Toronto Math Forum

a)
x²y''−2xy'+(x²+2)y=0
y''-2/xy'+(1+2/x²)y=0
W=ce^(∫(2/x)dx)=cx²
Let c=1, then W=x²

b)
y1(x)=x\cos(x)
y1'=\cosx-x\sinx
y1''=-\sinx-\sinx-x\cosx
x²(-\sinx-\sinx-x\cosx)-2x(\cosx-x\sinx)+(x²+2)(x\cos(x)=0
Therefore, y1(x)=x\cos(x) is a solution of this equation.
W=|y1  y2| = |x\cos(x)                  y2| = x²
     |y1' y2'|    |\cosx-x\sinx  y2'|
Thus, x\cos(x)y2'-(\cosx-x\sinx)y2 = x²
y2'-(\cosx-x\sinx)/(x\cosx)y2 = x²/(x\cosx)
y2'-(1/x-\tanx)y2 = x²/(x\cosx)
μ= e^∫[\tanx-(1/x)]dx = e^(ln(\secx))*e^(ln(1/x)) = (\secx)/x = 1/(x\cosx)
Multiply both sides by μ
y2'/(x\cosx)-[(\cosx-x\sinx)/(x\cosx)²]y2 = [x/(\cosx)]/(x*\cosx)= 1/cos²x
[y2*1/(x\cosx)]' = 1/cos²x
y2*1/(x\cosx) = ∫[1/(\cos²x)]dx = \tanx + C
y2 = \tanx*x*\cosx = x*\sinx

c)
General solution: Y = c1y1+c2y2 = c1x*\cosx + c2x*\sinx
Y(π/2)=c2*π/2=1
c2=2/π
Y'(π/2)=-c1π/2+c2=0
c1=4/π
Thus, Y = 4/π(x*\cosx) + 2/π(x*\sinx)

Q: Find the Wronskian of two solutions of the given differential equation without solving the equation
cos(t)y''+sin(t)y'-ty=0

Answer:
y''+sin(t)y'/cos(t)-ty/cos(t)=0
P(t)=sin(t)/cos(t)
W=ce^(∫p(t)dt)
=ce^(-∫(sin(t)/cos(t))dt)  let u=cos(t), du=-sin(t)dt
=ce^(∫1/u du)
=ce^(lnu)
=cu
=ccos(t)