Solving an Exact Differential Equation (#1)

[r-soy]

View attachment 682

 

[alyafey22]

Re: show exat or not and solve

the same idea here as well ,but i don't understand what you are doing exactly !

 

[I like Serena]

Re: show exat or not and solve

Welcome to MHB, rsoy! :)

Your problem seems to be that you assume $\dfrac{x+y}{x^2+y^2} = \dfrac 1 {x+y}$.
But this is not true.So instead your next step for the first part of the expression would be (edited):
$$\begin{aligned} \int^y \frac{x+y}{x^2+y^2}dy &= \int^y \frac{x}{x^2+y^2}dy &&+ \int^y \frac{y}{x^2+y^2}dy &\\
&= \arctan \left(\frac y x \right) &&+ \frac 1 2 \ln(x^2+y^2) &+ C(x) \end{aligned}$$

 

[alyafey22]

Re: show exat or not and solve

Welcome to MHB, rsoy! :)

Your problem seems to be that you assume $\dfrac{x+y}{x^2+y^2} = \dfrac 1 {x+y}$.
But this is not true.So instead your next step for the first part of the expression would be:
$$\begin{aligned} \int^y \frac{x+y}{x^2+y^2}dy &= \int^y \frac{x}{x^2+y^2}dy &&+ \int^y \frac{y}{x^2+y^2}dy &\\
&= \arctan \left(\frac y x \right) &&+ \frac 1 2 \ln(x^2+y^2) &+ C \end{aligned}$$

Shouldn't the resultant constant be a function of x !

 

[I like Serena]

Re: show exat or not and solve

Shouldn't the resultant constant be a function of x !

Good point!
Edited.

 

[r-soy]

thaaanks