Example Solve the following initial value problem. $(2x + 3y)dx + (3x + y^2 + 3)dy = 0, y(0) = 1$

Answer Note that $M_{y} = 3 = N_{x}$. Then the differential equation is exact. Now we write $M(x,y),N(x,y)$ in the following way.

$$\underbrace{2x dx}_{\mbox{function of } x} + \underbrace{(3ydx + ... ...,y \mbox{mix functions}} + \underbrace{(y^2 + 3)dy}_{\mbox{function of} y} = 0$$

Now we write this using total differential.

$$d(x^2) + d(3xy) + d(\frac{y^3}{3} + 3y) = d(c) .$$

From this, we have

$$x^2 + 3xy + \frac{y^3}{3} + 3y = c .$$

Now we apply the initial condition $y(0) = 1$. Then $0+0+\frac{1}{3}+3 = c$. Thus, $c = \frac{4}{3}$. Therefore,

$$x^2 + 3xy + \frac{y^3}{3} + 3y = \frac{4}{3}$$