The given differential equation is:
y=(x−ydydx)sin(yx)
We are provided with the initial condition
x(1)=2π and our goal is to determine the value of
cos(x(2)).
Observe the argument inside the sine function,
yx. This suggests that the equation can be simplified by expressing it in terms of this ratio. Dividing both sides by
y (assuming
y>0), we obtain:
sin(yx)1=yx−dydx
Recall the quotient rule for the derivative of
yx with respect to
y:
dyd(yx)=y2ydydx−x=−y1(yx−dydx)
Substituting this expression back into our equation yields:
1=−ydyd(yx)sin(yx)
Rearranging the terms to separate the variables, we get:
−ydy=sin(yx)d(yx)
Integrating both sides, we find:
−lny=−cos(yx)+C
Multiplying by
−1, the general solution is:
cos(yx)=lny+C
Using the condition
x(1)=2π, we substitute
y=1 and
x=2π into the general solution:
cos(1π/2)=ln(1)+C⇒0=0+C
Thus,
C=0, and our particular solution is
cos(yx)=lny.
To find
cos(x) when
y=2, we first note that
cos(2x)=ln2. We apply the double angle identity
cos(x)=2cos2(2x)−1:
cos(x)=2(ln2)2−1