\((dL)/(L-a)= k(x+b)dx\)

\((dL)/(L-a)= (kx+bk)dx\)

Now integrating both sides

\(int (dL)/(L-a)= int (kx+bk)dx\) \(ln(L-a)= (kx^2)/(2+bkx+c)\) (because \(int (1/(x+a)dx=ln(x+a))\) and \(int x^n dx= x^(n+1)/(n+1))\) where c is the constant of integration

So, the solution of the differential equation \(L'(x) = k(x+b)(L-a)\) will be \(ln(L-a)= (kx^2)/(2 + bkx + c)\)

\((dL)/(L-a)= (kx+bk)dx\)

Now integrating both sides

\(int (dL)/(L-a)= int (kx+bk)dx\) \(ln(L-a)= (kx^2)/(2+bkx+c)\) (because \(int (1/(x+a)dx=ln(x+a))\) and \(int x^n dx= x^(n+1)/(n+1))\) where c is the constant of integration

So, the solution of the differential equation \(L'(x) = k(x+b)(L-a)\) will be \(ln(L-a)= (kx^2)/(2 + bkx + c)\)