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How do you solve this IVP: dy/dx = x^2/(1+y^2) , y(0)=1?
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- Iggy RockoLv 75 years ago
dy/dx = x^2/(1 + y^2)
1 + y^2 dy = x^2 dx
∫ 1 + y^2 dy = ∫ x^2 dx
y + y^3/3 = x^3/3 + c
1 + 1^3/3 = 0^3/3 + c
4/3 = 0 + c
c = 4/3
y + y^3/3 = x^3/3 + 4/3
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