Problem 103
Question
The differential equation for the family of curves \(x^{2}+y^{2}-2 a y=0\), where \(a\) is an arbitrary constant is [2004] (A) \(2\left(x^{2}-y^{2}\right) y^{\prime}=x y\) (B) \(2\left(x^{2}+y^{2}\right) y^{\prime}=x y\) (C) \(\left(x^{2}-y^{2}\right) y^{\prime}=2 x y\) (D) \(\left(x^{2}+y^{2}\right) y^{\prime}=2 x y\)
Step-by-Step Solution
Verified Answer
The differential equation is \((x^2 + y^2) y' = 2xy\), option (D).
1Step 1: Recognize the curve equation
The given family of curves is represented by the equation \(x^2 + y^2 - 2ay = 0\).
2Step 2: Solve for y
Rewriting the equation, we isolate \(y\) by moving terms around:\[ y^2 - 2ay + x^2 = 0 \]Comparing this with the standard quadratic form \(y^2 - by + c = 0\), where \(b = 2a\) and \(c = x^2\), allows us to identify it as a quadratic in \(y\).
3Step 3: Find the derivative using implicit differentiation
To find \(y'\), we first differentiate the original curve equation implicitly with respect to \(x\):- Differentiate \(x^2\) to get \(2x\).- Differentiate \(y^2\) to get \(2y\cdot y'\).- Differentiate \(-2ay\) to get \(-2a y'\).Thus, we have:\[ 2x + 2yy' - 2a y' = 0 \]
4Step 4: Gather like terms and simplify
Simplify the equation from differentiation to solve for \(y'\):- Combine terms involving \(y'\):\[ 2yy' - 2a y' = -2x \] - Factor \(y'\) from the left side: \[ (2y - 2a)y' = -2x \]
5Step 5: Express y' in terms of x and y
Divide both sides by \((2y - 2a)\) to solve for \(y'\):\[ y' = \frac{-2x}{2y - 2a} \] Simplify further by factoring out the 2:\[ y' = \frac{-x}{y - a} \]
6Step 6: Substitute a from the original equation
From Step 1, rewrite \(x^2 + y^2 = 2ay\) as \(a = \frac{x^2 + y^2}{2y}\).Substitute \(a\) back into the expression for \(y'\):\[ y' = \frac{-x}{y - \frac{x^2 + y^2}{2y}} \]
7Step 7: Simplify the expression for y'
Simplify the expression for \(y'\):\[ y' = \frac{-x}{\frac{2y^2 - (x^2 + y^2)}{2y}} \] \[ y' = \frac{-2xy}{2y^2 - x^2 - y^2} \] Finally, invert both sides and distribute terms on the denominator:\[ 2(x^2 + y^2)y' = 2xy \]We recognize this form to match option (D).
Key Concepts
Implicit DifferentiationQuadratic EquationsFamily of Curves
Implicit Differentiation
When dealing with equations of curves that incorporate variables raised to various powers, it's often necessary to use implicit differentiation to find their derivatives. Implicit differentiation is a technique used when it's difficult or impossible to solve for one variable in terms of others within an equation.
It involves differentiating each term of the equation with respect to a certain variable, often 'x', while treating other variables, like 'y', as functions of 'x'. This process allows us to find derivatives without needing an explicit expression for 'y'.
In our exercise, the curve equation is given by:
It involves differentiating each term of the equation with respect to a certain variable, often 'x', while treating other variables, like 'y', as functions of 'x'. This process allows us to find derivatives without needing an explicit expression for 'y'.
In our exercise, the curve equation is given by:
- \(x^2 + y^2 - 2ay = 0\)
- The derivative of \(x^2\) is \(2x\).
- The derivative of \(y^2\) is \(2y \cdot y'\) because \(y\) is a function of \(x\).
- The derivative of \(-2ay\) is \(-2a \cdot y'\).
Quadratic Equations
Quadratic equations are fundamental in mathematics and frequently appear in various applications, from physics to finance. They typically take the form:
Working with quadratic forms allows us to identify important properties such as roots and parabola vertex positions, which describe the behavior of the curve's shape and orientation. By solving the quadratic equation, we can express one variable explicitly in terms of others or recognize patterns that help in further analysis, such as finding the derivative using implicit differentiation.
- \(ax^2 + bx + c = 0\)
- \(y^2 - 2ay + x^2 = 0\)
Working with quadratic forms allows us to identify important properties such as roots and parabola vertex positions, which describe the behavior of the curve's shape and orientation. By solving the quadratic equation, we can express one variable explicitly in terms of others or recognize patterns that help in further analysis, such as finding the derivative using implicit differentiation.
Family of Curves
Describing how multiple curves can be represented by a single equation, a family of curves is a set of related curves where each individual curve is defined by varying a constant parameter. The equation:
Studying a family of curves aids in understanding how changes to parameters like \(a\) can affect the geometry of the curve. By considering multiple instances of \(a\), patterns or trends within the family can be analyzed, and specific properties can be explored, such as distances or intersections with axes, helping better understand the essence of differential equations in dynamic systems.
- \(x^2 + y^2 - 2ay = 0\)
Studying a family of curves aids in understanding how changes to parameters like \(a\) can affect the geometry of the curve. By considering multiple instances of \(a\), patterns or trends within the family can be analyzed, and specific properties can be explored, such as distances or intersections with axes, helping better understand the essence of differential equations in dynamic systems.
Other exercises in this chapter
Problem 101
The degree and order of the differential equation of the family of all parabolas whose axis is \(x\)-axis, are respectively (A) 2,1 (B) 1,2 (C) 3,2 (D) 2,3
View solution Problem 102
The solution of the differential equation \(\left(1+y^{2}\right)\left(x-e^{2 \tan ^{-1} y}\right) \frac{d y}{d x}=0\), is \(\quad[\mathbf{2 0 0 3}]\) (A) \((x-2
View solution Problem 104
The solution of the differential equation \(y d x+(x+\) \(\left.x^{2 y}\right) d y=0\) is (A) \(-\frac{1}{x y}=C\) (B) \(-\frac{1}{x y}+\log y=C\) (C) \(\frac{1
View solution Problem 105
The differential equation representing the family of curves \(y^{2}=2 c(x+\sqrt{c})\) where \(c>0\), is a parameter, is of order and degree as follows: \([2005]
View solution