Integrating Factor for linear first order

[mares0909]

X(dy/dx) + (1+x)y = e^(-x) sin(2x)

What I have so far is I have divided by X so I can make it a linear equation. I get:

(dy/dx) + ((1+x)/x)y = (e^(-x) sin(2x)) / x

Then I use P(x) = (1+x)/x = S (1+x)/x dx = lnx + x.
looking for the integrating factor I get: x + e^x

The I plug in: Y (x + e^x) = S (x + e^x)*((e^(-x) sin(2x)) / x)

The problem is that when I try to integrate this I get an integral back, Im stuck integrating S (sin(2x)) / x dx.

Thank you!

 

[soroban]

Hello, mares0909!

\(\displaystyle x\,\frac{dy}{dx} + (1+x)y \:= \:e^{-x}\sin(2x)\)

I have divided by x to make it a linear equation.

\(\displaystyle \text{I get: }\;\frac{dy}{dx} + \frac{1+x}{x}y \:= \:\frac{(e^{-x}\sin(2x)}{x}\)

\(\displaystyle \text{Then I use: }\: P(x) \:= \:\frac{1+x}{x}\quad\Rightarrow\quad \int\frac{1+x}{x}\,dx \:=\:\ln x + x\)

\(\displaystyle \text{For the integrating factor I get: }\;x + e^x\) . . . . Careful!

\(\displaystyle \text{The integrating factor is: }\:e^{\ln x + x} \;=\;e^{\ln x}\cdot e^x \;=\;x\cdot e^x\)


\(\displaystyle \text{Then we have: }\;xe^x\cdot\frac{dy}{dx} + xe^x\cdot\frac{1+x}{x}\,u \;=\;xe^x\cdot\frac{e^{-x}\sin(2x)}{x}\)

\(\displaystyle \text{Hence: }\;\frac{d}{dx}\left(xe^xy\right) \;=\;\sin(2x) \quad\hdots\;\text{Got it?}\)