Question
The gradient of the curve passing through $$\left( {4,\,0} \right)$$ is given by $$\frac{{dy}}{{dx}} - \frac{y}{x} + \frac{{5x}}{{\left( {x + 2} \right)\left( {x - 3} \right)}} = 0$$ if the point $$\left( {5,\,a} \right)$$ 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 :
$$5\,\log \,\frac{7}{{12}}$$
Solution :
The differential equation is
$$\eqalign{
& \frac{{dy}}{{dx}} - \frac{y}{x} = - \frac{{5x}}{{\left( {x + 2} \right)\left( {x - 3} \right)}} \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\left( { - \frac{1}{x}} \right)dx} }} = {e^{ - \ln \,x}} = \frac{1}{x} \cr} $$
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 $$\left( {4,\,0} \right),\,{\text{so }}C = - \ln \,6$$
$$\eqalign{
& \therefore \,y = x\,\ln \left\{ {\frac{{\left( {x + 2} \right)}}{{6\left( {x - 3} \right)}}} \right\}{\text{ }}\,{\text{Putting }}\left( {5,\,a} \right) \cr
& {\text{we get }}a = 5\,\ln \,\left( {\frac{7}{{12}}} \right) \cr} $$