Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.9

Choose the Most Suitable Answer From the Given Four Alternatives:

Question 1. The order and degree of the differential equation \( \frac { d^2y }{ dx^2 } + (\frac { dy }{ dx })^{1/3} + x^{1/4} = 0 \) are respectively
(a) 2, 3
(b) 3, 3
(c) 2, 6
(d) 2, 4
Answer: (a) 2, 3
In simple words: To find the degree, we first need to remove any fractional powers from the derivatives. Rewrite the equation to isolate the term with the fractional power, then raise both sides to a power that removes the fraction. After doing this, the highest derivative is \( \frac{d^2y}{dx^2} \), which means the order is 2. The power of this highest derivative after removing fractions is 3, which is the degree. The degree of a differential equation is only defined when the equation is a polynomial in its derivatives.

🎯 Exam Tip: Remember to rationalize the differential equation to remove fractional powers of derivatives before determining the degree. This step is crucial for accurate degree identification.

Question 2. The differential equation representing the family of curves \( y = A \cos (x + B) \), where A and B are parameters, is
(a) \( \frac { d^2y }{ dx^2 } – y = 0 \)
(b) \( \frac { d^2y }{ dx^2 } + y = 0 \)
(c) \( \frac { d^2y }{ dx^2 } = 0 \)
(d) \( \frac { d^2x }{ dy^2 } = 0 \)
Answer: (b) \( \frac { d^2y }{ dx^2 } + y = 0 \)
In simple words: We have two unknown constants (A and B) in the given equation. So, we need to differentiate the equation two times to eliminate these constants. First, find \( \frac{dy}{dx} \), then find \( \frac{d^2y}{dx^2} \). You will notice that the second derivative is equal to \( -y \), which leads to the final differential equation \( \frac{d^2y}{dx^2} + y = 0 \). This type of equation is often seen in physics to model simple harmonic motion.

🎯 Exam Tip: The order of the differential equation will always be equal to the number of arbitrary constants in the original equation, as each differentiation helps eliminate one constant.

Question 3. The order and degree of the differential equation, \( \sqrt { \sin x } (dx + dy) = \sqrt { \cos x } (dx - dy) \) is
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
Answer: (c) 1, 1
In simple words: First, divide all parts of the equation by \( dx \) to get \( \frac{dy}{dx} \). Then, gather all terms containing \( \frac{dy}{dx} \) on one side and the rest on the other. After simplifying, you will find that the highest derivative present is \( \frac{dy}{dx} \) (first order), and it is raised to the power of 1 (first degree). This equation is a first-order, first-degree differential equation, meaning it involves only the first derivative and that derivative is raised to the power of one.

🎯 Exam Tip: Always simplify the differential equation fully to isolate the derivative term before determining its order and degree, ensuring no implicit powers remain.

Question 4. The order of the differential equation of all circles with centre at (h, k) and radius a
(a) 2
(b) 3
(c) 4
(d) 1
Answer: (a) 2
In simple words: The general equation for a circle is \( (x - h)^2 + (y - k)^2 = a^2 \). In this equation, \( h \) and \( k \) are the coordinates of the center, and \( a \) is the radius. The problem states "radius a," meaning \( a \) is a fixed value and not an arbitrary constant to be eliminated. Therefore, there are only two arbitrary constants, \( h \) and \( k \). The order of a differential equation is determined by the number of arbitrary constants present in the original family of curves. Thus, the order of the differential equation will be 2.

🎯 Exam Tip: Count the number of independent arbitrary constants in the equation of the family of curves; this count directly gives the order of the resulting differential equation.

Question 5. The differential equation of the family of curves \( y = Ae^x + Be^{-x} \), where A and B are arbitrary constants is
(a) \( \frac { d^2y }{ dx^2 } + y = 0 \)
(b) \( \frac { d^2y }{ dx^2 } -y=0 \)
(c) \( \frac { dy }{ dx } + y = 0 \)
(d) \( \frac { dy }{ dx } -y=0 \)
Answer: (b) \( \frac { d^2y }{ dx^2 } -y=0 \)
In simple words: Since there are two arbitrary constants, A and B, we need to differentiate the given equation two times. After the first differentiation, we get \( \frac{dy}{dx} = Ae^x - Be^{-x} \). Differentiating a second time gives \( \frac{d^2y}{dx^2} = Ae^x + Be^{-x} \). Notice that this second derivative is exactly equal to the original equation for \( y \). So, \( \frac{d^2y}{dx^2} = y \), which can be written as \( \frac{d^2y}{dx^2} - y = 0 \). This particular differential equation describes systems where the rate of change of growth or decay is proportional to the quantity itself, such as in population dynamics or radioactive decay.

🎯 Exam Tip: When eliminating arbitrary constants, differentiate the equation a number of times equal to the number of constants, then substitute back into the equations to eliminate them completely.

Question 6. The general solution of the differential equation \( \frac { dy }{ dx } = \frac { y }{ x } \) is
(a) \( xy = k \)
(b) \( y = k \log x \)
(c) \( y = kx \)
(d) \( \log y = kr \)
Answer: (c) \( y = kx \)
In simple words: This is a separable differential equation. First, rearrange the equation so that all terms involving \( y \) are on one side with \( dy \), and all terms involving \( x \) are on the other side with \( dx \). This gives \( \frac{dy}{y} = \frac{dx}{x} \). Next, integrate both sides. The integral of \( \frac{1}{y} \) is \( \log|y| \), and the integral of \( \frac{1}{x} \) is \( \log|x| \). After integrating, introduce a constant of integration, preferably as \( \log|k| \) to simplify the logarithm properties. Combining the logarithms gives \( \log|y| = \log|kx| \), which means \( y = kx \). This is a separable differential equation, which means you can arrange it so that all terms involving 'y' and 'dy' are on one side, and all terms involving 'x' and 'dx' are on the other.

🎯 Exam Tip: For separable differential equations, always remember to include the constant of integration after performing the indefinite integral, and choose its form (e.g., \( C \) or \( \log C \)) to simplify the final solution.

Question 7. The solution of the differential equation \( 2x\frac { dy }{ dx } – y = 3 \) represents
(a) straight lines
(b) circles
(c) parabola
(d) ellipse
Answer: (c) parabola
In simple words: Start by rearranging the differential equation to separate the variables. Move \( y \) and the constant term to one side and \( x \) terms to the other, making it \( \frac{dy}{3+y} = \frac{dx}{2x} \). Integrate both sides. This results in \( \log|3+y| = \frac{1}{2}\log|x| + \log|k| \). Simplify the logarithmic terms to get \( \log|3+y| = \log|k\sqrt{x}| \). After removing the logarithms, square both sides to eliminate the square root, giving \( (3+y)^2 = k^2x \). This equation matches the standard form of a parabola, which is \( Y^2 = AX \). A parabola is a conic section formed by slicing a cone with a plane parallel to its side, and its general form \( (y+c)^2 = kx \) or \( (x+c)^2 = ky \) characterizes its U-shape.

🎯 Exam Tip: After solving a differential equation, rearrange the solution into a standard form (e.g., \( y = mx+c \) for a line, \( (x-h)^2 + (y-k)^2 = r^2 \) for a circle) to easily identify the type of curve it represents.

Question 8. The solution of \( \frac { dy }{ dx } + p(x) y = 0 \) is
(a) \( y = c e^{\int pdx} \)
(b) \( y = c e^{-\int pdx} \)
(c) \( x = c e^{-\int pdy} \)
(d) \( x = c e^{\int pdy} \)
Answer: (b) \( y = c e^{-\int pdx} \)
In simple words: This is a first-order linear differential equation, which can be solved by separating variables. Move the \( p(x)y \) term to the right side, giving \( \frac{dy}{dx} = -p(x)y \). Then, separate \( y \) and \( dy \) from \( x \) and \( dx \): \( \frac{dy}{y} = -p(x)dx \). Integrate both sides. The integral of \( \frac{1}{y} \) is \( \log|y| \). After integration, you'll get \( \log|y| = -\int p(x)dx + \log|c| \). Rearrange to isolate \( y \), which results in \( y = c e^{-\int p(x)dx} \). This equation is a homogeneous first-order linear differential equation, which is a special case of \( \frac{dy}{dx} + P(x)y = Q(x) \) where \( Q(x) = 0 \).

🎯 Exam Tip: Remember the standard form and solution for a first-order linear differential equation, \( y = c e^{-\int P(x)dx} \), especially when the right-hand side \( Q(x) \) is zero.

Question 9. The integrating factor of the differential equation \( \frac { dy }{ dx } + y = \frac { 1+y }{ x } \) is
(a) \( \frac { x }{ e^x } + y = 0 \)
(b) \( \frac { e^x }{ x } - y = 0 \)
(c) \( \lambda e^x \)
(d) \( e^x \)
Answer: (b) \( \frac { e^x }{ x } - y = 0 \)
In simple words: To find the integrating factor, first rewrite the given differential equation into the standard linear form: \( \frac{dy}{dx} + P(x)y = Q(x) \). This involves rearranging the terms to get \( \frac{dy}{dx} + (1 - \frac{1}{x})y = \frac{1}{x} \). Here, \( P(x) = 1 - \frac{1}{x} \). The integrating factor (IF) is then calculated as \( e^{\int P(x) dx} \). Integrating \( P(x) \) gives \( x - \log|x| \). So, the IF is \( e^{x - \log|x|} \), which simplifies to \( \frac{e^x}{x} \). An integrating factor is a function that makes a non-exact differential equation exact, allowing it to be solved more easily.

🎯 Exam Tip: Carefully rearrange the given differential equation into the standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \) before identifying \( P(x) \) to calculate the integrating factor. Ensure all terms involving \( y \) are grouped correctly.

Question 10. The Integrating factor of the differential equation \( \frac { dy }{ dx } + p(x) y = Q(x) \) is x, then p(x)
(a) \( x \)
(b) \( \frac { x^2 }{ 2 } \)
(c) \( \frac { 1 }{ x } \)
(d) \( \frac { 1 }{ x^2 } \)
Answer: (c) \( \frac { 1 }{ x } \)
In simple words: The integrating factor (IF) is given by the formula \( e^{\int p(x) dx} \). We are told that the IF is \( x \). So, we set \( e^{\int p(x) dx} = x \). To find \( p(x) \), take the natural logarithm of both sides: \( \int p(x) dx = \log x \). Then, differentiate both sides with respect to \( x \). The derivative of \( \int p(x) dx \) is \( p(x) \), and the derivative of \( \log x \) is \( \frac{1}{x} \). Therefore, \( p(x) = \frac{1}{x} \). The integrating factor transforms a first-order linear differential equation into a form that can be integrated directly.

🎯 Exam Tip: When the integrating factor is given, use the relationship \( \text{IF} = e^{\int P(x) dx} \) and apply inverse operations (logarithm and differentiation) to efficiently find the function \( P(x) \).

Question 11. The degree of the differential equation \( y(x) = 1 + \frac { dy }{ dx } + \frac { 1 }{ 1.2 } (\frac { dy }{ dx })² + \frac { 1 }{ 1.2.3 } (\frac { dy }{ dx })³ + ....... \) is
(a) 2
(b) 3
(c) 1
(d) 4
Answer: (c) 1
In simple words: The expression on the right side of the equation is the Taylor series expansion for \( e^u \), where \( u \) is \( \frac{dy}{dx} \). So, the equation can be rewritten as \( y(x) = e^{\frac{dy}{dx}} \). To find the degree, we need to express the equation as a polynomial in its derivatives. Take the natural logarithm of both sides, which gives \( \log y = \frac{dy}{dx} \). In this simplified form, the highest order derivative is \( \frac{dy}{dx} \), and its power is 1. Therefore, the order is 1, and the degree is 1. Understanding common series expansions, like that for \( e^x \), can simplify complex-looking differential equations and reveal their true order and degree.

🎯 Exam Tip: The degree of a differential equation is defined only when it can be expressed as a polynomial in its derivatives. If it involves transcendental functions (like exponential or logarithm) of derivatives, transform it first.

Question 12. If p and q are the order and degree of the differential equation \( y \frac { dy }{ dx } +x³ (\frac { d^2y }{ dx^2 }) + xy = \cos x \), then
(a) p = q
(b) p < q
(c) p > q
(d) p exists and q does not exist
Answer: (c) p > q
In simple words: To find the order and degree of a differential equation, first identify the highest derivative present. In this equation, the highest derivative is \( \frac{d^2y}{dx^2} \). The order \( p \) is the order of this highest derivative, which is 2. The degree \( q \) is the power to which this highest derivative is raised. Here, \( \frac{d^2y}{dx^2} \) is raised to the power of 1. So, \( q = 1 \). Comparing \( p \) and \( q \), we find that \( 2 > 1 \), so \( p > q \). The order of a differential equation specifies the highest derivative present in the equation, while the degree refers to the power of that highest derivative once the equation is rationalized and free of radicals.

🎯 Exam Tip: When determining order and degree, always look for the highest derivative first; its order is the overall order of the equation, and its power (after rationalization and simplification) is the degree.

Question 13. The solution of the differential equation \( \frac { dy }{ dx } + \frac { 1 }{ \sqrt{1-x^2} } = 0 \) is
(a) \( y + \sin^{-1} x = c \)
(b) \( x + \sin^{-1} y = 0 \)
(c) \( y² + 2\sin^{-1} x = c \)
(d) \( x² + 2\sin^{-1} y = 0 \)
Answer: (a) \( y + \sin^{-1} x = c \)
In simple words: Rearrange the equation to separate the variables, which means getting all \( y \) terms with \( dy \) and \( x \) terms with \( dx \). This gives \( dy = - \frac{1}{\sqrt{1-x^2}} dx \). Now, integrate both sides. The integral of \( dy \) is \( y \), and the integral of \( - \frac{1}{\sqrt{1-x^2}} dx \) is \( -\sin^{-1} x \). Add a constant of integration, \( C \), to get \( y = -\sin^{-1} x + C \). Finally, move \( -\sin^{-1} x \) to the left side to get \( y + \sin^{-1} x = C \). This problem involves the standard integral form of \( \frac{1}{\sqrt{1-x^2}} \), which is directly related to the derivative of the inverse sine function.

🎯 Exam Tip: Be familiar with standard integral forms, especially those involving inverse trigonometric functions, as they frequently appear in solutions of differential equations.

Question 14. The solution of the differential equation \( \frac { dy }{ dx } = 2xy \) is
(a) \( y = Ce^{x^2} \)
(b) \( y = 2x² + C \)
(c) \( y = Ce^{-x^2} + C \)
(d) \( y = x² + C \)
Answer: (a) \( y = Ce^{x^2} \)
In simple words: This is a separable differential equation. First, separate the variables by moving \( y \) terms to the left side with \( dy \) and \( x \) terms to the right side with \( dx \). This gives \( \frac{dy}{y} = 2x dx \). Next, integrate both sides. The integral of \( \frac{1}{y} \) is \( \log|y| \), and the integral of \( 2x \) is \( x^2 \). After integration, introduce a constant, often written as \( \log|C| \) for convenience. So, \( \log|y| = x^2 + \log|C| \). Combine the logarithm terms and convert to exponential form to get \( y = Ce^{x^2} \). This type of solution, involving an exponential term, often arises when the rate of change of a quantity is directly proportional to the quantity itself, leading to exponential growth or decay.

🎯 Exam Tip: When integrating after separating variables, choose the form of the integration constant (e.g., C or log C) that simplifies the final expression most effectively, especially when logarithms are involved.

Question 15. The general solution of the differential equation \( \log(\frac { dy }{ dx }) = x + y \) is
(a) \( e^x + e^{-y} = C \)
(b) \( e^x + e^{-y} = C \)
(c) \( e^{-x} + e^{-y} = C \)
(d) \( e^x + e^y = C \)
Answer: (b) \( e^x + e^{-y} = C \)
In simple words: First, convert the logarithmic equation into an exponential one by applying \( e \) to both sides. This transforms \( \log(\frac{dy}{dx}) = x+y \) into \( \frac{dy}{dx} = e^{x+y} \). Then, use exponent rules to write \( e^{x+y} \) as \( e^x \cdot e^y \). Now, separate the variables: move \( e^y \) to the left side as \( e^{-y} \) with \( dy \), and keep \( e^x \) on the right side with \( dx \). Integrate both sides: \( \int e^{-y} dy = \int e^x dx \). This gives \( -e^{-y} = e^x + C' \). Rearrange to get the final solution \( e^x + e^{-y} = C \). Exponential functions are critical in modeling situations where the rate of growth or decay is proportional to the current amount, often appearing in population growth, compound interest, or radioactive decay models.

🎯 Exam Tip: When a differential equation involves a logarithm, convert it to exponential form first to simplify the separation of variables and make integration straightforward.

Question 16. The solution of \( \frac { dy }{ dx } = 2^{y-x} \) is
(a) \( 2x + 2y = C \)
(b) \( 2x – 2y = C \)
(c) \( \frac { 1 }{ 2^x } – \frac { 1 }{ 2^y } = C \)
(d) \( x + y = C \)
Answer: (c) \( \frac { 1 }{ 2^x } – \frac { 1 }{ 2^y } = C \)
In simple words: Begin by using exponent rules to rewrite \( 2^{y-x} \) as \( 2^y \cdot 2^{-x} \). Then, separate the variables by moving all terms with \( y \) to the left side with \( dy \) (as \( 2^{-y} dy \)) and all terms with \( x \) to the right side with \( dx \) (as \( 2^{-x} dx \)). Integrate both sides: \( \int 2^{-y} dy = \int 2^{-x} dx \). Remember that the integral of \( a^u \) is \( \frac{a^u}{\ln a} \) multiplied by the derivative of \( u \). After integrating and simplifying, you will get \( - \frac{2^{-y}}{\ln 2} = - \frac{2^{-x}}{\ln 2} + C' \). Multiply by \( -\ln 2 \) and rearrange to achieve \( \frac{1}{2^x} - \frac{1}{2^y} = C \). Differential equations involving exponential bases other than 'e' often require careful application of integration rules for \( a^u \), remembering to include the natural logarithm of the base in the denominator.

🎯 Exam Tip: When integrating \( a^u \) forms, remember that the integral is \( \frac{a^u}{\ln a} \), and correctly handle any chain rule factors from the exponent when integrating.

Question 17. The solution of the differential equation \( \frac { dy }{ dx } = \frac { y }{ x } + \frac { \phi(\frac { y }{ x }) }{ \phi'(\frac { y }{ x }) } \) is
(a) \( x\phi(\frac { y }{ x }) = k \)
(b) \( \phi(\frac { y }{ x }) = kx \)
(c) \( y\phi(\frac { y }{ x }) = k \)
(d) \( \phi(\frac { y }{ x }) = ky \)
Answer: (b) \( \phi(\frac { y }{ x }) = kx \)
In simple words: This is a homogeneous differential equation. To solve it, use the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Substitute these into the given equation. The \( v \) terms will cancel out, leaving \( x \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)} \). Now, separate the variables: \( \frac{\phi'(v)}{\phi(v)} dv = \frac{dx}{x} \). Integrate both sides. The left side is of the form \( \int \frac{f'(v)}{f(v)} dv = \log|f(v)| \), so it becomes \( \log|\phi(v)| \). The right side integrates to \( \log|x| + \log|k| \). Combine the logarithms and remove them, resulting in \( \phi(v) = kx \). Finally, substitute back \( v = \frac{y}{x} \) to get \( \phi(\frac{y}{x}) = kx \). Homogeneous differential equations are characterized by having all terms of the same degree, and they can be solved by a simple substitution that transforms them into separable equations.

🎯 Exam Tip: For homogeneous differential equations, the substitution \( y = vx \) (or \( x = vy \)) is key, as it transforms the equation into a separable form that is much easier to integrate.

Question 18. If \( \sin x \) is the integrating factor of the linear differential equation \( \frac { dy }{ dx } + Py = Q \), then P is
(a) \( \log \sin x \)
(b) \( \cos x \)
(c) \( \tan x \)
(d) \( \cot x \)
Answer: (d) \( \cot x \)
In simple words: The integrating factor (IF) for a linear differential equation in the form \( \frac{dy}{dx} + Py = Q \) is defined as \( e^{\int P dx} \). We are given that the IF is \( \sin x \). So, we can write \( e^{\int P dx} = \sin x \). To find \( P \), take the natural logarithm of both sides: \( \int P dx = \log (\sin x) \). Now, differentiate both sides with respect to \( x \). The derivative of \( \int P dx \) is simply \( P \), and the derivative of \( \log (\sin x) \) is \( \frac{1}{\sin x} \cdot \cos x \), which simplifies to \( \cot x \). So, \( P = \cot x \). The integrating factor is a crucial component in solving first-order linear differential equations, as it directly relates to the coefficient \( P(x) \) of the \( y \) term.

🎯 Exam Tip: Remember that the derivative of \( \log(\sin x) \) is \( \cot x \). Knowing common derivative and integral pairs for trigonometric and logarithmic functions can significantly speed up solving integrating factor problems.

Question 19. The number of arbitrary constants in the general solutions of order n and n + 1 is respectively.
(a) n - 1, n
(b) n, n + 1
(c) n + 1, n + 2
(d) n + 1, n
Answer: (b) n, n + 1
In simple words: The number of constant values in a differential equation's general answer always matches its 'order' (how many times it was differentiated). So, an equation of order 'n' has 'n' constants, and one of order 'n+1' has 'n+1' constants.

🎯 Exam Tip: Remember that each time you integrate a differential equation, a new constant of integration is introduced, which is why the number of arbitrary constants equals the order of the differential equation.

Question 20. The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) 0
Answer: (d) 0
In simple words: Once you find the 'particular' answer to a differential equation, all the changeable numbers (arbitrary constants) become fixed values. So, there are no more changeable numbers left.

🎯 Exam Tip: A particular solution is derived from a general solution by using initial or boundary conditions, which replace all arbitrary constants with specific numerical values.

Question 21. Integrating factor of the differential equation \( \frac { dy }{ dx } = \frac { x+y+1 }{ x+1 } \) is
(a) \( \frac { 1 }{ x+1 } \)
(b) \( x+1 \)
(c) \( \frac { 1 }{ \sqrt{x+1} } \)
(d) \( \sqrt {x+1} \)
Answer: (a) \( \frac { 1 }{ x+1 } \)
In simple words: We change the equation to a standard form, find a part called \( P(x) \), and then use a special formula with 'e' and 'log' to get a value. This value is the integrating factor, which helps solve the equation.

🎯 Exam Tip: Always rearrange the differential equation into the standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \) before identifying \( P(x) \) and calculating the integrating factor \( e^{\int P(x)dx} \).

Question 22. The population P in any year t is such that the rate of increase in the population is proportional to the population. Then
(a) \( P = Ce^{kt} \)
(b) \( P = Ce^{-kt} \)
(c) \( P = Ckt \)
(d) \( Pt = C \)
Answer: (a) \( P = Ce^{kt} \)
In simple words: If a population grows faster when there are more people, its size can be found using a special formula: \( P = Ce^{kt} \). This means it grows by multiplying, not just adding, over time.

🎯 Exam Tip: Recognize that "rate of increase proportional to the quantity itself" translates to the differential equation \( \frac{dP}{dt} = kP \), which leads to exponential growth or decay solutions.

Question 23. P is the amount of certain substance left in after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then
(a) \( P = Ce^{kt} \)
(b) \( P = Ce^{-kt} \)
(c) \( P = Ckt \)
(d) \( Pt = C \)
Answer: (b) \( P = Ce^{-kt} \)
In simple words: When something like water evaporates, and the faster it evaporates, the more there is, then the amount left can be found using \( P = Ce^{-kt} \). The minus sign means the amount goes down over time.

🎯 Exam Tip: A negative sign in the proportionality constant (\( -kP \)) indicates a decrease or decay of the quantity over time, as seen in evaporation or radioactive decay.

Question 24. If the solution of the differential equation \( \frac { dy }{ dx } = \frac { ax+3 }{ 2y+f} \) represents a circle, then the value of a is
(a) 2
(b) -2
(c) 1
(d) -1
Answer: (b) -2
In simple words: First, we solve the equation by doing 'integration'. The answer we get will be the equation of a shape. For this shape to be a circle, the number in front of \( x^2 \) must be the same as the negative of the number in front of \( y^2 \). When we do this, we find \( a \) should be \( -2 \).

🎯 Exam Tip: For an equation of the form \( Ax^2 + By^2 + ... = 0 \) to represent a circle, the coefficients of \( x^2 \) and \( y^2 \) must be equal and have the same sign (or be equal and opposite if one side of the equation is set to a constant). In this case, comparing \( \frac{a}{2}x^2 \) and \( -y^2 \), they must be \( A = -B \).

Question 25. The slope at any point of a curve y = f(x) is given by \( \frac { dy }{ dx } = 3x² \) and it passes through (-1, 1). Then the equation of the curve is
(a) \( y = x³ + 2 \)
(b) \( y = 3x² + 4 \)
(c) \( y = 3x³ + 4 \)
(d) \( y = x³ + 5 \)
Answer: (a) \( y = x³ + 2 \)
In simple words: We start with how fast the curve is changing and then 'integrate' it to find the basic equation of the curve, which includes a letter 'C'. Since we know the curve goes through the point (-1, 1), we use these numbers to figure out what 'C' is. Finally, we put 'C' back into the equation to get the exact answer for the curve.

🎯 Exam Tip: Remember to integrate the given slope expression (\( \frac{dy}{dx} \)) to find the general equation of the curve, then use the provided point to solve for the constant of integration (\( C \)) to get the particular solution.