Question

Find general solutions of the differential equation. Primes denote derivatives with respect to x.

6xy^3+2y^4+(9x^2y^2+8xy^3) yâ²=0

Answer 1

`\mathbf{3x^2y^3+2xy^4=C}`

Explanation:

From the differential equation given:

`6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0`

The equation above can be re-written as:

`6xy^3 +2y^4 +(9x^2y^2+8xy^3)(dy)/(dx)=0`

`(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0`

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

`\frac{{\partial M }}{{\partial y }}= (\partial N)/(\partial x)}} \ \ \text{at each point of R}`

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

`M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3`

So;

`(\partial M)/(\partial y )= 18xy^3 +8y^3`

`(\partial N)/(\partial y )`

Let's Integrate
`(\partial F)/(\partial x)= M(x,y)` with respect to x

Then;

`F(x,y) = \int (6xy^3 +2y^4) \ dx`

`F(x,y) = 3x^2 y^3 +2xy^4 +g(y)`

Now, we will have to differentiate the above equation with respect to y and set
`(\partial F)/(\partial x)= N(x,y)`; we have:

`(\partial F)/(\partial y) = (\partial)/(\partial y ) (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1`

Hence,
`F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1`

Finally; the general solution to the equation is:

`\mathbf{3x^2y^3+2xy^4=C}`