Author Topic: TUT0601 quiz 2  (Read 358 times)

xuanzhong

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TUT0601 quiz 2
« on: October 04, 2019, 07:25:01 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
\frac{x^2}{y^3}+x(1+y^2\ )\frac{dy}{dx} =0,\mu(x,y)=\frac{1}{xy^3}
$$
First, define $M(x,y)=\ x^2\ y^3$,$N(x,y)=x(1+y^2)$
$$
M_y=\partial y/\partial x\ (x^2\ y^3\ )=3x^2\ y^2
$$
$$
N_y=\partial y/\partial x[x(1+y^2\ )]=1+y^2
$$

Since $M_y\neq N_y$, this implies the given equation is not exact.
Now, multiplying both sides by the integrating factor $\mu(x,y)$, that is
$$
\frac{1}{xy^3}x^2y^3+\frac{1}{xy^3}x(1+y^2\ )\frac{dy}{dx}=x+(\frac{1}{y^3}+\frac{1}{y}) \frac{dy}{dx}=0
$$

Define $M^\prime\ (x,y)=\ x$,$N^\prime\ (x,y)=\frac{1}{y^3}+\frac{1}{y}$
$$
M_y=\partial y/\partial x\ (x)=0
$$
$$
N_y=\partial y/\partial x[\frac{1}{y^3}+\frac{1}{y}]=0
$$

Since $M_y$=$N_y$, this implies the given equation is exact.
Thus, we know there exists a function $\phi(x,y)$ such that $\partial\phi/\partial x=\ M^\prime\ (x,y)=\ x$
Then $\phi(x,y)=\int M^\prime\ dx=\frac{1}{2} x^2+h(y)$
Take derivative on both sides with respect to y we get
$$
\partial\phi/\partial y=h^\prime\ (y)=N^\prime\ (x,y)=\frac{1}{y^3}+\frac{1}{y}
$$

Integrating with respect to y we have
$$
h(y)=\frac{-1}{2}\frac{1}{y^2}+ln|y|+C
$$

Hence, $\phi(x,y)=\frac{1}{2} x^2-\frac{1}{2} \frac{1}{y^2}+ln|y|=C$ 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.


« Last Edit: October 11, 2019, 06:35:46 PM by xuanzhong »