Question

The gradient of the curve passing through (4, 0) is given by $\frac{dy}{dx} - \frac{y }{x} + \frac{5x}{\left(x+2\right)\left(x-3\right)} = 0$ if the point (5, a) lies on the curve, then the value of a is

• A. $\frac{67}{12}$
• B. $5\,sin\frac{7}{12}$
• C. $5\,log\frac{7}{12}$
• D. None of these

Answer 1

Answer: (C)

The differential equation is $\frac{dy}{dx} - \frac{y }{x} = - \frac{5x}{\left(x+2\right)\left(x-3\right)} $
$I. F = e^{\int\left(-\frac{1}{x}\right)dx} = e^{-ln\,x} = \frac{1}{x}$
Solution is $y\left(\frac{1}{x}\right) = \int\left(\frac{1}{x}\right) \times\frac{-5x}{\left(x+2\right)\left(x-3\right)}dx = ln\left(\frac{x+2}{x-3}\right) + C$
It passes through $\left(4, \,0\right)$, so $C = - ln\, 6$
$\therefore y = x\, ln \left\{\frac{\left(x+2\right)}{6\left(x-3\right)}\right\}$
Putting $\left(5, \,a\right)$, we get $a = 5\,ln\left(\frac{7}{12}\right)$