Question

Solve 6x^2*y^2*dx+4x^3*y*dy=0.

Answer 1

@phi

Answer 2

Do you know how to solve this exact equation?

Answer 3

I would factor out 2 x^2 y from each term

Answer 4

@SithsAndGiggles

Answer 5

are you saying you can't factor out 2 x^2 y ?
to get
3 y dx + 2 x dy = 0
which is separable into
\[ - \frac{3}{2} \frac{dx}{x} = \frac{dy}{y} \]

Answer 6

But the answer in the book says 2x^3*y^2=C and I did it your way by making it separable and I didn't get the same answer in the book. Is the book's answer correct?

Answer 7

take the square root of both sides
\[ \sqrt{2} x^\frac{3}{2} y = \sqrt{C} \\
y = \frac{\sqrt{C}}{\sqrt{2} x^\frac{3}{2} } \\
y = C' x^{-\frac{3}{2} }
\]

Answer 8

\[ - \frac{3}{2}\int \frac{dx}{x} = \int \frac{dy}{y} \\
- \frac{3}{2}\ln(x) + C = \ln(y) \\
\ln(y) = \ln(x^{- \frac{3}{2}}+C )
\]
make each side the exponent of e
\[ y = e^C x^{- \frac{3}{2}} \\
y = C' x^{- \frac{3}{2}} \]

Answer 9

** paren put in the wrong spot for
\[ \ln(y) = \ln(x^{-\frac{3}{2}}) + C \]

Answer 10

Here is how they might want you to solve this.
if you have
P dx + Q dy
First, show that you have an exact differential
d/dy P = d/dx Q
P = 6x^2 y^2 d/dy P = 12 x^2 y
Q= 4 x^3 y d/dx Q = 12 x^2 y
they are equal so it's exact.

we now integrate using the formula
\[ f(x,y) = \int_{x_0}^x P(x,y) dx + \int_{y_0}^y Q(x_0,y) \ dy = c \]
where \( (x_0, y_0) \) is a point in the region where the integral exists.
In this case (0,0) works (and is the simplest point to use)
in particular, Q(0,y) is 12 0^2 y = 0. i.e. we only have one integral to do
\[ f(x,y) = \int_{0}^x 6x^2 y^2 dx = C\\
2 x^3 y^2 = C
\]
which is probably how the text wants you to solve the problem.

Answer 11

\[2x ^{3}y ^{2}+g(y)\]