Question
The solution of the differential equation $$\frac{{dy}}{{dx}} + \frac{y}{x}\log \,y = \frac{y}{{{x^2}}}{\left( {\log \,y} \right)^2}$$ is :
A.
$$y = \log \left( {{x^2} + cx} \right)$$
B.
$$\log \,y = x\left( {c{x^2} + \frac{1}{2}} \right)$$
C.
$$x = \log \,y\left( {c{x^2} + \frac{1}{2}} \right)$$
D.
none of these
Answer :
$$x = \log \,y\left( {c{x^2} + \frac{1}{2}} \right)$$
Solution :
Divide the equation by $$y{\left( {\log \,y} \right)^2}$$
$$\eqalign{
& \frac{1}{{y{{\left( {\log \,y} \right)}^2}}}\frac{{dy}}{{dx}} + \frac{1}{{\log \,y}}.\frac{1}{x} = \frac{1}{{{x^2}}} \cr
& {\text{Put }}\frac{1}{{\log \,y}} = z \cr
& \Rightarrow \frac{{ - 1}}{{y{{\left( {\log \,y} \right)}^2}}}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}} \cr
& {\text{Thus, we get, }} - \frac{{dz}}{{dx}} + \frac{1}{x}.z = \frac{1}{{{x^2}}},{\text{ linear in }}z \cr
& \Rightarrow \frac{{dz}}{{dx}} + \left( { - \frac{1}{x}} \right)z = - \frac{1}{{{x^2}}}\,; \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{ - \int {\frac{1}{x}dx} }} = {e^{ - \log \,x}} = \frac{1}{x} \cr
& \therefore \,{\text{The solution is,}} \cr
& z\left( {\frac{1}{x}} \right) = \int {\frac{{ - 1}}{{{x^2}}}} \left( {\frac{1}{x}} \right)dx + c \cr
& \Rightarrow \frac{1}{{\log \,y}}\left( {\frac{1}{x}} \right) = \frac{{ - {x^{ - 2}}}}{{ - 2}} + c \cr
& \Rightarrow x = \log \,y\left( {c{x^2} + \frac{1}{2}} \right) \cr} $$