Find the general solution to the first order differential equation: xy2dy - (x^3+y^3)dx = 0?

Question

This is an AP calculus question... I will be so grateful if you can help me solve this question. PLEASE AND THANK YOU!
Is the solution...
a) 3x^4 + 8xy^3 = C
b) y^3 = lnCx
c) y^3 -x^3 + y^3lnx^3 = C
d) y^3 = 3x^3lnCx
e) None of these

Answer 1

xy² dy - (x³+ y³) dx = 0

First, we divide by dx to turn this into a proper differential equation:

xy²y' - x³ - y³ = 0

Now, we make the substitution u=y³, u'=3y²y'. So this becomes:

xu'/3 - x³ - u = 0

This is a linear equation in u. So first move x³ to the other side, and multiply by 3/x:

u' - 3u/x = 3x²

Now, we find an integrating factor of the form e^(∫-3/x dx) = e^(-3 ln |x|) = x^(-3). Multiplying by x^(-3):

u'/x³ - 3u/x⁴ = 3/x

Now this is an equation of the form:

(u/x³)' = 3/x

Integrating both sides:

u/x³ = 3 ln |x| + C

Multiply by x³:

u=3x³ ln |x| + Cx³

Finally, since u=y³:

y³ = 3x³ ln |x| + Cx³

Or letting K=e^C/3:

y³ = 3x³ ln |x| + 3x³ ln |K|

Which using the laws of logarithms:

y³ = 3x³ ln |Kx|

Which is option D.

Answer 2

Like whitesox stated, there comes a element the place you in basic terms can no longer get a closed variety answer. The (3x^2)^(x^2) is fairly difficult to combine. yet the place did maximum of those folk come from. you haven't any longer any contacts.

Answer 3

is it xy^2dy or 2xydy for the first term?