the given equation can be written as
y1dxdy+2(1−x2)2xy=x .....(1)
this is of the form f'(y) y' + P f(y) = Q
where P = 2(1−x2)x
Q= x
f(y) = 2 y
let v= 2y
differentiating,
y1dxdy=dxdv
(1) becomes
dxdv+2(1−x2)xv=x
this is of the form dxdv+Av=B
A=2(1−x2)xandB=x
integrating factor = e∫Pdx=e∫2(1−x2)xdx=(1−x2)(−1/4)
the general solution is given by:
v (1-x2)-1/4 = ∫ x. (1-x2)-1/4 dx + c
2y = -2/3 (1-x2) + c(1-x2)1/4
now, given y(0)=1
2 = -2/3 + c
c= 8/3
so, the solution is
2y = -2/3 (1-x2) + 8/3(1-x2)1/4
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