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Question: The solution of \[ydx - xdy = 0\] A) \[{y^2} = cx\] B) \[y = c{x^2}\] C) \[y = cx\] D) \...

The solution of ydxxdy=0ydx - xdy = 0
A) y2=cx{y^2} = cx
B) y=cx2y = c{x^2}
C) y=cxy = cx
D) x2=cy{x^2} = cy

Explanation

Solution

Here we have to solve the given differential equation. For that, we will equate the terms and divide the terms in such a way that the equation comes in dydx\dfrac{{dy}}{{dx}} form. Then we will use integration to solve the equation. On further simplification, we will get the solution of the differential equation.

Complete step by step solution:
The given differential equation is ydxxdy=0ydx - xdy = 0.
Taking the term xdyxdy to right side of equation, we get
ydx=xdyydx = xdy
Now, we will divide all the terms on both sides of the equation by the term xyxy.
ydxxy=xdyxy\Rightarrow \dfrac{{ydx}}{{xy}} = \dfrac{{xdy}}{{xy}}
On further simplification, we get
1xdx=1ydy\Rightarrow \dfrac{1}{x}dx = \dfrac{1}{y}dy
Now, we will integrate both the terms.
1xdx=1ydy\Rightarrow \int {\dfrac{1}{x}dx} = \int {\dfrac{1}{y}dy}
On integrating the terms, we get
logx=logy+logC\log x = \log y + \log C
We have added constant logC\log C because it is an indefinite integral.
We know by the property of logarithmic function loga+logb=logab\log a + \log b = \log ab.
Now, we will be using the same property of logarithmic function for the term logy+logc\log y + \log c.
Thus, the above equation becomes;
logx=logyC\Rightarrow \log x = \log yC
Rewriting the equation, we get
x=yC\Rightarrow x = yC
Dividing CC on both the side, we get
1Cx=y\Rightarrow \dfrac{1}{C}x = y
As 1C\dfrac{1}{C} is also a constant we can denote it as cc.
Thus, the final equation becomes;
y=cx\Rightarrow y = cx

Hence, the correct answer is option C.

Note:
Here, we need to keep basic integration property in mind. A logarithmic function is defined as a function, which is inverse of the exponential function.
Some important properties of logarithmic function are:-
The logarithm of a product of two or more terms is equal to the sum of the logarithm of each term.
The logarithm of a division of two terms is equal to the difference of the logarithm of these two terms.