Author Topic: TUT0601 Quiz2  (Read 3941 times)

Mingzhu Yu

  • Newbie
  • *
  • Posts: 3
  • Karma: 0
    • View Profile
TUT0601 Quiz2
« on: October 05, 2019, 02:05:06 AM »
find the value of b for which the given equation is exact, and then solve it using that value of b.
$$\therefore M(x,y)=ye^{2x}+x$$
since the differential equation to be exact
we get b=1, then put b=1 in the differential equation.
$$M=ye^{2xy}+x, N=xe^{2xy}$$
Since the differential equation is exact then there is a function $\varphi(x,y)$ such that $\varphi_x(x,y)=M(x,y)$ and $\varphi_y(x,y)=N(x,y)$
$$\because\varphi_y=N\qquad\therefore xe^{2xy}+h'(y)=xe^{2xy}$$
$$\therefore h'(y)=0\qquad\therefore h(y)=c$$
General solution  $\varphi=\dfrac{e^{2xy}}{2}+\dfrac{x^2}{2}=c$

« Last Edit: October 05, 2019, 02:34:30 AM by Mingzhu Yu »