MHB B.2.1.4 trig w/ integrating factor

[karush]

$\begin{array}{rl}
\textit{Find } \mu(x): &\mu(x) =\exp\left(\int \dfrac{1}{x}\,dx\right)=e^{\ln{x}}=x\\
\textit{multiply thru by x} &xy^\prime+y=3x\cos 2x\\
\textit{rewrite as } &(xy)'=3x\cos 2x \\
\textit{}integrate &xy=\int 3x\cos 2x \, dx=\dfrac{3}{2}x\sin(2x)+\dfrac{3}{4}\cos(2x)+c\\
\textit{divide thru by x} &y=\dfrac{3}{2}\sin(2x)+\dfrac{3}{4}\dfrac{\cos(2x)}{x}+\dfrac{c}{x}\\
\textit{re-order} &y=\dfrac{c}{x}+\dfrac{3}{4}\dfrac{\cos 2x}{x}+\dfrac{3}{2}\sin 2x
\end{array}$

ok quite sure there are some oops in this one
thot I would try array to do the steps so...

hopefully there kinda
Mahalo for inputhttps://dl.orangedox.com/6rStfn4eMFHuHvAKuX

 

[HOI]

It would help if you would tell us what the original problem was!

I think it is to solve the differential equation y'+ y/x= 3 cos(2x).

Personally, I wouldn't use the "integrating factor" method at all. I would instead look at this as a linear, non-homogeneous equation. The corresponding homogeneous equation is y'+ y/x= 0 or dy/dx= -y/x so that dy/y= -dx/x. Integrating both sides ln(y)= = -ln(x)+ c and taking the exponential of both sides, y= Cx^-1= C/x (C= e^c). To find a solution to the entire equation, look for a function of the form y= z(x)/x for some function, z. y'= z'/x- z/x^2 so the equation is z'/x- z/x^2+ z/x^2= z'/x= 3 cos(2x).

z'= 3 x cos(2x) and now use integration by parts: let u= 3x, dv= cos(2x). Then du= 3 dx and v= (1/2) sin(2x).
$z= (3/2) x sin(2x)- (3/2)\int sin(2x)dx= (3/2) x sin(2x)+ (3/4) cos(x)$.
I've dropped the "constant of integration" since we only need one such function.

z(x)/x= (3/2) sin(2X)+ (3/4) cos(x)/x so the full solution is

y(x)= C/x+ (3/2) sin(2X)+ (3/4) cos(x)/x

That's exactly what you have! Well done! (To both of us!)