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MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Zuwei Zhao on October 04, 2019, 02:01:16 PM

Title: MAT24f4 TUT5103 Quiz2
Post by: Zuwei Zhao on October 04, 2019, 02:01:16 PM
$$
(ye^{2xy}+x)+bxe^{2xy}y^{\prime}=0.
$$
$$
\begin{aligned}
M&=ye^{2xy}+x\\
M_y&=e^{2xy}+2xye^{2xy}\\
N_x&=bxe^{2xy}+2bxye^{2xy}\\
e^{2xy}&+2xye^{2xy}=bxe^{2xy}+2bxye^{2xy}\\
&b=1\\
\end{aligned}
$$

Substitute $(ye^{2xy}+x)+xe^{2xy}y^{\prime}=0$




$$
\begin{aligned}
M&=ye^{2xy}+x\qquad N=xe^{2xy}\\
M_y&=N_x\qquad \therefore \text{the function now is exact.}\\
\varphi_x&=M\\
\varphi&=\int ye^{2xy}+xdx\\
&=\frac{ye^{2xy}}{2y}+\frac{x^2}{2}+h(y)=\frac{e^{2xy}}{2}\frac{x^2}{2}+h(y)\\
\varphi_y&=\frac{2xe^{2xy}}{2}+h^{\prime}(y)\\
&=xe^{2xy}+h^{\prime}(y)\\
h^{\prime}(y)&=0\\
h(y)&=\text{constant}\\
\varphi&=\frac{e^{2xy}}{2}+\frac{x^2}{2}=c
\end{aligned}
$$