### Author Topic: TUT0602 QUIZ 2  (Read 508 times)

#### Fenglun Wu

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##### TUT0602 QUIZ 2
« 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
$$h(y) = -\frac{1}{2}y^{-2} + ln|y| + C$$
Therefore, we have
$$\phi (x, y) = \frac{1}{2}x^{2} - \frac{1}{2}y^{-2} + ln|y| = C$$
is the general solution to the given differential equation.