linear differential!

**lawochekel** · September 17th 2012, 05:11 AM

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

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

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

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

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

thanks

**Prove It** · September 17th 2012, 05:40 AM

Originally Posted by lawochekel

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

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

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

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

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

thanks

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

**HallsofIvy** · September 17th 2012, 05:47 AM

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

**Prove It** · September 17th 2012, 05:48 AM

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...

**lawochekel** · September 17th 2012, 06:34 AM

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

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

**HallsofIvy** · September 17th 2012, 07:52 AM

I would recommend $u= e^v$ .

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

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