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y(0) =1
[x/(1+y) ]dx -[y/(1+x) ]dy =0
[x/(1+y) ]dx =[y/(1+x) ]dy
∫y(1+y) dy =∫ x(1+x) dx
(1/2)y^2 + (1/3)y^3 = (1/2)x^2 +(1/3)x^3 +C
y(0)=1
1/2+1/3 =C
C= 5/6
ie
(1/2)y^2 + (1/3)y^3 = (1/2)x^2 +(1/3)x^3 +5/6
ans : C
[x/(1+y) ]dx -[y/(1+x) ]dy =0
[x/(1+y) ]dx =[y/(1+x) ]dy
∫y(1+y) dy =∫ x(1+x) dx
(1/2)y^2 + (1/3)y^3 = (1/2)x^2 +(1/3)x^3 +C
y(0)=1
1/2+1/3 =C
C= 5/6
ie
(1/2)y^2 + (1/3)y^3 = (1/2)x^2 +(1/3)x^3 +5/6
ans : C
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