Question

Solve: $ \left( x-y\ln y+y\ln x \right)dx+x\left( \ln y-\ln x \right)dy=0 $
A. $ x\ln \left( \dfrac{y}{x} \right)-y+x\ln x+Cy=0 $
B. $ y\ln \left( \dfrac{x}{y} \right)-y+x\ln x+Cx=0 $
C. $ x\ln \left( \dfrac{x}{y} \right)-y+x\ln x+Cy=0 $
D. $ y\ln \left( \dfrac{y}{x} \right)-y+x\ln x+Cx=0 $

Answer 1

Hint: The given differential equation can be converted into a variable separable form by making the appropriate substitution $ y=vx $ . Separate the variables x and v and then integrate both sides to obtain the solution.
Recall some properties of the natural logarithm: $ \ln \left( \dfrac{a}{b} \right)=\ln a-\ln b $ , $ \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} $ and $ \int{\left( \ln x \right)dx}=x\left( \ln x \right)-x $ .
The rule for derivative of a product of two functions is: $ d(uv)=udv+vdu $ .

Complete step-by-step answer:
The given equation $ \left( x-y\ln y+y\ln x \right)dx+x\left( \ln y-\ln x \right)dy=0 $ is not yet variable separable.
Taking out y as common in the first term, we get:
⇒ $ \left[ x-y\left( \ln y-\ln x \right) \right]dx+x\left( \ln y-\ln x \right)dy=0 $
Using $ \ln \left( \dfrac{a}{b} \right)=\ln a-\ln b $ , we get:
⇒ $ \left[ x-y\ln \left( \dfrac{y}{x} \right) \right]dx+x\ln \left( \dfrac{y}{x} \right)dy=0 $
Since y is the dependent variable, let's substitute $ y=vx $ , where v is a function of x, and try to separate the variables:
⇒ $ \left[ x-vx\ln \left( \dfrac{vx}{x} \right) \right]dx+x\ln \left( \dfrac{vx}{x} \right)d(vx)=0 $
Using the rule of multiplication for derivatives $ d(uv)=udv+vdu $ , we get:
⇒ $ \left[ x-vx\ln v \right]dx+x\left( \ln v \right)(xdv+vdx)=0 $
On multiplying the terms and separating the variables, we get:
⇒ $ xdx-vx(\ln v)dx+{{x}^{2}}(\ln v)dv+vx(\ln v)dx=0 $
⇒ $ xdx+{{x}^{2}}(\ln v)dv=0 $
Dividing by $ {{x}^{2}} $ , we get:
⇒ $ \dfrac{dx}{x}+(\ln v)dv=0 $
The variables are separated. Now, integrating both sides, we get:
⇒ $ \ln x+\left( v\ln v-v \right)+C=0 $
⇒ $ \ln x+v\left( \ln v-1 \right)+C=0 $
Back substitution $ y=vx $ , we get:
⇒ $ \ln x+\left( \dfrac{y}{x} \right)\left[ ln\left( \dfrac{y}{x} \right)-1 \right]+C=0 $
Multiplying by x, we get:
⇒ $ x\ln x+y\ln \left( \dfrac{y}{x} \right)-y+Cx=0 $
The correct answer is D. $ y\ln \left( \dfrac{y}{x} \right)-y+x\ln x+Cx=0 $ .

Note: The linear differential equations are identified as variable separable, ordinary, homogenous or exact which aids in getting to know the required steps for solving them.
In general, a differential equation of the variable separable form, $ f(x)dx=g(y)dy $ , can be solved by simply integrating both the sides. Other types of differential equations can be converted into this form by either substitution or by multiplying with the integrating factors.