates all together as the quotient x%5E3%2Fy, and, the right side 
integrates as y%5E4%2F4 and we must add an arbitrary constant:

x%5E3%2Fy%22%22=%22%22y%5E4%2F4%22%22%2B%22%22c%5B1%5D

Multiply through by 4y

4x%5E3%22%22=%22%22y%5E5%22%22%2B%22%22Cy

[We just write C because 4 times an arbitrary constant is just another arbitrary
constant].

Edwin