### Author Topic: quiz2 0502  (Read 901 times)

#### yueyangyu

• Jr. Member
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• Karma: 0 ##### quiz2 0502
« on: October 04, 2019, 02:52:24 PM »
Find an integrating factor and solve the given equation:
$$(3x+\frac{6}{y})+(\frac{x^{2}}{y}+3\frac{y}{x})\frac{dy}{dx}=0$$

Firstly, find an integrating factor $\mu$ as a function of xy s.t.
$$(\mu M)_y=(\mu N)_x$$
Let $$z=xy$$.
$$\mu M_y+xM\frac{d\mu}{dz}=\mu N_x+yN\frac{d\mu}{dz}$$

$$\frac{d\mu}{dz}=\mu(\frac{N_x-M_y}{xM-yN})$$
Therefore,
$$\mu(z)=exp(\int(R(z))dz)$$
where $$R(z)=R(xy)=\frac{N_x-M_y}{xM-yN}$$
Let $$M=3x+\frac{6}{y}$$ $$N=\frac{x^{2}}{y}+3\frac{y}{x}$$
Then,
$$M_y=\frac{-6}{y^{2}} \quad N_x=\frac{2x}{y}-\frac{37}{x^{2}}$$
We can see that this equation is not exact
$$\frac{N_x-M_y}{xM-yN}=\frac{\frac{2x}{y}-\frac{3y}{x^{2}}+\frac{6}{y^{2}}}{2x^{2}+\frac{6x}{y}-\frac{3y^{2}}{x}}=\frac{1}{xy}$$
Thus, we have an integrating factor
$$\mu(xy)=exp(\int\frac{1}{z}dz)=z=xy$$
Multiplying the original equation by the integrating factor, we have
$$(3x^{2}y+6x)+(x^{3}+3y{2})\frac{dy}{dx}=0$$
This equation is exact because
$$M_y=N_x=3x^{2}$$
Thus, there exists a function $$\Psi(x,y)$$ such that $$\Psi_x(x,y)=3x^{2}y+6x$$
$$\Psi_y(x,y)=x^{3}+3y{2}$$
$$\Psi(x,y)=x^{3}y+3x^{2}+h(y)$$
Differentiating with respect to y, we get
$$\Psi_y(x,y)=x^{3}+h'(y)$$
$$h'(y)=3y^{2} \quad h(y)=y^{3}$$
and we have
$$\Psi(x,y)=x^{3}y+3x^{2}+y^{3}$$
Thus the solutions of the differential equation are given implicitly by
$$x^{3}y+3x^{2}+y^{3}=C$$