Answer :
Answer:
The answer is " [tex]C= \frac{5}{2x^2}+ 4xy - 2y^4.[/tex]"
Step-by-step explanation:
It given that, [tex]\frac {\partial(5x + 4y)}{\partial y} = \frac{\partial(4x- 8y^3)} {\partial x} \\\\[/tex] it is the differential equation.
Solve the above equation.
[tex]\frac {\partial M (x, y)}{\partial y} = (5x+ 4y)[/tex]
So, [tex]M (x, y) = \frac{5}{2x^2}+4xy +\varphi (y).[/tex]
[tex]\frac{\partial M (x, y)}{ \partial y }= 4x- 8y^3 \ and \ \ 4x + \varphi ' (y) = 4x- 8y^3 \\\\\ so, \ \\varphi ' (y) = -8y^3[/tex]
or [tex]\varphi (y) = -2y^4+c[/tex]
To find the general solution
[tex]M(x, y )= c \\C= \frac{5}{2x^2}+ 4xy - 2y^4.[/tex]