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Messages - Xuefen luo

Pages: 1 [2]
Quiz-3 / TUT0301
« on: October 11, 2019, 02:14:55 PM »
Dividing both sides by $x^2$, then we have:
$y''+ \frac{1}{x}y'+\frac{(x^2-v^2)}{x^2}y=0$
Since $W=ce^{\int -p(x)dx}$, and $p(x)=\frac{1}{x}$ in this case, we have:
$W=ce^{-\int \frac{1}{x}dx}=ce^{-ln(x)+C}=ce^{ln(x^{-1})+C}=cx^{-1}e^C$
We know $e^C$ is just a constant, so we can just subsume it into $c$. Then the Wronskian is $W=\frac{c}{x}$

Quiz-2 / TUT0301
« on: October 04, 2019, 02:01:48 PM »
$1+(\frac{x}{y} - sin(y))y' = 0$
Let $M(x,y)=1$ and $N(x,y)=(\frac{x}{y} - sin(y))$
Then, $M_y = 0$ and $N_x = \frac{1}{y}$
We can see that this equation is not exact.
However, we know $\frac{N_x-M_y}{M}=\frac{1}{y}$, and $\mu =y$
Multiplying the original equation by $\mu(y)$, we have
Since $M_y=1$ and $N_x=1$, this equation is exact.
Thus, there exists a function $\psi(x,y)$ such that
$\psi_x(x,y)=y$ and $\psi_y(x,y)=x-ysin(y)$
Hence, $\psi(x,y) = \int y dx = xy + h(y)$
then $\psi_y(x,y)= x+h'(y) = x-ysin(y)$
We have $h'(y)=-ysin(y) \Rightarrow h(y)=- \int ysin(y)dy = ycos(y)-sin(y)$
Therefore, $\psi(x,y)=xy+ycos(y)-sin(y)$, and the solutions of the equation are given implicitly by $xy+ycos(y)-sin(y)=C$

Pages: 1 [2]