1. linear differential!

$\displaystyle (\exp^{\frac{y}{x}}-\frac{y}{x}\exp^{\frac{y}{x}}+\frac{1}{1+x^2})dx+\ exp^{\frac{y}{x}}dy=0$

have reduce $\displaystyle (1-\frac{y}{x}+\frac{\exp^{\frac{y}{x}}}{1+x^2})dx+dy =0$

$\displaystyle \frac{dy}{dx}=\frac{y}{x}-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1$

$\displaystyle \frac{dy}{dx}-\frac{y}{x}=-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1$

pls am stock here, whether am on the right part or not, help!

thanks

2. Re: linear differential!

Originally Posted by lawochekel
$\displaystyle (\exp^{\frac{y}{x}}-\frac{y}{x}\exp^{\frac{y}{x}}+\frac{1}{1+x^2})dx+\ exp^{\frac{y}{x}}dy=0$

have reduce $\displaystyle (1-\frac{y}{x}+\frac{\exp^{\frac{y}{x}}}{1+x^2})dx+dy =0$

$\displaystyle \frac{dy}{dx}=\frac{y}{x}-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1$

$\displaystyle \frac{dy}{dx}-\frac{y}{x}=-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1$

pls am stock here, whether am on the right part or not, help!

thanks
When you divide by \displaystyle \displaystyle \begin{align*} e^{\frac{y}{x}} \end{align*} you should actually get \displaystyle \displaystyle \begin{align*} \left[ 1 - \frac{y}{x} + \frac{1}{\left( 1 + x^2 \right) e^{\frac{y}{x}}} \right] dx + dy = 0 \end{align*}.

3. Re: linear differential!

You understand, do you not, that this is NOT a "linear" equation?

4. Re: linear differential!

Originally Posted by HallsofIvy
You understand, do you not, that this is NOT a "linear" equation?
At least not yet. With the right substitution it may become one...

5. Re: linear differential!

$\displaystyle -\frac{(1+x^2)\exp^v}{1+(1+x^2)\exp^v}dv=\frac{1}{x }dx$

by substitution, i got that above. Do i need further substitution?

6. Re: linear differential!

I would recommend $\displaystyle u= e^v$.

(Notation: either "$\displaystyle e^v$" or "exp(v)" but not "$\displaystyle exp^v$".)