Answer:
[tex]\mathbf{3x^2y^3+2xy^4=C}[/tex]
Step-by-step explanation:
From the differential equation given:
[tex]6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0[/tex]
The equation above can be re-written as:
[tex]6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0[/tex]
[tex](6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0[/tex]
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:
[tex]\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}[/tex]
relating with equation M(x,y)dx + N(x,y) dy = 0
Then;
[tex]M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3[/tex]
So;
[tex]\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3[/tex]
[tex]\dfrac{\partial N}{\partial y }[/tex]
Let's Integrate [tex]\dfrac{\partial F}{\partial x}= M(x,y)[/tex] with respect to x
Then;
[tex]F(x,y) = \int (6xy^3 +2y^4) \ dx[/tex]
[tex]F(x,y) = 3x^2 y^3 +2xy^4 +g(y)[/tex]
Now, we will have to differentiate the above equation with respect to y and set [tex]\dfrac{\partial F}{\partial x}= N(x,y)[/tex]; we have:
[tex]\dfrac{\partial F}{\partial y} = \dfrac{\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[/tex]
Hence, [tex]F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1[/tex]
Finally; the general solution to the equation is:
[tex]\mathbf{3x^2y^3+2xy^4=C}[/tex]
