Respuesta :

carlosego
(x − a)(x − b)y' − (y −c)= 0
 Rewriting the equation we have
 (x − a)(x − b)y' = (y −c)
 (x − a)(x − b)(dy/dx) = (y −c)
 separating the variables
 dy/(y −c)=dx/((x − a)(x − b))
 integrating both sides of the equation
 Left side:
 Ln(y-c)
 Right side:
 Simple fractions
 (1/((x − a)(x − b)))=(A/(x − a)) + (B/(x − b))
 A=(1/(a-b))
 B=(1/(b-a))
 I[dx/((x − a)(x − b))]= I[((1/(a-b))dx/(x − a))] + I[((1/(b-a))dx/(x − b))]
 I[dx/((x − a)(x − b))]= (1/(a-b))*Ln(x-a) + (1/(b-a))*Ln(x-b) + C
 Then the solution is
 Ln(y-c)= (1/(a-b))*Ln(x-a) + (1/(b-a))*Ln(x-b) + C
 Rewriting
 Exp[Ln(y-c)]= exp[(1/(a-b))*Ln(x-a) + (1/(b-a))*Ln(x-b) + C]
 y-c= exp[(1/(a-b))*Ln(x-a)]*exp[(1/(b-a))*Ln(x-b)]*exp[C]
 y-c=[(x-a)^(1/(a-b))]*[(x-b)^(1/(b-a))]*C’
 final answer:
 y=C’*[(x-a)^(1/(a-b))]*[(x-b)^(1/(b-a))]* + c

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