2(1+x) dy/dx = 1-y^2?

Question

also when x = 5 y = 0.5

thanks

solve the differential equation

Answer 1

∫ dy / (1 - y).(1 + y) = ∫ dx / (2 (1 + x))
Consider LHS
1 / (1-y).(1 + y) = A / (1 - y) + B / (1 + y)
1 = A.(1 + y) + B.(1 - y)
1 = (A - B).y + (A + B)
A - B = 0
A + B = 1
A = 1/2 , B = 1/2
I = (1/2).∫1 / (1 - y).dy + (1/2).∫1 / (1 + y).dy
I = (- 1/2).log (1 - y) + (1/2).log (1 + y)
I = (1/2) [ log (1 + y) - log (1 - y) ]
I = (1/2).log [ (1 + y) / (1 - y) ]
Consider RHS
I = (1/2).∫dx / (1 + x)
I = (1/2).log (1 + x)
Thus:-
log [ (1 + y) / (1 - y) ] = log (1 + x) + ln C
log [ (1 + y) / (1 - y) ] - log (1 + x) = log C
log (1 + y) / [(1 - y).(1 + x) ] = log C
(1 + y) / (1 - y).(1 + x) = C
(1.5) / [ (0 . 5).(6) ] = C
C = 0.5
(1 + y) = 0.5 (1 - y).(1 + x)
(1 + y) / (1 - y) = 0 .5 (1 + x)
2.(1 + y) / (1 - y) = 1 + x

Answer 2

2(1+x)y' = 1-y^2
y'/(1-y^2) = (1/2)/(1+x)
y'/(1-y)+y'/(1+y) = (1/4)/(1+x)
-ln(1-y)+ln(1+y) = (1/4)ln(1+x) +c
(1+y)/(1-y) = C(1+x)^(1/4)
at(5,0.5): (3/2)(1/2) = C 6^4 so C = (3/4)/6^4
(1+y) = C(1+x)^(1/4) (1-y)
y = [ C(1+x)^(1/4) -1 ] / [ C(1+x)^(1/4) +1 ]
y = [ 3( (1+x)/6) ^4 - 4 ] / [ 3( (1+x)/6) ^4 + 4 ]