### Author Topic: TUT0801 Quiz2  (Read 503 times)

#### bella

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##### TUT0801 Quiz2
« on: October 04, 2019, 02:04:47 PM »
$$x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3}$$

First, let's show the given DE $x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0$ is not exact.

Define $M(x, y)=x^{2} y^{3}, N(x, y)=x\left(1+y^{2}\right)$

$$M_{y}=\frac{\partial}{\partial y}\left[x^{2} y^{3}\right]=3 x^{2} y^{2}$$

$$N_{x}=\frac{\partial}{\partial x}\left[x\left(1+y^{2}\right)\right]=1+y^{2}$$

Since $3 x^{2} y^{2} \neq 1+y^{2}$, this implies the given DE is not exact.

Now, let's show that the given DE multipled by the integrating factor $\mu(x, y)=\frac{1}{x y^{3}}$ is exact.

That is to show

$$\frac{1}{x y^{3}} x^{2} y^{3}+\frac{1}{x y^{3}} x\left(1+y^{2}\right) y^{\prime}=x+\left(y^{-3}+y^{-1}\right) y^{\prime}=0$$

is exact.

Define $M^{\prime}(x, y)=x, N^{\prime}(x, y)=y^{-3}+y^{-1}$

Since

$$M_{y}^{\prime}=\frac{\partial}{\partial y}(x)=0$$

$$N_{x}^{\prime}=\frac{\partial}{\partial x}\left[y^{-3}+y^{-1}\right]=0$$

By theorem in the book, we can conclude that $x+\left(y^{-3}+y^{-1}\right) y^{\prime}=0$ is exact.

Thus, we know there exists a function $\phi(x, y)=C$ which satisfies the given DE.

Also,

$$\frac{\partial \phi}{\partial x}=x$$

$$\frac{\partial \phi}{\partial y}=y^{-3}+y^{-1}$$

Integrate $\frac{\partial \phi}{\partial x}=x$ with respect to $x$ we have

$$\phi(x, y)=\frac{1}{2} x^{2}+g(y)$$

Take derivative on both sides with respect to $y$ we get

$$\frac{\partial \phi}{\partial y}=g^{\prime}(y)$$

Since we know that $\frac{\partial \phi}{\partial y}=y^{-3}+y^{-1}$

Then $g^{\prime}(y)=y^{-3}+y^{-1}$

Integrate with respect to $y$ we have

$g(y)=-\frac{1}{2} y^{-2}+\ln |y|+C$

Altogether, we have $\phi(x, y)=\frac{1}{2} x^{2}-\frac{1}{2} y^{-2}+\ln |y|=C$, which means

$$\frac{1}{2} x^{2}-\frac{1}{2} y^{-2}+\ln |y|=C$$

is the general solution to the given DE.

Besides, notice that the constant function $y(x)=0 \forall x$ is also a solution to the given DE.