Question

Solve the given differential equation. (x − a)(x − b)y' − (y −

c.= 0, where a, b, c are constants. (assume a ≠
b.)

Answer 1

(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