Lecture Notes in Differential Equations - Bruce E. Shapiro

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Example 3.10. Find a general solution of
dy
dt = 1 + t2 + y 3 + t 2 y 3 (3.60)
From (3.59) we have
dy
dt = (1 + y3 )(1 + t 2 ) (3.61)
hence the equation can be re-written as
∫
∫
dy
1 + y 3 = (1 + t 2 )dt (3.62)
The integral on the left can be solved using the method of partial fractions:
1
1 + y 3 = 1
(y + 1)(y 2 − y + 1) = A
1 + y + By + C
y 2 − y + 1
(3.63)
Cross-multiplying gives
Substituting y = −1 gives
1 = A(y 2 − y + 1) + (By + C)(1 + y) (3.64)
1 = A(1 + 1 + 1) =⇒ A = 1 3
(3.65)
Sbstituting y = 0
1 = A(0 − 0 + 1) + (0 + C)(1 + 0) = 1 3 + C =⇒ C = 2 3
(3.66)
Using y = 1
1 = A(1 − 1 + 1) + (B(1) + C)(1 + 1) = 1 3 + 2B + 4 3
(3.67)
Hence
2B = 1 − 5 3 = −2 3 =⇒ B = −1 3
(3.68)