Exact Differential Equation - Could someone check if I did this right?

Question

$$ \begin{cases} \text{ b) } \cos(x)+ye^{xy}+xe^{xy}\frac {dy}{dx}=0 \\ \text { c) } e^x+e^y\frac {dy}{dx}=0 \\ \end{cases} $$

I think I did the problem correctly, but I'm not sure if I wrote the answer in correct form, could someone check?

Answer 1

Looks good to me; $f(x,y)=C$ is the standard way to write solutions to exact differential equations. You could always try to solve explicitly for $y$, and both of those answers are nice enough that you could do that here if you wanted to, but that's usually not necessary.

Answer 2

$$ \text{ b) } \cos(x)+ye^{xy}+xe^{xy}\frac {dy}{dx}=0 $$

I got for this one : $f(x,y)=\sin(x)+e^{xy}=K$

$$ \text { c) } e^x+e^y\frac {dy}{dx}=0 $$

And here $f(x,y)=e^x+e^y=K$

So you did very well for both equations.

Answer 3

$$\cos(x)+ye^{xy}+xe^{xy}\frac {dy}{dx}=0$$ $$\cos(x)+e^{xy}\left(y+x\frac {dy}{dx}\right)=0$$ $\frac{d}{dx}e^{xy}=e^{xy}\left(y+x\frac {dy}{dx}\right)$ $$\cos(x)+\frac{d}{dx}e^{xy}=0$$ The integration leads to $$\sin(x)+e^{xy}=C$$ So, your solution is correct. But you can express it explicitly :

$e^{xy}=C-\sin(x)\quad$ so $\quad C-\sin(x)>0 \quad\implies\quad xy=\ln(C-\sin(x))$ $$y=\frac{\ln(C-\sin(x))}{x}$$

$$ $$ Second ODE :

$$e^x+e^y\frac {dy}{dx}=0$$ $\frac {d}{dx}e^y=e^y\frac {dy}{dx}$ $$e^x+\frac {d}{dx}e^y=0$$ The integration leads to $$e^x+e^y=C$$ So, your solution is correct. But you can express it explicitly :

$e^{y}=C-e^x\quad$ so $\quad C-e^x>0 \quad\implies\quad $ $$y=\ln(C-e^x)$$