Question
The solution of the differential equation $$\frac{{dy}}{{dx}} = \frac{{x + y}}{x}$$ satisfying the condition $$y\left( 1 \right) = 1$$ is-
A.
$$y = \ln \,x + x$$
B.
$$y = x\ln \,x + {x^2}$$
C.
$$y = x{e^{\left( {x\, - \,1} \right)}}$$
D.
$$y = x\ln \,x + x$$
Answer :
$$y = x\ln \,x + x$$
Solution :
$$\frac{{dy}}{{dx}} = \frac{{x + y}}{x} = 1 + \frac{y}{x}$$
Putting $$y=vx$$ and $$\frac{{dv}}{{dx}} = v + x\frac{{dv}}{{dx}}$$
we get
$$\eqalign{
& v + x\frac{{dv}}{{dx}} = 1 + v\,\,\, \Rightarrow \int {\frac{{dx}}{x} = \int {dv} } \cr
& \Rightarrow v = \ln \,x + c\,\, \Rightarrow y = x\,\ln \,x + cx \cr
& {\text{As }}y\left( 1 \right) = 1 \cr
& \therefore c = 1 \cr} $$
So solution is $$y=x \ln x + x$$