eExactDFQ

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.

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

Now we write this using total differential.

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

From this, we have

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

Now we apply the initial condition $y(0) = 1$

. Then $0+0+\frac{1}{3}+3 = c$ . Thus, $c = \frac{4}{3}$ . Therefore,

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

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