Question

Solve the following given equation: $y\log ydx-xdy=0$ .

Answer 1

Hint: Rearrange the terms of the equation to get all the terms related to x on one side of the equation and all the terms related to y to the other side of the equation. Now integrate both the sides and let $\log y$ to be t. Now differentiate log y with respect to t and convert the equation in terms of x and t. Now use the identity $\int{\dfrac{1}{x}dx=\log x+c}$ and finally substitute the value of t according to the assumption to get the answer.

Complete step-by-step answer:
Let us start the solution to the above question by rearranging the terms, such that the terms related to x on one side of the equation and all the terms related to y to the other side of the equation.
$y\log ydx-xdy=0$
$\Rightarrow y\log ydx=xdy$
$\Rightarrow \dfrac{dx}{x}=\dfrac{dy}{y\log y}$
Now as the given equation is a differential equation, we will integrate both sides of the equation to get the equation in algebraic form.
$\int{\dfrac{dx}{x}}=\int{\dfrac{dy}{y\log y}}$
Now, to solve the integrals, we will let $\log y$ to be t.
$\therefore \log y=t$
$\Rightarrow \dfrac{d\log y}{dt}=\dfrac{dt}{dt}$
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dt}=1$
$\Rightarrow dy=ydt$
Now we will substitute the value of dy and logy in our integral. On doing so, we get
$\int{\dfrac{dx}{x}}=\int{\dfrac{ydt}{yt}}$
$\Rightarrow \int{\dfrac{dx}{x}}=\int{\dfrac{dt}{t}}$
Now, we know that $\int{\dfrac{1}{x}dx=\log x+c}$ . So, our integral becomes:
$\log x+{{c}_{1}}=\operatorname{logt}+{{c}_{2}}$
$\Rightarrow \log x=\operatorname{logt}+{{c}_{2}}-{{c}_{1}}$
$\Rightarrow \log x-\operatorname{logt}={{c}_{2}}-{{c}_{1}}$
Now, we know $\log x-\operatorname{logt}=\log \dfrac{x}{t}$ .
$\therefore \log \dfrac{x}{t}={{c}_{2}}-{{c}_{1}}$
$\Rightarrow \dfrac{x}{t}={{e}^{{{c}_{2}}-{{c}_{1}}}}$
We can replace ${{e}^{{{c}_{2}}-{{c}_{1}}}}$ with another constant c, as the value of ${{e}^{{{c}_{2}}-{{c}_{1}}}}$ is a constant. So, our equation becomes:
$\dfrac{x}{t}=c$
$x=ct$
Now, we will substitute the value of t as logy, as we have assumed in the solution part. In doing so, our final answer is equal to $x=c\log y$ .

Note: The general mistake that a student makes in a problem related to integrations is that they forget about the constant term, which makes the equation that we get after integration is completely wrong. Also, you need to learn all the formulas related to integration and differentiation, as they are used very often in such questions. Identities related to logarithmic and exponential functions are also important.