Samacheer Kalvi Class 12 Business Maths Solutions Chapter 4 Differential Equations Exercise 4.6

Question 1. The degree of the differential equation \( \frac { d^2y }{dx^4} – (\frac { d^2y }{dx^2}) + \frac { dy }{dx} = 3 \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
In simple words: The degree of a differential equation is the power of the highest order derivative, once the equation is made free from radicals and fractions. In this equation, the highest order derivative is \( \frac{d^4y}{dx^4} \), and its power is 1.

🎯 Exam Tip: To find the degree, first ensure the differential equation is a polynomial in its derivatives (no roots or fractions of derivatives). Then, identify the highest derivative's exponent.

Question 2. The order and degree of the differential equation \( \sqrt{\frac { d^2y }{dx^2}} = \sqrt{\frac { dy }{dx}+5} \) are respectively.
(a) 2 and 2
(b) 3 and 2
(c) 2 and 1
(d) 2 and 3
Answer: (c) 2 and 1
In simple words: First, square both sides to remove the square roots. This gives \( \frac { d^2y }{dx^2} = \frac { dy }{dx} + 5 \). The highest derivative is \( \frac{d^2y}{dx^2} \), which means the order is 2. The power of this highest derivative is 1, so the degree is 1.

🎯 Exam Tip: Always clear radicals (like square roots) and fractions involving derivatives before determining the degree of a differential equation. Squaring both sides is a common first step.

Question 3. The order and degree of the differential equation \( (\frac { d^2y }{dx^2})^{3/2} – \sqrt{(\frac { dy }{dx})} – 4 = 0 \)
(a) 2 and 6
(b) 3 and 6
(c) 2 and 3
(d) 2 and 4
Answer: (a) 2 and 6
In simple words: To find the degree, we first need to remove all fractional powers. Start by isolating one fractional term and squaring both sides repeatedly until all fractional exponents are cleared. Here, isolate \( (\frac{d^2y}{dx^2})^{3/2} \) and square it, then isolate \( \sqrt{\frac{dy}{dx}} \) and square it again. The highest order derivative will be \( \frac{d^2y}{dx^2} \) (order 2), and after clearing the fractions, its power will be 6.

🎯 Exam Tip: For equations with fractional powers, raise both sides to appropriate powers to eliminate fractions. Isolate the most complex fractional derivative term first to simplify the process.

Question 4. The differential equation \( (\frac { dx }{dy})^3 + 2y^{1/2} = x \)
(a) of order 2 and degree 1
(b) of order 1 and degree 3
(c) of order 1 and degree 6
(d) of order 1 and degree 2
Answer: (b) of order 1 and degree 3
In simple words: In this equation, the derivative is \( \frac{dx}{dy} \). This means we are differentiating x with respect to y. The highest and only derivative present is \( \frac{dx}{dy} \), so the order is 1. The power of this highest derivative is 3, which makes the degree 3.

🎯 Exam Tip: Pay close attention to whether the derivative is \( \frac{dy}{dx} \) or \( \frac{dx}{dy} \). This tells you which variable is dependent and which is independent, which is crucial for determining order and degree.

Question 5. The differential equation formed by eliminating a and b from \( y = ae^x + be^{-x} \) is
(a) \( \frac { d^2y }{dx^2} - y = 0 \)
(b) \( \frac { d^2y }{dx^2} – \frac { dy }{dx}y = 0 \)
(c) \( \frac { d^2y }{dx^2} = 0 \)
(d) \( \frac { d^2y }{dx^2} − x = 0 \)
Answer: (a) \( \frac { d^2y }{dx^2} - y = 0 \)
In simple words: To remove the constants 'a' and 'b', we differentiate the original equation twice. First, find \( \frac{dy}{dx} \), then find \( \frac{d^2y}{dx^2} \). You will notice that the second derivative simplifies back to the original function 'y'. This means \( \frac{d^2y}{dx^2} = y \), which can be rewritten as \( \frac{d^2y}{dx^2} - y = 0 \).

🎯 Exam Tip: When eliminating arbitrary constants, differentiate the given function as many times as there are constants. Then, use the original equation and its derivatives to eliminate the constants and form the differential equation.

Question 6. If \( y = ex + c − c³ \) then its differential equation is
(a) \( y = x\frac { dy }{dx} + \frac { dy }{dx} – (\frac { dy }{dx})³ \)
(b) \( y + (\frac { dy }{dx})³ = x \frac { dy }{dx} – \frac { dy }{dx} \)
(c) \( \frac { dy }{dx} + (\frac { dy }{dx})³ – x\frac { dy }{dx} \)
(d) \( \frac { d^3y }{dx^3} = 0 \)
Answer: (a) \( y = x\frac { dy }{dx} + \frac { dy }{dx} – (\frac { dy }{dx})³ \)
In simple words: This equation is a Clairaut's equation. To find the differential equation, first differentiate \( y = ex + c - c^3 \) with respect to \( x \). You will find that \( \frac{dy}{dx} = e \). Replace 'c' in the original equation with \( \frac{dy}{dx} \) to get the final differential equation.

🎯 Exam Tip: For equations of the form \( y = Px + f(P) \), where \( P = \frac{dy}{dx} \), the differential equation is formed by replacing 'c' with 'P' (or \( \frac{dy}{dx} \)).

Question 7. The integrating factor of the differential equation \( \frac { dy }{dx} + Px = Q \) is
(a) \( e^{\int Pdx} \)
(b) \( e^{\int Pdx} \)
(c) \( e^{\int Pdy} \)
(d) \( e^{\int Pdy} \)
Answer: (d) \( e^{\int Pdy} \)
In simple words: For a linear differential equation of the form \( \frac{dx}{dy} + P(y)x = Q(y) \), where \( P(y) \) and \( Q(y) \) are functions of \( y \), the integrating factor is \( e^{\int P(y)dy} \). Although the question writes \( \frac{dy}{dx} \), the options suggest the equation is being treated as \( \frac{dx}{dy} + Px = Q \) where P is a function of y, making x the dependent variable.

🎯 Exam Tip: The integrating factor depends on the form of the linear differential equation. If the equation is \( \frac{dy}{dx} + P(x)y = Q(x) \), the IF is \( e^{\int P(x)dx} \). If it's \( \frac{dx}{dy} + P(y)x = Q(y) \), the IF is \( e^{\int P(y)dy} \).

Question 8. The complementary function of \( (D² + 4) y = e^{2x} \) is
(a) \( (Ax+B)e^{2x} \)
(b) \( (Ax+B)e^{-2x} \)
(c) \( A \cos 2x + B \sin 2x \)
(d) \( Ae^{-2x} + Be^{2x} \)
Answer: (c) \( A \cos 2x + B \sin 2x \)
In simple words: To find the complementary function (CF), we set the right-hand side of the differential equation to zero and solve the auxiliary equation. Here, the auxiliary equation is \( m^2 + 4 = 0 \), which gives imaginary roots \( m = \pm 2i \). For imaginary roots of the form \( \alpha \pm i\beta \), the complementary function is \( e^{\alpha x} (A \cos \beta x + B \sin \beta x) \). Since \( \alpha = 0 \) and \( \beta = 2 \), the CF becomes \( e^{0x} (A \cos 2x + B \sin 2x) \), which simplifies to \( A \cos 2x + B \sin 2x \).

🎯 Exam Tip: Remember the different forms of complementary functions based on the roots of the auxiliary equation: real and distinct, real and repeated, or complex conjugate roots.

Question 9. The differential equation of \( y = mx + c \) is (m and c are arbitrary constants)
(a) \( \frac { d^2y }{dx^2} = 0 \)
(b) \( y = x\frac { dy }{dx} + o \)
(c) \( xdy + ydx = 0 \)
(d) \( ydx – xdy = 0 \)
Answer: (a) \( \frac { d^2y }{dx^2} = 0 \)
In simple words: The equation \( y = mx + c \) represents a straight line. Since 'm' and 'c' are arbitrary constants, we need to eliminate them by differentiating. First, \( \frac{dy}{dx} = m \). Since 'm' is also an arbitrary constant, differentiating again with respect to 'x' eliminates it. So, \( \frac{d^2y}{dx^2} = 0 \).

🎯 Exam Tip: The order of a differential equation formed by eliminating constants is equal to the number of arbitrary constants in the original relation.

Question 10. The particular integral of the differential equation \( \frac { d^2y }{dx^2} - 8\frac { dy }{dx} + 16y = 2e^{4x} \) is
(a) \( \frac {x^2e^{4x} }{2!} \)
(b) \( y = x\frac { e^{4x} }{2!} \)
(c) \( x²e^{4x} \)
(d) \( xe^{4x} \)
Answer: (c) \( x²e^{4x} \)
In simple words: We need to find the Particular Integral (PI) for \( f(D)y = 2e^{4x} \) where \( f(D) = D^2 - 8D + 16 \). This can be written as \( f(D) = (D-4)^2 \). Since the exponential term is \( e^{4x} \) and `4` is a root of `f(D)` with multiplicity 2, the formula for PI is \( \frac{x^2}{2!} e^{4x} \times 2 \) (because the RHS is \( 2e^{4x} \)). This simplifies to \( x^2 e^{4x} \).

🎯 Exam Tip: When finding the particular integral for \( f(D)y = e^{ax} \) and \( f(a) = 0 \), check the multiplicity of the root 'a'. If it's 'n', the formula involves \( \frac{x^n}{n!} \).

Question 11. Solution of \( \frac { dx }{dy} + Px = 0 \)
(a) \( x = ce^{Py} \)
(b) \( x = ce^{-Py} \)
(c) \( x = py + c \)
(d) \( x = cy \)
Answer: (b) \( x = ce^{-Py} \)
In simple words: This is a linear first-order differential equation where 'x' is the dependent variable and 'y' is the independent variable. We can rewrite it as \( \frac{dx}{x} = -P dy \). Integrating both sides gives \( \ln|x| = -Py + \ln|c| \). Exponentiating both sides yields \( x = e^{-Py + \ln|c|} \), which simplifies to \( x = e^{-Py} \cdot e^{\ln|c|} \), or \( x = ce^{-Py} \).

🎯 Exam Tip: For variable separable equations, first separate the variables \( x \) and \( y \) (and their differentials) to opposite sides of the equation. Then, integrate each side to find the general solution.

Question 12. If \( \sec^2 x \) is an integrating factor of the differential equation \( \frac { dx }{dy} + Px = Q \) then P =
(a) \( 2 \tan x \)
(b) \( \sec x \)
(c) \( \cos 2 x \)
(d) \( \tan 2 x \)
Answer: (a) \( 2 \tan x \)
In simple words: For a linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is \( e^{\int P(x)dx} \). Given the options and the solution, we interpret the question as if the differential equation was \( \frac{dy}{dx} + P(x)y = Q(x) \) and the integrating factor is given in terms of \( x \). If \( e^{\int Pdx} = \sec^2 x \), then \( \int Pdx = \ln(\sec^2 x) = 2 \ln(\sec x) \). Differentiating both sides with respect to \( x \) gives \( P = \frac{d}{dx}(2 \ln(\sec x)) = 2 \frac{1}{\sec x} (\sec x \tan x) = 2 \tan x \).

🎯 Exam Tip: When the integrating factor is provided, you can work backward by taking the natural logarithm and then differentiating to find the function P. Ensure you apply the chain rule correctly if the argument of the logarithm is complex.

Question 13. The integrating factor of the differential equation is \( x \frac { dy }{dx} – y = x² \)
(a) \( \frac {-1}{x} \)
(b) \( \frac { 1 }{x} \)
(c) \( \log x \)
(d) \( x \)
Answer: (b) \( \frac { 1 }{x} \)
In simple words: First, convert the given equation into the standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \). Divide the entire equation by \( x \) to get \( \frac{dy}{dx} - \frac{1}{x}y = x \). Here, \( P(x) = -\frac{1}{x} \). The integrating factor is then \( e^{\int P(x)dx} = e^{\int -\frac{1}{x}dx} = e^{-\ln|x|} = e^{\ln|x|^{-1}} = x^{-1} = \frac{1}{x} \).

🎯 Exam Tip: Always make sure the coefficient of \( \frac{dy}{dx} \) is 1 before identifying \( P(x) \) in a linear differential equation. This often involves dividing the entire equation by the coefficient.

Question 14. The solution of the differential equation where P and Q are the function of x is
(a) \( y = \int Q e^{\int Pdx} dx + c \)
(b) \( y = \int Q e^{-\int Pdx} dx + c \)
(c) \( ye^{\int Pdx} = \int Q e^{-\int Pdx} dx + c \)
(d) \( ye^{\int Pdx} = \int Q e^{\int Pdx} dx + c \)
Answer: (d) \( ye^{\int Pdx} = \int Q e^{\int Pdx} dx + c \)
In simple words: For a first-order linear differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \), the general solution is found by multiplying both sides by the integrating factor \( I.F. = e^{\int P(x)dx} \). After multiplication, the left side becomes \( \frac{d}{dx}(y \cdot I.F.) \), and integrating both sides gives \( y \cdot I.F. = \int (Q \cdot I.F.) dx + C \).

🎯 Exam Tip: Memorize the general solution formula for linear first-order differential equations. It is a direct application of the integrating factor method and a frequently tested concept.

Question 15. The differential equation formed by eliminating A and B from \( y = e^{-2x} (A \cos x + B \sin x) \) is
(a) \( y_2 – 4y_1 + 5 = 0 \)
(b) \( y_2 + 4y – 5 = 0 \)
(c) \( y_2 – 4y_1 + 5 = 0 \)
(d) \( y_2 + 4y_1 − 5 = 0 \)
Answer: (d) \( y_2 + 4y_1 − 5 = 0 \)
In simple words: To eliminate the arbitrary constants A and B, we need to differentiate the given equation twice. Start by moving \( e^{-2x} \) to the left side: \( ye^{2x} = A \cos x + B \sin x \). Differentiate this equation once and then again. After finding the first and second derivatives, substitute back to eliminate A and B. The characteristic equation corresponding to \( y = e^{-2x} (A \cos x + B \sin x) \) has roots \( m = -2 \pm i \), which means \( (m+2)^2 = -1 \implies m^2+4m+4 = -1 \implies m^2+4m+5 = 0 \). This implies the differential equation should be \( y'' + 4y' + 5y = 0 \).

🎯 Exam Tip: When eliminating arbitrary constants from solutions involving complex exponentials, it's often helpful to first derive the characteristic equation from the roots implied by the solution. This directly gives the differential equation's structure.

Question 16. The particular integral of the differential equation \( f (D) y = e^{ax} \) where \( f(D) = (D – a)² \)
(a) \( \frac {x^2}{2} e^{ax} \)
(b) \( xe^{ax} \)
(c) \( \frac { x }{2} e^{ax} \)
(d) \( x² e^{ax} \)
Answer: (a) \( \frac {x^2}{2} e^{ax} \)
In simple words: When finding the particular integral (PI) for an equation like \( f(D)y = e^{ax} \) and the exponential term's coefficient 'a' is a root of \( f(D) \) with multiplicity 'n', the PI is found using the formula \( \frac{x^n}{n!} e^{ax} \). In this case, \( f(D) = (D-a)^2 \), so 'a' is a root with multiplicity \( n=2 \). Thus, the PI is \( \frac{x^2}{2!} e^{ax} = \frac{x^2}{2} e^{ax} \).

🎯 Exam Tip: Always identify the multiplicity of the root 'a' in \( f(D) \) when solving for the particular integral of \( e^{ax} \). If \( f(a) \neq 0 \), a simpler formula applies.

Question 17. The differential equation of \( x² + y² = a² \)
(a) \( xdy + ydx = 0 \)
(b) \( ydx – xdy = 0 \)
(c) \( xdx – ydx = 0 \)
(d) \( xdx + ydy = 0 \)
Answer: (d) \( xdx + ydy = 0 \)
In simple words: To find the differential equation, we differentiate the given equation implicitly with respect to 'x'. Differentiating \( x^2 \) gives \( 2x \), and differentiating \( y^2 \) gives \( 2y \frac{dy}{dx} \). Since \( a^2 \) is a constant, its derivative is 0. So, \( 2x + 2y \frac{dy}{dx} = 0 \). Dividing by 2 and rearranging gives \( x + y \frac{dy}{dx} = 0 \), which can be written as \( x dx + y dy = 0 \). This equation describes a family of circles centered at the origin.

🎯 Exam Tip: Remember that the differential of a constant is always zero. When differentiating implicitly, use the chain rule for terms involving the dependent variable.

Question 18. The complementary function of \( \frac { d^2y }{dx^2} - \frac { dy }{dx} = 0 \) is
(a) \( A + Be^x \)
(b) \( (A + B)e^x \)
(c) \( (Ax + B)e^x \)
(d) \( Ae^x + B \)
Answer: (a) \( A + Be^x \)
In simple words: To find the complementary function, we first write the auxiliary equation by replacing \( \frac{d^2y}{dx^2} \) with \( m^2 \) and \( \frac{dy}{dx} \) with \( m \). This gives \( m^2 - m = 0 \). Factoring out 'm', we get \( m(m-1) = 0 \). The roots are \( m = 0 \) and \( m = 1 \). For real and distinct roots \( m_1 \) and \( m_2 \), the complementary function is \( Ae^{m_1x} + Be^{m_2x} \). So, it becomes \( Ae^{0x} + Be^{1x} \), which simplifies to \( A + Be^x \).

🎯 Exam Tip: Be careful when one of the roots of the auxiliary equation is zero. Remember that \( e^{0x} = 1 \), so the corresponding term in the complementary function is just a constant.

Question 19. The P.I of \( (3D² + D – 14) y = 13e^{2x} \) is
(a) \( \frac { 1 }{2} e^x \)
(b) \( xe^{2x} \)
(c) \( \frac { x^2 }{2} e^{2x} \)
(d) \( Ae^x + B \)
Answer: (b) \( xe^{2x} \)
In simple words: To find the particular integral (PI) for \( f(D)y = e^{ax} \), we first substitute 'a' into \( f(D) \). Here \( a=2 \), and \( f(D) = 3D^2 + D - 14 \). \( f(2) = 3(2)^2 + 2 - 14 = 12 + 2 - 14 = 0 \). Since \( f(2)=0 \), we use the rule \( PI = x \cdot \frac{1}{f'(D)} e^{ax} \). First, find \( f'(D) = 6D + 1 \). Then, evaluate \( f'(2) = 6(2) + 1 = 13 \). So, \( PI = x \cdot \frac{1}{13} (13e^{2x}) = xe^{2x} \).

🎯 Exam Tip: If \( f(a) = 0 \) but \( f'(a) \neq 0 \), apply the rule \( PI = x \cdot \frac{1}{f'(a)} e^{ax} \). If \( f'(a) \) is also zero, you'll need to differentiate \( f(D) \) further.

Question 20. The general solution of the differential equation \( \frac { dy }{dx} = \cos x \) is
(a) \( y = \sin x + 1 \)
(b) \( y = \sin x – 2 \)
(c) \( y = \sin x + C, C \) is an arbitary constant
(d) \( y = \sin x + C, C \) is an arbitary constant
Answer: (d) \( y = \sin x + C, C \) is an arbitary constant
In simple words: To find the general solution, we need to integrate both sides of the equation. Rewrite \( \frac{dy}{dx} = \cos x \) as \( dy = \cos x dx \). Integrating \( \int dy \) gives \( y \), and integrating \( \int \cos x dx \) gives \( \sin x \). Since it's an indefinite integral, we must add an arbitrary constant of integration, C. Therefore, the solution is \( y = \sin x + C \).

🎯 Exam Tip: Always remember to add the constant of integration 'C' when performing indefinite integration to find the general solution of a differential equation. Without it, the solution is incomplete.

Question 21. A homogeneous differential equation of the form \( \frac { dy }{dx} = f(\frac { y }{x}) \) can be solved by making substitution.
(a) \( y = v x \)
(b) \( y = y x \)
(c) \( x = v y \)
(d) \( x = v \)
Answer: (a) \( y = v x \)
In simple words: A homogeneous differential equation is one where all terms have the same total degree. When the equation can be written in the form \( \frac{dy}{dx} = f(\frac{y}{x}) \), the standard way to solve it is by substituting \( y = vx \). This substitution transforms the equation into a variable separable form, making it easier to integrate.

🎯 Exam Tip: When using the substitution \( y = vx \), remember to also replace \( \frac{dy}{dx} \) with \( v + x\frac{dv}{dx} \) before simplifying and separating variables.

Question 22. A homogeneous differential equation of the form \( \frac { dy }{dx} = f(\frac { x }{y}) \) can be solved by making substitution.
(a) \( x = v y \)
(b) \( y = v x \)
(c) \( y = v \)
(d) \( x = v \)
Answer: (a) \( x = v y \)
In simple words: If a homogeneous differential equation is given in the form \( \frac{dy}{dx} = f(\frac{x}{y}) \), it's often more convenient to rearrange it into \( \frac{dx}{dy} = g(\frac{x}{y}) \). In this case, the appropriate substitution is \( x = vy \). This will transform the equation into a form that is separable with respect to \( v \) and \( y \).

🎯 Exam Tip: For homogeneous equations, if the function depends on \( \frac{y}{x} \), use \( y = vx \). If it depends on \( \frac{x}{y} \), use \( x = vy \). This choice simplifies the algebra significantly.

Question 23. The variable separable form of \( \frac { dy }{dx} = \frac { y(x-y) }{x(x+y)} \) by taking \( y = v x \) and \( \frac { dy }{dx} = v + x \frac { dv }{dx} \) is
(a) \( \frac {2v^2 }{1+v} dv = \frac { dx }{x} \)
(b) \( \frac {2v^2}{1+v} dv = -\frac { dx }{x} \)
(c) \( \frac { 2v^2 }{1-v} dv = \frac { dx }{x} \)
(d) \( \frac {1+v }{2v^2} dv = -\frac { dx }{x} \)
Answer: (d) \( \frac {1+v }{2v^2} dv = -\frac { dx }{x} \)
In simple words: This is a homogeneous differential equation because all terms have the same degree (degree 2 in this case). To solve it, substitute \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). After substitution and simplification, the equation can be rearranged so that all terms involving \( v \) and \( dv \) are on one side, and all terms involving \( x \) and \( dx \) are on the other side. This process isolates the variables, allowing for direct integration. The simplified form is \( \frac{1+v}{2v^2} dv = -\frac{dx}{x} \).

🎯 Exam Tip: When performing the substitution for homogeneous equations, be meticulous with algebraic simplifications to correctly separate the variables for integration.

Question 24. Which of the following is the homogeneous differential equation?
(a) \( (3x – 5) dx = (4y – 1) dy \)
(b) \( xy dx – (x³ + y³) dy = 0 \)
(c) \( y²dx + (x² – xy – y²) dy = 0 \)
(d) \( (x² + y) dx (y² + x) dy \)
Answer: (c) \( y²dx + (x² – xy – y²) dy = 0 \)
In simple words: A differential equation is homogeneous if all its terms have the same total degree. For option (c), we can write \( \frac{dy}{dx} = \frac{-y^2}{x^2 - xy - y^2} \). In the numerator, \( y^2 \) has degree 2. In the denominator, \( x^2 \), \( xy \), and \( y^2 \) all have degree 2. Since all terms in the numerator and denominator have the same degree, this equation is homogeneous.

🎯 Exam Tip: To quickly check if an equation is homogeneous, examine the degree of each term. If all terms have the same total degree (sum of powers of x and y in each term), the equation is homogeneous.

Question 25. The solution of the differential equation \( \frac { dy }{dx} = \frac { y }{x} + \frac { f(\frac { y }{x}) }{ f(\frac { y }{x}) } \) is
(a) \( f\frac { y }{x} = k x \)
(b) \( x f\frac { y }{x} = k \)
(c) \( f\frac { y }{x} = ky \)
(d) \( x f\frac { y }{x} = k \)
Answer: (a) \( f\frac { y }{x} = k x \)
In simple words: The given equation simplifies to \( \frac{dy}{dx} = \frac{y}{x} + 1 \) because \( \frac{f(\frac{y}{x})}{f(\frac{y}{x})} = 1 \). This is a linear differential equation. To solve it, we can use the integrating factor method or by substitution \( y = vx \). Using \( y = vx \), we get \( v + x\frac{dv}{dx} = v + 1 \). This simplifies to \( x\frac{dv}{dx} = 1 \), or \( dv = \frac{dx}{x} \). Integrating both sides gives \( v = \ln|x| + C \). Substituting \( v = \frac{y}{x} \) back, we get \( \frac{y}{x} = \ln|x| + C \). While the answer format uses \( f(\frac{y}{x}) = kx \), the principle involves finding a relation between \( y/x \) and \( x \).

🎯 Exam Tip: Always simplify the given differential equation first. If a term like \( \frac{f(u)}{f(u)} \) appears, simplify it to 1 to reveal the true form of the equation.

Choose the most suitable answer from the given four alternatives:

Question 1. The degree of the differential equation \( \frac { d^2y }{dx^4} - (\frac { d^2y }{dx^2}) + \frac { dy }{dx} = 3 \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
In simple words: The degree of a differential equation is the highest power of the highest order derivative after it has been made free from radicals and fractions. In this equation, the highest order is 4 (from \( \frac{d^4y}{dx^4} \)), and its power is 1. This means how many times the variable has been differentiated and its highest power.

🎯 Exam Tip: To find the degree, first identify the highest order derivative. Then, check its power after making sure there are no roots or fractions involving derivatives.

Question 2. The order and degree of the differential equation \( \sqrt{\frac { d^2y }{dx^2}} = \sqrt{\frac { dy }{dx}+5} \) are respectively.
(a) 2 and 2
(b) 3 and 2
(c) 2 and 1
(d) 2 and 3
Answer: (c) 2 and 1
In simple words: First, remove the square roots by squaring both sides of the equation. After doing this, the highest order derivative is 2 (the second derivative), and its power will be 1. The order tells us the highest derivative, and the degree tells us its power.

🎯 Exam Tip: Always clear fractional powers or radicals from derivatives before determining the degree of a differential equation. Squaring both sides is a common first step.

Question 3. The order and degree of the differential equation \( (\frac { d^2y }{dx^2})^{3/2} – \sqrt{(\frac { dy }{dx})} – 4 = 0 \)
(a) 2 and 6
(b) 3 and 6
(c) 3 and 2
(d) 2 and 4
Answer: (a) 2 and 6
In simple words: To find the degree, we first need to remove the fractional powers. Square both sides to get rid of the \( \sqrt{} \) term, and then raise to the power of 2 again to remove the \( 3/2 \) fractional power. The highest derivative is 2 (second derivative), and its highest power will become 6 after these steps. The order is simply the highest derivative in the equation.

🎯 Exam Tip: To clear fractional powers like \( \frac{3}{2} \) and square roots, you might need to raise both sides of the equation to a suitable power multiple times. Always isolate one fractional term before clearing it.

Question 4. The differential equation \( (\frac { dx }{dy})^3 + 2y^{1/2} = x \)
(a) of order 2 and degree 1
(b) of order 1 and degree 3
(c) of order 1 and degree 6
(d) of order 1 and degree 2
Answer: (b) of order 1 and degree 3
In simple words: This equation involves \( \frac{dx}{dy} \), which is a first-order derivative. Its highest power is 3, making the degree 3. The order is simply 1 because it's a first derivative.

🎯 Exam Tip: When the differential is \( \frac{dx}{dy} \), the order refers to the differentiation with respect to y, and the degree is the power of that highest order derivative.

Question 5. The differential equation formed by eliminating a and b from \( y = ae^x + be^{-x} \)
(a) \( \frac { d^2y }{dx^2} - y = 0 \)
(b) \( \frac { d^2y }{dx^2} – \frac { dy }{dx}y = 0 \)
(c) \( \frac { d^2y }{dx^2} = 0 \)
(d) \( \frac { d^2y }{dx^2} − x = 0 \)
Answer: (a) \( \frac { d^2y }{dx^2} - y = 0 \)
In simple words: We start with the given equation and find its first and second derivatives. Then, we substitute these back into the original equation to remove the constants 'a' and 'b'. The goal is to create an equation that only has 'y' and its derivatives.

🎯 Exam Tip: To eliminate constants, differentiate the given equation as many times as there are arbitrary constants. Then, form a determinant or substitute back to remove the constants.

Question 6. If \( y = ex + c - c^3 \) then its differential equation is
(a) \( y = x\frac { dy }{dx} + \frac { dy }{dx} - (\frac { dy }{dx})^3 \)
(b) \( y + (\frac { dy }{dx})³ = x \frac { dy }{dx} – \frac { dy }{dx} \)
(c) \( \frac { dy }{dx} + (\frac { dy }{dx})³ – x\frac { dy }{dx} \)
(d) \( \frac { d^3y }{dx^3} = 0 \)
Answer: (a) \( y = x\frac { dy }{dx} + \frac { dy }{dx} – (\frac { dy }{dx})³ \)
In simple words: This equation is a Clairaut's form. We find the derivative of y with respect to x. Since \( \frac{dy}{dx} \) equals 'c', we can substitute 'c' back into the original equation using \( \frac{dy}{dx} \) to get the differential equation.

🎯 Exam Tip: For equations of the form \( y = px + f(p) \) where \( p = \frac{dy}{dx} \), the differential equation is found by replacing 'c' with 'p' (i.e., \( \frac{dy}{dx} \)) in the given equation.

Question 7. The integrating factor of the differential equation \( \frac { dy }{dx} + Px = Q \) is
(a) \( e^{\int pdx} \)
(b) \( e^{\int Pdx} \)
(c) \( e^{\int Pdy} \)
(d) \( e^{\int pdy} \)
Answer: (d) \( e^{\int pdy} \)
In simple words: The integrating factor helps us solve linear differential equations. For an equation in the form \( \frac{dx}{dy} + Px = Q \), where P and Q are functions of y, the integrating factor is found by raising 'e' to the power of the integral of 'P' with respect to 'y'. This form is different from the usual \( \frac{dy}{dx} + Py = Q \).

🎯 Exam Tip: Pay close attention to whether the equation is \( \frac{dy}{dx} + Py = Q \) (integrating factor \( e^{\int Pdx} \)) or \( \frac{dx}{dy} + Px = Q \) (integrating factor \( e^{\int Pdy} \)). The variable of integration depends on the denominator of the derivative.

Question 8. The complementary function of \( (D^2 + 4) y = e^{2x} \) is
(a) \( (Ax+B)e^{2x} \)
(b) \( (Ax+B)e^{-2x} \)
(c) \( A \cos 2x + B \sin 2x \)
(d) \( Ae^{-2x} + Be^{2x} \)
Answer: (c) \( A \cos 2x + B \sin 2x \)
In simple words: The complementary function is part of the general solution for a non-homogeneous differential equation. To find it, we solve the auxiliary equation \( m^2 + 4 = 0 \), which gives imaginary roots \( m = \pm 2i \). For imaginary roots of the form \( \alpha \pm i\beta \), the complementary function is \( e^{\alpha x} (A \cos \beta x + B \sin \beta x) \). Here, \( \alpha = 0 \) and \( \beta = 2 \).

🎯 Exam Tip: Remember the three cases for auxiliary equation roots: real and distinct, real and equal, and complex conjugate. Each case has a specific form for the complementary function.

Question 9. The differential equation of \( y = mx + c \) is (m and c are arbitrary constants)
(a) \( \frac { d^2y }{dx^2} = 0 \)
(b) \( y = x\frac { dy }{dx} + c \)
(c) \( xdy + ydx = 0 \)
(d) \( ydx – xdy = 0 \)
Answer: (a) \( \frac { d^2y }{dx^2} = 0 \)
In simple words: We start with the equation \( y = mx + c \). Differentiate it once to remove 'c', giving \( \frac{dy}{dx} = m \). Then differentiate it a second time to remove 'm', resulting in \( \frac{d^2y}{dx^2} = 0 \). This is the differential equation without any constants.

🎯 Exam Tip: For an equation of a straight line, which has two arbitrary constants (m and c), the differential equation will be of the second order.

Question 10. The particular integral of the differential equation \( \frac { d^2y }{dx^2} - 8\frac { dy }{dx} + 16y = 2e^{4x} \) is
(a) \( \frac {x^2e^{4x} }{2!} \)
(b) \( y = x\frac { e^{4x} }{2!} \)
(c) \( x^2e^{4x} \)
(d) \( xe^{4x} \)
Answer: (a) \( \frac {x^2e^{4x} }{2!} \)
In simple words: To find the particular integral for \( e^{ax} \) type functions when the denominator becomes zero, we multiply by \( x \) and differentiate the denominator with respect to D. We repeat this process until the denominator is non-zero. The expression \( \frac{x^2}{2!} \) arises because the root \( D=4 \) is repeated twice in the auxiliary equation \( (D-4)^2=0 \), matching the \( e^{4x} \) term.

🎯 Exam Tip: When \( f(D) = 0 \) for the particular integral with \( e^{ax} \), apply the rule \( PI = x^k \frac{1}{f^{(k)}(D)} e^{ax} \), where 'k' is the order of the root \( D=a \) in \( f(D) \).

Question 11. Solution of \( \frac { dx }{dy} + Px = 0 \)
(a) \( x = ce^{py} \)
(b) \( x = ce^{-py} \)
(c) \( x = py + c \)
(d) \( x = cy \)
Answer: (b) \( x = ce^{-py} \)
In simple words: This is a linear differential equation. We can rewrite it as \( \frac{dx}{dy} = -Px \). Then, separate the variables to get \( \frac{dx}{x} = -Pdy \). Integrating both sides gives \( \ln|x| = -Py + \ln|c| \), which simplifies to \( x = ce^{-py} \). This type of equation is often solved by variable separation and integration.

🎯 Exam Tip: For simple first-order differential equations, always check if variable separation is possible before resorting to more complex methods like integrating factors.

Question 12. If \( \sec^2x \) is an integrating factor of the differential equation \( \frac { dx }{dy} + Px = Q \) then P =
(a) 2 tan x
(b) sec x
(c) cos 2 x
(d) tan 2 x
Answer: (a) 2 tan x
In simple words: The integrating factor (IF) for an equation like \( \frac{dx}{dy} + Px = Q \) is \( e^{\int Pdy} \). However, the given IF is \( \sec^2x \), which suggests that the original equation might actually be in the form \( \frac{dy}{dx} + Py = Q \), where P is a function of x. If IF is \( e^{\int Pdx} \), then \( e^{\int Pdx} = \sec^2x \). Taking natural log on both sides and differentiating reveals that \( P = 2 \tan x \). This matches the form for the variable x.

🎯 Exam Tip: When given the integrating factor, remember that \( IF = e^{\int Pdx} \) for equations in \( \frac{dy}{dx} \) form, and \( IF = e^{\int Pdy} \) for equations in \( \frac{dx}{dy} \) form. Differentiate \( \ln(IF) \) to find P.

Question 13. The integrating factor of the differential equation \( x \frac { dy }{dx} – y = x^2 \) is
(a) \( \frac {-1}{x} \)
(b) \( \frac { 1 }{x} \)
(c) \( \log x \)
(d) \( x \)
Answer: (b) \( \frac { 1 }{x} \)
In simple words: First, divide the entire equation by 'x' to bring it into the standard linear form: \( \frac{dy}{dx} + (-\frac{1}{x})y = x \). Here, P is \( -\frac{1}{x} \). The integrating factor is then \( e^{\int Pdx} = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = e^{\ln x^{-1}} = x^{-1} = \frac{1}{x} \). This is a common pattern for linear differential equations.

🎯 Exam Tip: Always convert the differential equation into the standard linear form \( \frac{dy}{dx} + Py = Q \) before identifying P and calculating the integrating factor.

Question 14. The solution of the differential equation where P and Q are the function of x is
(a) \( y = \int Q e^{\int Pdx} dx + c \)
(b) \( y = \int Q e^{-\int Pdx} dx + c \)
(c) \( y e^{\int Pdx} = \int Q e^{\int Pdx} dx + c \)
(d) \( y e^{\int Pdx} = \int Q e^{-\int Pdx} dx + c \)
Answer: (c) \( y e^{\int Pdx} = \int Q e^{\int Pdx} dx + c \)
In simple words: For a first-order linear differential equation in the form \( \frac{dy}{dx} + Py = Q \), the solution is found by multiplying the integrating factor \( (IF = e^{\int Pdx}) \) by 'y' on the left side, and by 'Q' on the right side, then integrating. The integral on the right is then taken with respect to x.

🎯 Exam Tip: Memorize the standard solution formula for linear differential equations, which involves the integrating factor. This is a fundamental result used frequently.

Question 15. The differential equation formed by eliminating A and B from \( y = e^{-2x} (A \cos x + B \sin x) \) is
(a) \( y_2 – 4y_1 + 5 = 0 \)
(b) \( y_2 + 4y – 5 = 0 \)
(c) \( y_2 – 4y_1 + 5y = 0 \)
(d) \( y_2 + 4y_1 + 5y = 0 \)
Answer: (d) \( y_2 + 4y_1 + 5y = 0 \)
In simple words: We start by multiplying both sides by \( e^{2x} \). Then, we differentiate the equation twice and try to substitute back to eliminate the constants A and B. The goal is to express \( y_2 \), \( y_1 \), and \( y \) in a relationship that does not contain A or B.

🎯 Exam Tip: When eliminating constants, multiply by appropriate exponential terms to simplify differentiation. The number of constants usually indicates the order of the resulting differential equation.

Question 16. The particular integral of the differential equation \( f(D) y = e^{ax} \) where \( f(D) = (D – a)^2 \) is
(a) \( \frac {x^2}{2} e^{ax} \)
(b) \( xe^{ax} \)
(c) \( \frac { x }{2} e^{ax} \)
(d) \( x^2 e^{ax} \)
Answer: (a) \( \frac {x^2}{2} e^{ax} \)
In simple words: When finding the particular integral for \( e^{ax} \) and the operator \( f(D) \) has \( (D-a)^k \) as a factor (meaning D=a is a root of order k), the particular integral is \( \frac{x^k}{k!} e^{ax} \). In this case, \( k=2 \) because \( f(D) = (D-a)^2 \). This is a special rule for resonant cases.

🎯 Exam Tip: Remember this specific rule for finding the particular integral when \( f(a) = 0 \) for the \( e^{ax} \) term. The power of x and the factorial in the denominator depend on the multiplicity of the root 'a'.

Question 17. The differential equation of \( x^2 + y^2 = a^2 \)
(a) \( xdy + ydx = 0 \)
(b) \( ydx – xdy = 0 \)
(c) \( xdx – ydx = 0 \)
(d) \( xdx + ydy = 0 \)
Answer: (d) \( xdx + ydy = 0 \)
In simple words: We differentiate the given equation \( x^2 + y^2 = a^2 \) with respect to x. The derivative of \( x^2 \) is \( 2x \), and the derivative of \( y^2 \) is \( 2y \frac{dy}{dx} \). The derivative of \( a^2 \) (a constant) is 0. So we get \( 2x + 2y \frac{dy}{dx} = 0 \). Dividing by 2 and rearranging terms gives \( x + y \frac{dy}{dx} = 0 \), which can also be written as \( xdx + ydy = 0 \). This shows how the equation relates to its derivative.

🎯 Exam Tip: Implicit differentiation is key for equations involving both x and y. Remember that \( \frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx} \).

Question 18. The complementary function of \( \frac { d^y }{dx^2} - \frac { dy }{dx} = 0 \) is
(a) \( A + Be^x \)
(b) \( (A + B)e^x \)
(c) \( (Ax + B)e^x \)
(d) \( Ae^x + B \)
Answer: (a) \( A + Be^x \)
In simple words: We write the auxiliary equation by replacing \( \frac{d^2y}{dx^2} \) with \( m^2 \) and \( \frac{dy}{dx} \) with \( m \). This gives \( m^2 - m = 0 \), which factors into \( m(m-1) = 0 \). The roots are \( m=0 \) and \( m=1 \). For real and distinct roots, the complementary function is \( C_1e^{m_1x} + C_2e^{m_2x} \). So, it's \( Ae^{0x} + Be^{1x} = A + Be^x \).

🎯 Exam Tip: Always factorize the auxiliary equation completely to find all roots. Remember that \( e^{0x} = 1 \), which often simplifies one term of the complementary function.

Question 19. The P.I of \( (3D^2 + D – 14) y = 13e^{2x} \) is
(a) \( \frac { 1 }{2} e^x \)
(b) \( xe^{2x} \)
(c) \( \frac { x^2 }{2} e^{2x} \)
(d) \( Ae^x + B \)
Answer: (b) \( xe^{2x} \)
In simple words: To find the particular integral for \( 13e^{2x} \), we substitute \( D=2 \) into \( 3D^2 + D - 14 \). If the result is zero, we must use a special rule: multiply by 'x' and differentiate the denominator with respect to D. Differentiating \( 3D^2 + D - 14 \) gives \( 6D + 1 \). Substituting \( D=2 \) into this new denominator gives \( 6(2)+1 = 13 \). So, the PI is \( x \frac{13}{13} e^{2x} = xe^{2x} \). This method is used when the exponential term is a root of the auxiliary equation.

🎯 Exam Tip: For particular integrals with \( e^{ax} \) terms, first substitute \( D=a \). If \( f(a)=0 \), apply the rule \( PI = x \frac{1}{f'(D)} e^{ax} \), and then substitute \( D=a \) into \( f'(D) \).

Question 20. The general solution of the differential equation \( \frac { dy }{dx} = \cos x \) is
(a) \( y = \sin x + 1 \)
(b) \( y = \sin x – 2 \)
(c) \( y = \sin x \), C is an arbitrary constant
(d) \( y = \sin x + C \), C is an arbitrary constant
Answer: (d) \( y = \sin x + C \), C is an arbitrary constant
In simple words: To solve \( \frac{dy}{dx} = \cos x \), we separate the variables to get \( dy = \cos x \, dx \). Then, we integrate both sides. The integral of \( dy \) is \( y \), and the integral of \( \cos x \, dx \) is \( \sin x \). We must also add an arbitrary constant of integration, C, because there are many functions whose derivative is \( \cos x \).

🎯 Exam Tip: Always remember to add the constant of integration, C, when performing indefinite integration to find the general solution of a differential equation.

Question 21. A homogeneous differential equation of the form \( \frac { dy }{dx} = f(\frac { y }{x}) \) can be solved by making substitution.
(a) \( y = v x \)
(b) \( y = y x \)
(c) \( x = v y \)
(d) \( x = v \)
Answer: (a) \( y = v x \)
In simple words: For a homogeneous differential equation where the right side is a function of \( \frac{y}{x} \), we substitute \( y = vx \). This means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). This substitution helps transform the homogeneous equation into a variable separable form, making it easier to solve.

🎯 Exam Tip: The key to solving homogeneous differential equations is the correct substitution. Always use \( y=vx \) when the function is in terms of \( y/x \).

Question 22. A homogeneous differential equation of the form \( \frac { dy }{dx} = f(\frac { x }{y}) \) can be solved by making substitution.
(a) \( x = v y \)
(b) \( y = v x \)
(c) \( y = v \)
(d) \( x = v \)
Answer: (a) \( x = v y \)
In simple words: When a homogeneous differential equation's right side is a function of \( \frac{x}{y} \), the correct substitution is \( x = vy \). This is different from the previous case and changes \( \frac{dx}{dy} \) to \( v + y \frac{dv}{dy} \), which helps to separate the variables for integration.

🎯 Exam Tip: Distinguish between \( f(y/x) \) and \( f(x/y) \) forms for homogeneous equations. The substitution \( y=vx \) is for the former, and \( x=vy \) is for the latter, directly simplifying the ratio within the function.

Question 23. The variable separable form of \( \frac { dy }{dx} = \frac { y(x-y) }{x(x+y)} \) by taking \( y = v x \) and \( \frac { dy }{dx} = v + x \frac { dv }{dx} \)
(a) \( \frac {2v^2 }{1+v} dv = \frac { dx }{x} \)
(b) \( \frac {2v^2}{1+v} dv = –\frac { dx }{x} \)
(c) \( \frac { 2v^2 }{1-v} dv = \frac { dx }{x} \)
(d) \( \frac {1+v }{2v^2} dv = -\frac { dx }{x} \)
Answer: (d) \( \frac {1+v }{2v^2} dv = -\frac { dx }{x} \)
In simple words: Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) into the given equation. After substitution and algebraic manipulation, terms involving 'x' will move to one side with 'dx', and terms involving 'v' will move to the other side with 'dv'. This process transforms the homogeneous equation into a variable separable form, which is then ready for integration.

🎯 Exam Tip: When converting a homogeneous equation to variable separable form, ensure all 'x' terms (including 'dx') are on one side and all 'v' terms (including 'dv') are on the other, with no mixing.

Question 24. Which of the following is the homogeneous differential equation?
(a) \( (3x – 5) dx = (4y – 1) dy \)
(b) \( xy dx – (x³ + y³) dy = 0 \)
(c) \( y²dx + (x² – xy – y²) dy = 0 \)
(d) \( (x² + y) dx (y² + x) dy \)
Answer: (c) \( y²dx + (x² – xy – y²) dy = 0 \)
In simple words: A differential equation is homogeneous if, after rearranging it into \( \frac{dy}{dx} = f(x,y) \), the function \( f(x,y) \) can be expressed as \( f(\frac{y}{x}) \) or \( f(\frac{x}{y}) \). This means that every term in \( f(x,y) \) has the same degree. In option (c), all terms (like \( y^2 \), \( x^2 \), \( xy \)) are of degree 2, making it homogeneous.

🎯 Exam Tip: To check for homogeneity, replace 'x' with 'kx' and 'y' with 'ky'. If the degree of 'k' cancels out and the function remains the same, it is homogeneous. Alternatively, check if all terms have the same total degree.

Question 25. The solution of the differential equation \( \frac { dy }{dx} = \frac { y }{x} + \frac { f(\frac { y }{x}) }{ f(\frac { y }{x}) } \) is
(a) \( f\frac { y }{x} = k x \)
(b) \( x f\frac { y }{x} = k \)
(c) \( f\frac { y }{x} = ky \)
(d) \( x f\frac { y }{x} = k \)
Answer: (a) \( f\frac { y }{x} = k x \)
In simple words: The given equation simplifies to \( \frac{dy}{dx} = \frac{y}{x} + 1 \). This is a homogeneous equation. By substituting \( y = vx \) and simplifying, we get a separable equation \( \frac{dv}{1} = \frac{dx}{x} \). Integrating both sides leads to \( v = \ln x + k \). Replacing 'v' with \( \frac{y}{x} \) gives \( \frac{y}{x} = \ln x + k \). This doesn't directly match the options with \( f(\frac{y}{x}) \). However, if the question implied an equation of the form \( \frac{dy}{dx} = \frac{y}{x} + G(\frac{y}{x}) \) where \( G(\frac{y}{x}) = \frac{f(\frac{y}{x})}{f(\frac{y}{x})} = 1 \), then the solution should be obtained by integration. Given the options, and typical forms for solutions of separable homogeneous equations, the form \( f(\frac{y}{x}) = kx \) implies a specific relationship after integrating. Without the exact derivation for \( f(y/x) \), this answer usually comes from a separable form where \( \frac{1}{f(y/x)} d(y/x) = \frac{1}{x} dx \).

🎯 Exam Tip: For problems involving arbitrary functions like \( f(y/x) \), focus on the structure of the equation and how substitution simplifies it. The exact form of the solution depends on the specific function. Treat \( \frac{f(y/x)}{f(y/x)} \) as 1 unless otherwise specified.