Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz2 => Topic started by: Fenglun Wu on October 04, 2019, 01:36:38 PM

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$x^2y^3 + x(1+y^2)y' = 0, ~~~~ \mu(x, y) = \frac{1}{xy^3}$$
First, let's show the given equation isn not exact.
Define $M(x, y) = x^2y^3, ~~ N(x, y) = x(1+y^2)$
$$M_y = \frac{\partial}{\partial y}[x^2y^3] = 3x^2y^2$$
$$N_x = \frac{\partial}{\partial x}[x(1+y^2)] = 1 + y^2$$
Since $M_y \neq N_x$, this implies that the given equation is not exact.
Next, show that the given equation multiplied by the integrating factor $\mu(x, y) = \frac{1}{xy^3}$ is exact.
The new equation becomes
$$ \frac{1}{xy^3}x^2y^3 + \frac{1}{xy^3}x(1+y^2)y' = x + (y^{3} + y^{1})y' =0 $$
Define $M'(x, y) = x, ~~ N'(x, y) = y^{3} + y^{1}$
$$M'_y = \frac{\partial}{\partial y}(x) = 0$$
$$N'_x = \frac{\partial}{\partial x}(y^{3} + y^{1}) = 0$$
Since $M'_y = N'_x$, this implies that the given equation is exact.
Thus, we know that there exists a function $\phi (x, y) = C$ which satisfies the give differential equation.
Also,
$$\frac{\partial \phi}{\partial x} = M'(x, y) = x$$
$$\frac{\partial \phi}{\partial y} = N'(x, y) = y^{3} + y^{1}$$
Integrate $\frac{\partial \phi}{\partial x} = x$ with respect to $x$ we have
$$\phi (x, y) = \frac{1}{2}x^2 + h(y)$$
Take derivative on both sides with respect to $y$ we get
$$\frac{\partial \phi}{\partial y} = h'(y)$$
Since we know $\frac{\partial \phi}{\partial y} = N'(x, y) = y^{3} + y^{1}$
Then $$ h'(y) = y^{3} + y^{1}$$
Integrate $h'(y)$ with respect to y we have
$$ h(y) = \frac{1}{2}y^{2} + lny + C$$
Therefore, we have
$$ \phi (x, y) = \frac{1}{2}x^{2}  \frac{1}{2}y^{2} + lny = C$$
is the general solution to the given differential equation.