B.2.1.4 trig w/ integrating factor

In summary, the problem is to solve the differential equation y'+ y/x= 3 cos(2x). The solution involves using the integrating factor method and integrating by parts to find the solution y= C/x+ (3/2) sin(2X)+ (3/4) cos(x)/x.
  • #1
karush
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$\begin{array}{rl}
\textit{Find } \mu(x): &\mu(x) =\exp\left(\int \dfrac{1}{x}\,dx\right)=e^{\ln{x}}=x\\
\textit{multiply thru by x} &xy^\prime+y=3x\cos 2x\\
\textit{rewrite as } &(xy)'=3x\cos 2x \\
\textit{}integrate &xy=\int 3x\cos 2x \, dx=\dfrac{3}{2}x\sin(2x)+\dfrac{3}{4}\cos(2x)+c\\
\textit{divide thru by x} &y=\dfrac{3}{2}\sin(2x)+\dfrac{3}{4}\dfrac{\cos(2x)}{x}+\dfrac{c}{x}\\
\textit{re-order} &y=\dfrac{c}{x}+\dfrac{3}{4}\dfrac{\cos 2x}{x}+\dfrac{3}{2}\sin 2x
\end{array}$

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
 
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  • #2
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!)
 
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FAQ: B.2.1.4 trig w/ integrating factor

What is B.2.1.4 trig w/ integrating factor?

B.2.1.4 trig w/ integrating factor is a mathematical concept that involves using trigonometric functions and an integrating factor to solve differential equations.

Why is integrating factor used in trigonometric equations?

Integrating factor is used in trigonometric equations to simplify the process of solving differential equations involving trigonometric functions. It allows for the use of basic integration techniques to find the solution.

What are some common applications of B.2.1.4 trig w/ integrating factor?

B.2.1.4 trig w/ integrating factor is commonly used in physics, engineering, and other fields to model and solve problems involving oscillations, waves, and other periodic phenomena.

How does one determine the integrating factor in B.2.1.4 trig w/ integrating factor?

The integrating factor in B.2.1.4 trig w/ integrating factor is determined by multiplying the original equation by a specific function, which is usually chosen to cancel out the trigonometric terms and simplify the equation.

Can B.2.1.4 trig w/ integrating factor be used to solve all trigonometric equations?

No, B.2.1.4 trig w/ integrating factor is specifically used for solving differential equations involving trigonometric functions. It may not be applicable to all types of trigonometric equations.

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