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

Answer 1

Answer: (C)

The differential equation is $\frac{dy}{dx}-\frac{y}{x}=-\frac{5x}{\left(x+2\right)\left(x-3\right)}$
$I.F=e^{\int \left(-\frac{1}{x}\right)dx}=e^{-lnx}=\frac{1}{x}$
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 $C=-ln6$
$\therefore y=xln\left\{\frac{\left(x+2\right)}{6\left(x-3\right)}\right\}$
Putting $\left(5,a\right)$, we get $a=5ln\left(\frac{7}{12}\right)$