solution of the Differential equation
7. (1+y² + xy²)dx + (x²y + y + 2xy)dy =0
8. 2ydx + x(x² Iny-1)dy = 0
9. y(2x+y²)dx + x(y²-x)dy = 0
10.(3x²-2y2)y' = 2xy; x = 1, y = 1
11.(2x-3y)dx + (2y-3x)dy; x = 1, y = 1
The first consideration is to determine if it is an exact DE.
d_/dy_ (1+y^2 +xy^2) = 2y + 2xy
d_/dx_ (x^2*y + y + 2xy) = 2xy + 2y. The two are equal, therefore, the Differential equation is exact.
The solution (known as the potential function) F(x,y) is derived from
F(x,y) = INTEG {1 + y^2 + xy^2 } dx + g(y) = x + x*y^2 + x^2* y^2 /2 + g(y) and
F(x,y) = INTEG{ x^2* y + y + 2xy } dy = x^ 2* y^2 /2 + y^2/ 2 + x*y^2 + h(x).
On comparison of the two, g(y) = y^2/ 2 and h(x) = x.
Therefore, the solution is :
F(x,y) = C →
x^ 2* y^2 /2 + y^2/ 2 + x*y^2 + x = C
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