ABSTRACT

This is a special case of equation 6.8.3 with f (y) = ky2. Solution in an implicit form:∫ y

[ 4Au – 2kF (u) + B2 – 4AC]–1/2 du = ±(x – a), F (u) = 13

( u3 – y30

) , y0 = Aa

2 + Ba + C.

3. y(x) + A ∫ x

tλy2(t) dt = Bxλ+1 + C.