
The differential equation of all straight lines passing through the origin is
A. \[y = \sqrt {x\frac{{dy}}{{dx}}} \]
B. \[\frac{{dy}}{{dx}} = y + x\]
C. \[\frac{{dy}}{{dx}} = \frac{y}{x}\]
D. None of these
Answer
161.1k+ views
Hint:
First, use the formula \[(y - {y_1}) = m(x - {x_1})\] to write the equation of the line passing through the origin. Then, with respect to \[x\], differentiate both sides, and finally, substitute the value of \[m\] in it. The differential equation of the family of all straight lines passing through the origin must be found here. We know that in order to find the differential equation of a family of curves, we must first find the equation of the curve.
Formula use:
Equation of line passing through the point \[({x_1},{y_1})\]
\[(y - {y_1}) = m(x - {x_1})\]
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
Complete step-by-step solution:
We know that the equation of the line with slope\[m\]and the point \[({x_1},{y_1})\] which it passes through is\[(y - {y_1}) = m(x - {x_1})\]
So, by substituting \[{y_1} = 0\] and \[{x_1} = 0\] in the above equation, we get the equation of the line passing through the origin as,
\[(y - 0) = m(x - 0)\]
This can be also written as,
\[y = mx \ldots ....(1)\]
Now, we divide the either side of the equation (1) by m:
\[\frac{y}{x} = m.........(2)\]
We know that, according to product rule of differentiation:
Since, \[\frac{d}{{dx}}(f.g) = g\left( {\frac{{df}}{{dx}}} \right) + f.\left( {\frac{{dg}}{{dx}}} \right)\]
Now, by differentiating both sides of equation (1) with respect to \[x\], we get
\[\frac{d}{{dx}} = m\left( {\frac{{dx}}{{dy}}} \right) + x\left( {\frac{{dm}}{{dx}}} \right)\]
For a particular value of x and y, we know that\[m\]is constant.
So, \[\left( {\frac{{dm}}{{dx}}} \right) = 0\].
Thus, we obtain
\[\frac{{dy}}{{dx}} = m(1) + x(0)\]
Now, by substituting the value of \[m\] from equation (2), we obtain
\[\frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right)\]
We have to subtract \[\frac{y}{x}\] from both sides of the above equation, we obtain
\[\frac{{dy}}{{dx}} - \left( {\frac{y}{x}} \right) = 0\]
Rewrite the above equation explicitly by having all the terms on one side:
\[\frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right)\]
Therefore, the differential equation of all straight lines passing through the origin is \[\frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right)\]
Hence, the option C is correct.
Note:
Students must remember that in order to find the differential equation of any curve, they must remove all constants from the equation, just as we did in the above solution. Students can also test this differential equation by changing the value of \[\frac{{dy}}{{dx}}\] in the differential equation and determining whether or not the original equation of the curve is obtained.
First, use the formula \[(y - {y_1}) = m(x - {x_1})\] to write the equation of the line passing through the origin. Then, with respect to \[x\], differentiate both sides, and finally, substitute the value of \[m\] in it. The differential equation of the family of all straight lines passing through the origin must be found here. We know that in order to find the differential equation of a family of curves, we must first find the equation of the curve.
Formula use:
Equation of line passing through the point \[({x_1},{y_1})\]
\[(y - {y_1}) = m(x - {x_1})\]
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
Complete step-by-step solution:
We know that the equation of the line with slope\[m\]and the point \[({x_1},{y_1})\] which it passes through is\[(y - {y_1}) = m(x - {x_1})\]
So, by substituting \[{y_1} = 0\] and \[{x_1} = 0\] in the above equation, we get the equation of the line passing through the origin as,
\[(y - 0) = m(x - 0)\]
This can be also written as,
\[y = mx \ldots ....(1)\]
Now, we divide the either side of the equation (1) by m:
\[\frac{y}{x} = m.........(2)\]
We know that, according to product rule of differentiation:
Since, \[\frac{d}{{dx}}(f.g) = g\left( {\frac{{df}}{{dx}}} \right) + f.\left( {\frac{{dg}}{{dx}}} \right)\]
Now, by differentiating both sides of equation (1) with respect to \[x\], we get
\[\frac{d}{{dx}} = m\left( {\frac{{dx}}{{dy}}} \right) + x\left( {\frac{{dm}}{{dx}}} \right)\]
For a particular value of x and y, we know that\[m\]is constant.
So, \[\left( {\frac{{dm}}{{dx}}} \right) = 0\].
Thus, we obtain
\[\frac{{dy}}{{dx}} = m(1) + x(0)\]
Now, by substituting the value of \[m\] from equation (2), we obtain
\[\frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right)\]
We have to subtract \[\frac{y}{x}\] from both sides of the above equation, we obtain
\[\frac{{dy}}{{dx}} - \left( {\frac{y}{x}} \right) = 0\]
Rewrite the above equation explicitly by having all the terms on one side:
\[\frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right)\]
Therefore, the differential equation of all straight lines passing through the origin is \[\frac{{dy}}{{dx}} = \left( {\frac{y}{x}} \right)\]
Hence, the option C is correct.
Note:
Students must remember that in order to find the differential equation of any curve, they must remove all constants from the equation, just as we did in the above solution. Students can also test this differential equation by changing the value of \[\frac{{dy}}{{dx}}\] in the differential equation and determining whether or not the original equation of the curve is obtained.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
