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Solving an Exact Differential Equation (#1)

rsoy

New member
Mar 3, 2013
14
866864662.jpg
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: show exat or not and solve

the same idea here as well ,but i dont understand what you are doing exactly !
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,874
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):
$$\begin{aligned} \int^y \frac{x+y}{x^2+y^2}dy &= \int^y \frac{x}{x^2+y^2}dy &&+ \int^y \frac{y}{x^2+y^2}dy &\\
&= \arctan \left(\frac y x \right) &&+ \frac 1 2 \ln(x^2+y^2) &+ C(x) \end{aligned}$$
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 !
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,874

rsoy

New member
Mar 3, 2013
14
thaaanks