I lied, one more (first-order DE). :)

**cinder** · June 27th 2007, 12:23 AM

$x\frac{dy}{dx}+y=\sin(x)$ , $x>0$ , initial condition: $y(\frac{\pi}{2})=1$

What I have so far...

$\frac{dy}{dx}+\frac{1}{x}y=\frac{\sin(x)}{x}<br /> \Rightarrow \int \frac{1}{x}dx=\ln|x| \Rightarrow e^{\ln|x|}=x<br />$

I'm not sure if I did the previous part right. And after that I'm unsure how to proceed.

**CaptainBlack** · June 27th 2007, 03:09 AM

Originally Posted by cinder

$x\frac{dy}{dx}+y=\sin(x)$ , $x>0$ , initial condition: $y(\frac{\pi}{2})=1$

What I have so far...

$\frac{dy}{dx}+\frac{1}{x}y=\frac{\sin(x)}{x}<br /> \Rightarrow \int \frac{1}{x}dx=\ln|x| \Rightarrow e^{\ln|x|}=x<br />$

I'm not sure if I did the previous part right. And after that I'm unsure how to proceed.

$x\frac{dy}{dx}+y=\sin(x)$

may be written as:

$\frac{d}{dx} (x~y) =\sin(x)$

so:

$x~y = -\cos(x) +C$

and the rest should be simple.

RonL

**galactus** · June 27th 2007, 03:11 AM

You're OK so far. You have your integrating factor.

$\frac{d}{dx}(xy)=sin(x)$

Integrate:

$\int\frac{d}{dx}(xy)=\int{sin(x)}dx$

$xy=-cos(x)+C$

$y=\frac{-cos(x)}{x}+\frac{C}{x}$

IC is $y(\frac{\pi}{2})=1$

Using this and solving for $C=\frac{\pi}{2}$

Therefore, you have:

$y=\frac{\pi}{2x}-\frac{cos(x)}{x}$

**cinder** · June 27th 2007, 09:05 AM

I guess I'm not seeing how it can be rewritten as $\frac{d}{dx}(xy)=sin(x)$ .

**topsquark** · June 27th 2007, 09:48 AM

Originally Posted by cinder

I guess I'm not seeing how it can be rewritten as $\frac{d}{dx}(xy)=sin(x)$ .

Think of it this way:
$\frac{d}{dx}(xy) = 1 \cdot y + x \frac{dy}{dx}$
by the product rule.

So
$\frac{d}{dx}(xy) = y + x \frac{dy}{dx} = sin(x)$
which is your initial equation.

-Dan

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