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

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.