Samacheer Kalvi Class 12 Business Maths Solutions Chapter 4 Differential Equations Exercise 4.1

Question 1. Find the order and degree of the following differential equations.
(i) \( \frac { dy }{dx} + 2y = x^3 \)
Answer:
(i) \( \frac { dy }{dx} + 2y = x^3 \)
The highest order derivative in this equation is \( \frac { dy }{dx} \).
So, the order of the differential equation is 1.
The power of this highest order derivative \( \frac { dy }{dx} \) is 1.
Thus, the degree of the differential equation is 1. The order and degree are key properties of differential equations.
In simple words: This equation has a first derivative. Its order is 1, and its power is also 1, so the degree is 1.

🎯 Exam Tip: To find the degree, the differential equation must be a polynomial in its derivatives. If not, the degree is undefined.

Question 1. Find the order and degree of the following differential equations.
(ii) \( \frac { d^3y }{dx^3} + 3(\frac { dy }{dx})^3+ 2\frac { dy }{dx} = 0 \)
Answer:
(ii) \( \frac { d^3y }{dx^3} + 3(\frac { dy }{dx})^3+ 2\frac { dy }{dx} = 0 \)
The highest order derivative in this equation is \( \frac { d^3y }{dx^3} \).
So, the order of the differential equation is 3.
The power of the highest order derivative \( \frac { d^3y }{dx^3} \) is 1.
Therefore, the degree of the differential equation is 1. Higher order derivatives describe how a function changes multiple times.
In simple words: The highest derivative is \( \frac { d^3y }{dx^3} \), making its order 3. Its power is 1, so the degree is 1.

🎯 Exam Tip: Always look for the derivative with the most differentiation steps (like d³y/dx³) to find the order first. Then find its power for the degree.

Question 1. Find the order and degree of the following differential equations.
(iii) \( \frac { d^2y }{dx^2} = \sqrt{y – \frac { dy }{dx}} \)
Answer:
(iii) \( \frac { d^2y }{dx^2} = \sqrt{y – \frac { dy }{dx}} \)
First, we need to remove the square root to make it a polynomial in terms of derivatives. We do this by squaring both sides:
\[ \left( \frac { d^2y }{dx^2} \right)^2 = y – \frac { dy }{dx} \]
The highest order derivative in the equation is \( \frac { d^2y }{dx^2} \).
So, the order of the differential equation is 2.
The power of the highest order derivative \( \frac { d^2y }{dx^2} \) is 2.
Hence, the degree of the differential equation is 2. Eliminating radicals helps in defining the degree clearly.
In simple words: Square both sides to remove the root. The highest derivative is \( \frac { d^2y }{dx^2} \), so the order is 2. Its power is 2, making the degree 2.

🎯 Exam Tip: When radicals (like square roots) are present, always raise both sides to a power to remove them before determining the degree. This ensures the equation is a polynomial in its derivatives.

Question 1. Find the order and degree of the following differential equations.
(iv) \( \frac { d^3y }{dx^3} = 0 \)
Answer:
(iv) \( \frac { d^3y }{dx^3} = 0 \)
The highest order derivative in this equation is \( \frac { d^3y }{dx^3} \).
So, the order of the differential equation is 3.
The power of the highest order derivative \( \frac { d^3y }{dx^3} \) is 1.
Therefore, the degree of the differential equation is 1. Even simple equations like this have an order and degree.
In simple words: The highest derivative is \( \frac { d^3y }{dx^3} \), so the order is 3. Its power is 1, making the degree 1.

🎯 Exam Tip: For equations set to zero, the order and degree rules still apply normally to the highest derivative present.

Question 1. Find the order and degree of the following differential equations.
(v) \( \frac { d^3y }{dx^3} + y + \left[ \frac { dy }{dx} – \frac { d^3y }{dx^3} \right]^{3/2} = 0 \)
Answer:
(v) \( \frac { d^3y }{dx^3} + y + \left[ \frac { dy }{dx} – \frac { d^3y }{dx^3} \right]^{3/2} = 0 \)
First, rearrange the equation to isolate the term with the fractional power:
\( \frac { d^3y }{dx^3} + y = - \left[ \frac { dy }{dx} – \frac { d^3y }{dx^3} \right]^{3/2} \)
Now, square both sides to remove the fractional exponent (since \( (A^{3/2})^2 = A^3 \)):
\[ \left( \frac { d^3y }{dx^3} + y \right)^2 = \left( - \left[ \frac { dy }{dx} – \frac { d^3y }{dx^3} \right]^{3/2} \right)^2 \]
\[ \left( \frac { d^3y }{dx^3} + y \right)^2 = \left[ \frac { dy }{dx} – \frac { d^3y }{dx^3} \right]^3 \]
The highest order derivative in the equation is \( \frac { d^3y }{dx^3} \).
So, the order of the differential equation is 3.
The highest power of the highest order derivative \( \frac { d^3y }{dx^3} \) on the left side is 2, and on the right side, it's 3. We take the higher power.
Therefore, the degree of the differential equation is 3. Clearing fractional powers is crucial for finding the degree correctly.
In simple words: Move the term with the 3/2 power to one side, then square both sides to remove the fraction. The highest derivative is \( \frac { d^3y }{dx^3} \), so the order is 3. Its highest power after squaring is 3, making the degree 3.

🎯 Exam Tip: Always make sure the differential equation is free from radicals and fractional powers of derivatives before determining the degree. This usually involves squaring or cubing both sides.

Question 1. Find the order and degree of the following differential equations.
(vi) \( (2 – y'')^2 = (y'')^2 + 2y' \)
Answer:
(vi) \( (2 – y'')^2 = (y'')^2 + 2y' \)
Expand the left side of the equation:
\( 2^2 - 2(2)(y'') + (y'')^2 = (y'')^2 + 2y' \)
\( 4 - 4y'' + (y'')^2 = (y'')^2 + 2y' \)
Subtract \( (y'')^2 \) from both sides:
\( 4 - 4y'' = 2y' \)
The highest order derivative in this simplified equation is \( y'' \) (which is \( \frac{d^2y}{dx^2} \)).
So, the order of the differential equation is 2.
The power of this highest order derivative \( y'' \) is 1. (Even though \( (y'')^2 \) appeared initially, it cancelled out, so the highest power of the remaining highest order derivative is 1).
Therefore, the degree of the differential equation is 1. Simplifying the equation helps to correctly identify the actual highest order derivative and its power.
In simple words: Expand the equation and simplify. The highest derivative left is \( y'' \), so the order is 2. Its power is 1, so the degree is 1.

🎯 Exam Tip: Simplify the equation as much as possible, expanding all terms and combining like derivatives, before determining the order and degree. Sometimes terms with higher powers cancel out.

Question 1. Find the order and degree of the following differential equations.
(vii) \( \left(\frac { dy }{dx}\right)^3 + y = x – \frac { dx }{dy} \)
Answer:
(vii) \( \left(\frac { dy }{dx}\right)^3 + y = x – \frac { dx }{dy} \)
First, rewrite \( \frac { dx }{dy} \) as \( \frac { 1 }{ (dy/dx) } \):
\( \left(\frac { dy }{dx}\right)^3 + y = x – \frac { 1 }{ (dy/dx) } \)
Multiply the entire equation by \( \frac { dy }{dx} \) to remove the fraction involving the derivative:
\( \frac { dy }{dx} \left(\frac { dy }{dx}\right)^3 + y \frac { dy }{dx} = x \frac { dy }{dx} – \frac { 1 }{ (dy/dx) } \frac { dy }{dx} \)
This simplifies to:
\( \left(\frac { dy }{dx}\right)^4 + y \frac { dy }{dx} = x \frac { dy }{dx} – 1 \)
The highest order derivative in this equation is \( \frac { dy }{dx} \).
So, the order of the differential equation is 1.
The highest power of the highest order derivative \( \frac { dy }{dx} \) is 4.
Therefore, the degree of the differential equation is 4. Making the equation a polynomial in derivatives is essential for finding the degree.
In simple words: Change \( \frac{dx}{dy} \) to \( \frac{1}{(dy/dx)} \). Then multiply everything by \( \frac{dy}{dx} \) to remove the fraction. The highest derivative is \( \frac{dy}{dx} \), so the order is 1. Its highest power is 4, making the degree 4.

🎯 Exam Tip: If the derivative appears in the denominator, multiply the entire equation by that derivative to clear it. This allows you to identify the highest power of the highest order derivative correctly.

Question 2. Find the differential equation of the following
(i) \( y = cx + c - c^3 \)
Answer:
(i) \( y = cx + c - c^3 \) ....... (1)
We need to remove the arbitrary constant 'c' from this equation. We start by differentiating with respect to \( x \).
\( \frac{dy}{dx} = c \cdot 1 + 0 - 0 \)
\( \implies \frac{dy}{dx} = c \) ....... (2)
Now, substitute the value of \( c \) from equation (2) into equation (1). This eliminates 'c'.
\( y = \left( \frac{dy}{dx} \right) x + \left( \frac{dy}{dx} \right) - \left( \frac{dy}{dx} \right)^3 \)
This is the required differential equation. The process of eliminating arbitrary constants leads to the differential equation.
In simple words: First, find the derivative of \( y \) to get \( c \). Then, put this \( c \) back into the original equation. This removes \( c \) and gives you the differential equation.

🎯 Exam Tip: To find a differential equation, differentiate the given equation with respect to x. Then, use this derivative to substitute and eliminate any arbitrary constants from the original equation.

Question 2. Find the differential equation of the following
(ii) \( y = c (x - c)^2 \)
Answer:
(ii) \( y = c (x - c)^2 \) ....... (1)
We need to eliminate the constant 'c'. Expand the equation:
\( y = c (x^2 - 2cx + c^2) \)
\( y = cx^2 - 2c^2x + c^3 \)
Now, differentiate equation (1) with respect to \( x \):
\( \frac{dy}{dx} = c \cdot 2(x - c) \cdot 1 \)
\( \implies \frac{dy}{dx} = 2c(x - c) \) ....... (2)
From equation (1), we have \( \frac{y}{(x-c)^2} = c \).
From equation (2), we have \( \frac{dy/dx}{2(x-c)} = c \).
Equating these two expressions for \( c \):
\( \frac{y}{(x-c)^2} = \frac{dy/dx}{2(x-c)} \)
Since \( (x-c) \neq 0 \), we can multiply both sides by \( (x-c) \):
\( \frac{y}{x-c} = \frac{dy}{dx} \frac{1}{2} \)
\( \implies 2y = (x-c) \frac{dy}{dx} \)
\( \implies x-c = \frac{2y}{(dy/dx)} \) ....... (3)
Now substitute (3) into equation (2) to find 'c':
\( \frac{dy}{dx} = 2c \left( \frac{2y}{(dy/dx)} \right) \)
\( \frac{dy}{dx} = c \frac{4y}{(dy/dx)} \)
\( c = \frac{(dy/dx)^2}{4y} \) ....... (4)
Substitute (3) and (4) back into equation (1):
\( y = c (x - c)^2 \)
\( y = \left( \frac{(dy/dx)^2}{4y} \right) \left( \frac{2y}{(dy/dx)} \right)^2 \)
\( y = \left( \frac{(dy/dx)^2}{4y} \right) \left( \frac{4y^2}{(dy/dx)^2} \right) \)
\( y = y \)
This equality shows our substitution is consistent, but it doesn't give the differential equation directly. Let's try substituting `c` from (4) into (3) and then using (3) in the original equation (1). From (3): \( x - c = \frac{2y}{dy/dx} \) So, \( c = x - \frac{2y}{dy/dx} \) Now substitute this `c` back into the original equation (1): \( y = \left( x - \frac{2y}{dy/dx} \right) \left( \frac{2y}{dy/dx} \right)^2 \)
\( y = \left( x - \frac{2y}{dy/dx} \right) \frac{4y^2}{(dy/dx)^2} \)
\( y = \frac{4y^2x}{(dy/dx)^2} - \frac{8y^3}{(dy/dx)^3} \)
Divide by y (assuming y ≠ 0):
\( 1 = \frac{4yx}{(dy/dx)^2} - \frac{8y^2}{(dy/dx)^3} \)
Multiply by \( (dy/dx)^3 \):
\( \left( \frac{dy}{dx} \right)^3 = 4xy \left( \frac{dy}{dx} \right) - 8y^2 \)
\( \implies \left( \frac{dy}{dx} \right)^3 - 4xy \frac{dy}{dx} + 8y^2 = 0 \). This is the required differential equation. Eliminating constants often involves algebraic manipulation.
In simple words: Take the first equation and find its derivative. Use both the original equation and its derivative to find "c" in terms of \( y \) and \( \frac{dy}{dx} \). Replace "c" in the original equation with this expression. Simplify the result to get the final differential equation.

🎯 Exam Tip: When eliminating a constant, you might need to use both the original equation and its derivative to form expressions for the constant or related terms. Sometimes, algebraic simplification can be extensive.

Question 2. Find the differential equation of the following
(iii) \( xy = c^2 \)
Answer:
(iii) \( xy = c^2 \) ....... (1)
We need to eliminate the constant 'c'. Differentiate equation (1) with respect to \( x \). Use the product rule for \( xy \).
\( x \frac{dy}{dx} + y \cdot 1 = 0 \)
\( \implies x \frac{dy}{dx} + y = 0 \)
This is the required differential equation. For equations with easily isolated constants, differentiation might directly yield the answer.
In simple words: Differentiate the equation with respect to \( x \). Since \( c^2 \) is a constant, its derivative is zero. The resulting equation is the differential equation.

🎯 Exam Tip: Remember the product rule for differentiation when variables are multiplied, and that the derivative of a constant is always zero. This simplifies the process of elimination.

Question 2. Find the differential equation of the following
(iv) \( x^2 + y^2 = a^2 \)
Answer:
(iv) \( x^2 + y^2 = a^2 \) ....... (1)
We need to eliminate the arbitrary constant 'a'. Differentiate equation (1) with respect to \( x \).
\( 2x + 2y \frac{dy}{dx} = 0 \)
Divide the entire equation by 2:
\( x + y \frac{dy}{dx} = 0 \)
This is the required differential equation. This equation represents all circles centered at the origin.
In simple words: Take the derivative of both sides. Since \( a^2 \) is a constant, it becomes zero. Then, simplify the equation to get the final answer.

🎯 Exam Tip: For equations like \( x^2 + y^2 = a^2 \), differentiating directly will often eliminate the constant if it's not intricately linked with the variables after differentiation. Always simplify the derivative before concluding.

Question 3. Form the differential equation by eliminating \( \alpha \) and \( \beta \) from \( (x – \alpha)^2 + (y – \beta)^2 = r^2 \)
Answer:
Given the equation: \( (x – \alpha)^2 + (y – \beta)^2 = r^2 \) ....... (1)
Here, \( \alpha \) and \( \beta \) are arbitrary constants to be eliminated. The constant \( r \) is also present. This equation represents a family of circles with radius \( r \).
Differentiate equation (1) with respect to \( x \):
\( 2(x – \alpha) \cdot 1 + 2(y – \beta) \frac{dy}{dx} = 0 \)
Divide by 2:
\( (x – \alpha) + (y – \beta) \frac{dy}{dx} = 0 \) ....... (2)
Now, differentiate equation (2) again with respect to \( x \):
\( 1 + \left( (y – \beta) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{dy}{dx} \right) = 0 \) (using product rule for the second term)
\( \implies 1 + (y – \beta) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0 \) ....... (3)
From equation (3), we can find \( (y – \beta) \):
\( (y – \beta) \frac{d^2y}{dx^2} = - \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] \)
\( \implies (y – \beta) = - \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] }{ \frac{d^2y}{dx^2} } \) ....... (4)
Now, substitute \( (y – \beta) \) from (4) into (2) to find \( (x – \alpha) \):
\( (x – \alpha) + \left( - \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] }{ \frac{d^2y}{dx^2} } \right) \frac{dy}{dx} = 0 \)
\( \implies (x – \alpha) = \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] }{ \frac{d^2y}{dx^2} } \frac{dy}{dx} \) ....... (5)
Finally, substitute the values of \( (x – \alpha) \) from (5) and \( (y – \beta) \) from (4) into the original equation (1):
\( \left( \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] }{ \frac{d^2y}{dx^2} } \frac{dy}{dx} \right)^2 + \left( - \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] }{ \frac{d^2y}{dx^2} } \right)^2 = r^2 \)
\[ \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^2 \left( \frac{dy}{dx} \right)^2 }{ \left( \frac{d^2y}{dx^2} \right)^2 } + \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^2 }{ \left( \frac{d^2y}{dx^2} \right)^2 } = r^2 \] Take \( \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^2 }{ \left( \frac{d^2y}{dx^2} \right)^2 } \) as a common factor:
\[ \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^2 }{ \left( \frac{d^2y}{dx^2} \right)^2 } \left( \left( \frac{dy}{dx} \right)^2 + 1 \right) = r^2 \] \[ \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^3 }{ \left( \frac{d^2y}{dx^2} \right)^2 } = r^2 \] This is the required differential equation. Eliminating multiple constants requires differentiation multiple times.
In simple words: Differentiate the original equation two times to get rid of \( \alpha \) and \( \beta \). Use the resulting equations to find expressions for \( (x-\alpha) \) and \( (y-\beta) \). Put these back into the first equation and simplify. This gives you the differential equation without the constants.

🎯 Exam Tip: When eliminating two constants, you will generally need to differentiate the equation twice. For three constants, differentiate three times, and so on.

Question 4. Find the differential equation of the family of all straight lines passing through the origin.
Answer:
The general equation for a family of lines passing through the origin is:
\( y = mx \) ....... (1)
Here, 'm' is the arbitrary constant (the slope) which needs to be eliminated. Straight lines passing through the origin are characterized by having no y-intercept.
Differentiate equation (1) with respect to \( x \):
\( \frac{dy}{dx} = m \cdot 1 \)
\( \implies \frac{dy}{dx} = m \) ....... (2)
Now, substitute the value of 'm' from equation (2) into equation (1):
\( y = \left( \frac{dy}{dx} \right) x \)
This is the required differential equation. It describes all lines that pass through (0,0).
In simple words: Start with the equation for a line through the origin, \( y = mx \). Take its derivative to find \( m \). Then put this \( m \) back into the original equation to get the differential equation.

🎯 Exam Tip: For simple linear equations, one differentiation is usually enough to eliminate a single arbitrary constant, leading directly to the differential equation.

Question 5. Form the differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to the x-axis.
Answer:
The equation of a parabola with latus rectum 4a and whose axis is parallel to the x-axis is given by:
\( (y – k)^2 = 4a (x – h) \) ....... (1)
Here, 'h' and 'k' are arbitrary constants (the coordinates of the vertex), which need to be eliminated. 'a' is a fixed value related to the latus rectum, not an arbitrary constant in this context.
Differentiate equation (1) with respect to \( x \):
\( 2(y – k) \frac{dy}{dx} = 4a \cdot 1 \)
\( \implies (y – k) \frac{dy}{dx} = 2a \) ....... (2)
Now, differentiate equation (2) again with respect to \( x \):
Apply the product rule for \( (y-k) \frac{dy}{dx} \):
\( (y – k) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{dy}{dx} = 0 \)
\( \implies (y – k) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0 \) ....... (3)
From equation (3), we can find \( (y – k) \):
\( (y – k) \frac{d^2y}{dx^2} = - \left( \frac{dy}{dx} \right)^2 \)
\( \implies (y – k) = - \frac{ \left( \frac{dy}{dx} \right)^2 }{ \frac{d^2y}{dx^2} } \) ....... (4)
Substitute \( (y – k) \) from equation (4) into equation (2):
\( \left( - \frac{ \left( \frac{dy}{dx} \right)^2 }{ \frac{d^2y}{dx^2} } \right) \frac{dy}{dx} = 2a \)
\( \implies - \frac{ \left( \frac{dy}{dx} \right)^3 }{ \frac{d^2y}{dx^2} } = 2a \) ....... (5)
Substitute the value of \( 2a \) from (5) back into equation (2):
\( (y – k) \frac{dy}{dx} = - \frac{ \left( \frac{dy}{dx} \right)^3 }{ \frac{d^2y}{dx^2} } \)
Now, substitute \( (y – k) \) from (4) into this equation:
\( \left( - \frac{ \left( \frac{dy}{dx} \right)^2 }{ \frac{d^2y}{dx^2} } \right) \frac{dy}{dx} = - \frac{ \left( \frac{dy}{dx} \right)^3 }{ \frac{d^2y}{dx^2} } \)
This simplifies to \( - \frac{ \left( \frac{dy}{dx} \right)^3 }{ \frac{d^2y}{dx^2} } = - \frac{ \left( \frac{dy}{dx} \right)^3 }{ \frac{d^2y}{dx^2} } \). This means our derived expressions are consistent.
The problem asks for the differential equation that *represents* the parabolas. Since \( 2a \) is a constant, equation (5) already holds the key. We can rewrite it as:
\( \left( \frac{dy}{dx} \right)^3 + 2a \left( \frac{d^2y}{dx^2} \right) = 0 \)
This equation describes the family of parabolas. The elimination of 'h' and 'k' is complete. 'a' is part of the problem definition. *(Self-correction: The prompt implied 4a is a constant, so the final differential equation should contain 'a'. The provided steps lead to a relation between derivatives and 'a'.)*
In simple words: Start with the standard equation for these parabolas. Differentiate it twice to get rid of the constants 'h' and 'k'. The final equation will link the derivatives and the constant 'a', which defines the latus rectum.

🎯 Exam Tip: When given an equation with multiple arbitrary constants, you will need to differentiate the equation as many times as there are constants to eliminate them. Look for opportunities to substitute back into previous equations.

Question 6. Find the differential equation of all circles passing through the origin and having their centers on the y-axis, \( x^2 + (y – k)^2 = r^2 \)
Answer:
The equation of a circle with center \( (h, k) \) and radius \( r \) is \( (x – h)^2 + (y – k)^2 = r^2 \).
Since the center is on the y-axis, the x-coordinate of the center is 0. So, \( h = 0 \).
The equation becomes: \( (x – 0)^2 + (y – k)^2 = r^2 \)
\( \implies x^2 + (y – k)^2 = r^2 \) ....... (1)
Also, the circle passes through the origin \( (0, 0) \). Substitute \( x=0, y=0 \) into equation (1):
\( 0^2 + (0 – k)^2 = r^2 \)
\( \implies k^2 = r^2 \)
Since \( r \) is a radius, it's positive, so \( r = |k| \). We can replace \( r^2 \) with \( k^2 \) in equation (1):
\( x^2 + (y – k)^2 = k^2 \)
Now, expand this equation:
\( x^2 + y^2 - 2yk + k^2 = k^2 \)
Subtract \( k^2 \) from both sides:
\( x^2 + y^2 - 2yk = 0 \)
\( \implies x^2 + y^2 = 2yk \) ....... (2)
Now, we need to eliminate the single arbitrary constant 'k' from equation (2). Differentiate equation (2) with respect to \( x \):
\( 2x + 2y \frac{dy}{dx} = 2k \frac{dy}{dx} \)
Divide the entire equation by 2:
\( x + y \frac{dy}{dx} = k \frac{dy}{dx} \)
\( \implies k = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx}} \) ....... (3)
Substitute the value of 'k' from equation (3) back into equation (2):
\( x^2 + y^2 = 2y \left( \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx}} \right) \)
\( x^2 + y^2 = 2y \left( \frac{x}{\frac{dy}{dx}} + y \right) \)
\( x^2 + y^2 = \frac{2xy}{\frac{dy}{dx}} + 2y^2 \)
Multiply the entire equation by \( \frac{dy}{dx} \):
\( (x^2 + y^2) \frac{dy}{dx} = 2xy + 2y^2 \frac{dy}{dx} \)
Now, rearrange terms to group \( \frac{dy}{dx} \):
\( x^2 \frac{dy}{dx} + y^2 \frac{dy}{dx} - 2y^2 \frac{dy}{dx} = 2xy \)
\( x^2 \frac{dy}{dx} - y^2 \frac{dy}{dx} = 2xy \)
Factor out \( \frac{dy}{dx} \):
\( (x^2 - y^2) \frac{dy}{dx} = 2xy \)
This is the required differential equation. Understanding geometric properties helps simplify the initial equation.
In simple words: First, use the given conditions (center on y-axis, passes through origin) to simplify the circle's equation and eliminate 'r'. This leaves one constant 'k'. Then, differentiate this new equation and use it to remove 'k', leading to the final differential equation.

🎯 Exam Tip: When given conditions (like "passes through origin" or "center on y-axis"), use them to simplify the initial equation by eliminating constants before you even start differentiating. This reduces the number of constants you need to eliminate later.

Question 7. Find the differential equation of the family of parabola with foci at the origin and axis along the x-axis, \( y^2 = 4a (x + a) \)
Answer:
The equation of the parabola with foci at the origin and axis along the x-axis is:
\( y^2 = 4a (x + a) \) ....... (1)
Here, 'a' is the arbitrary constant that needs to be eliminated. This form of parabola is unique for its focal point and axis alignment.
Differentiate equation (1) with respect to \( x \):
\( 2y \frac{dy}{dx} = 4a (1 + 0) \)
\( \implies 2y \frac{dy}{dx} = 4a \)
Divide by 2:
\( y \frac{dy}{dx} = 2a \) ....... (2)
Now, we have a value for \( 2a \). Substitute \( 2a \) from equation (2) back into equation (1):
\( y^2 = (2a) (2x + 2a) \)
From (2), \( 2a = y \frac{dy}{dx} \). Substitute this into the previous step:
\( y^2 = \left( y \frac{dy}{dx} \right) \left( 2x + y \frac{dy}{dx} \right) \)
Assuming \( y \neq 0 \), we can divide both sides by \( y \):
\( y = \left( \frac{dy}{dx} \right) \left( 2x + y \frac{dy}{dx} \right) \)
Now, distribute \( \frac{dy}{dx} \) on the right side:
\( y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2 \)
This is the required differential equation. It describes all parabolas fitting the given criteria.
In simple words: Start with the given equation for the parabola. Differentiate it to get an expression for 'a'. Then, put this 'a' back into the original equation. Simplify it to get the differential equation, which will not contain 'a'.

🎯 Exam Tip: For problems involving specific families of curves, always start by writing down the standard equation and identifying the arbitrary constants to be eliminated. Careful substitution is key after differentiation.