### Author Topic: TUT0601 Quiz2  (Read 823 times)

#### Mingzhu Yu

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• Karma: 0 ##### 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.
$$(ye^{2xy}+x)+bxe^{2xy}y'=0$$
$$\therefore M(x,y)=ye^{2x}+x$$
$$N(x,y)=bxe^{2xy}$$
$$M_y=e^{2xy}+2xye^{2xy}$$
$$N_x=be^{2xy}+2bxye^{2xy}$$
since the differential equation to be exact
$$M_y=N_x$$
$$e^{2xy}+2xye^{2xy}=be^{2xy}+2bxye^{2xy}$$
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 $\varphi_y(x,y)=N(x,y)$
$$\therefore\varphi_x(x,y)=ye^{2xy}+x$$
$$\therefore\varphi(x,y)=\int(ye^{2xy}+x)dx=\dfrac{ye^{2xy}}{2y}+\dfrac{x^2}{2}+h(y)$$
$$\therefore\varphi_y=xe^{2xy}+h'(y)$$
$$\because\varphi_y=N\qquad\therefore xe^{2xy}+h'(y)=xe^{2xy}$$
$$\therefore h'(y)=0\qquad\therefore h(y)=c$$
General solution  $\varphi=\dfrac{e^{2xy}}{2}+\dfrac{x^2}{2}=c$

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