# Solving an Exact Differential Equation (#1)

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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}

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
Re: show exat or not and solve

Shouldn't the resultant constant be a function of x !
Good point!
Edited.

thaaanks