CBSE Class 12 Mathematics Differential Equations MCQs Set 11

Answer: (b) 4

Explanation:
1. Each differentiation of the family equation lets us eliminate one arbitrary constant.
2. With four independent arbitrary constants, we must differentiate four times to eliminate all of them.
3. So the highest derivative in the resulting equation is \( \frac{d^4y}{dx^4} \) and the order is 4.

Teacher's Note:
1. The order of the differential equation of a family equals the number of independent arbitrary constants in it.
2. Before counting, merge constants that can be combined, such as \( c_1e^{c_2} \), which is really just one constant.

Answer: (c) 2, 2

Explanation:
1. The highest derivative is \( \frac{d^2y}{dx^2} \), so the order is 2.
2. Square both sides to remove the radical: \( \left(\frac{d^2y}{dx^2}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2 \).
3. The power of \( \frac{d^2y}{dx^2} \) is now 2, so the degree is 2.
4. Order and degree are 2, 2.

Teacher's Note:
1. Degree is read only after the equation is free from radicals and fractions in the derivatives.
2. Squaring here changes the power of the highest derivative from 1 to 2, which is the usual trap in such questions.

Answer: (a) 2, 2

Explanation:
1. The highest derivative present is \( \frac{d^2s}{dt^2} \), so the order is 2.
2. The equation is already a polynomial in derivatives and \( \frac{d^2s}{dt^2} \) appears with power 2.
3. Hence order = 2 and degree = 2.

Teacher's Note:
1. The degree is the power of the highest order derivative only, not the largest power seen anywhere in the equation.
2. Here \( \left(\frac{ds}{dt}\right)^3 \) has power 3, but it is a lower order derivative, so it does not decide the degree.

Answer: (a) 4, 1

Explanation:
1. The highest derivative present is \( \frac{d^4y}{dx^4} \), so the order is 4.
2. It appears with power 1 and the equation is a polynomial in the derivatives.
3. Hence order = 4 and degree = 1.

Teacher's Note:
1. A linear differential equation always has degree 1, because every derivative appears to the first power.
2. Read the question carefully: "respectively" means the first number is the order and the second is the degree.

Answer: (d) 1 and 2

Explanation:
1. Write p = \( \frac{dy}{dx} \). The equation is \( \sqrt{p} = 4p + 7x \).
2. Squaring both sides removes the radical: p = (4p + 7x)², which is a polynomial in p.
3. Expanding, the highest power of p is p², so the degree is 2.
4. The only derivative is the first, so the order is 1. Order and degree are 1 and 2.

Teacher's Note:
1. Always isolate the radical term first and then square, otherwise you cannot read the degree.
2. A degree such as 1/2 is never possible, because degree must be a positive integer once radicals are removed.

Answer: (d) 2 and 3

Explanation:
1. The highest derivative present is \( \frac{d^2y}{dx^2} \), so the order is 2.
2. The power of \( \frac{d^2y}{dx^2} \) is 3; the power 4 belongs to the lower derivative \( \frac{dy}{dx} \).
3. Hence order = 2 and degree = 3.

Teacher's Note:
1. Find the highest order derivative first and then look only at its power to get the degree.
2. Option (a) uses the power of \( \frac{dy}{dx} \), which is the most common mistake in this type.

Answer: (a) 1, 3

Explanation:
1. Opening the bracket gives \( \frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + 1 = 0 \).
2. The highest derivative is \( \frac{d^3y}{dx^3} \), so the order is 3.
3. It appears to the first power, so the degree is 1.
4. The question asks for degree and order in that sequence, so the answer is 1, 3.

Teacher's Note:
1. Note the order in which the question asks: here degree comes first, so the pair is (1, 3), not (3, 1).
2. The bracket only contains a second derivative, so it does not change the order or degree.

Answer: (c) 2 and 1

Explanation:
1. The highest derivative is \( \frac{d^2y}{dx^2} \), so the order is 2.
2. It appears with power 1 (the coefficient x does not matter), so the degree is 1.
3. Order and degree are 2 and 1.

Teacher's Note:
1. Coefficients that are functions of x, like the x in front of \( \frac{d^2y}{dx^2} \), never affect order or degree.
2. The square on \( \frac{dy}{dx} \) makes the equation non-linear, but the degree is still 1.

Answer: (b) 4

Explanation:
1. \( c_1e^{2x + c_2} = (c_1e^{c_2})e^{2x} = Ae^{2x} \), so \( c_1 \) and \( c_2 \) merge into one constant A.
2. \( c_4\sin(x + c_5) \) has two genuine constants (amplitude and phase), and \( c_3e^x \) has one.
3. So y = \( Ae^{2x} + c_3e^x + c_4\sin(x + c_5) \) has 4 independent constants.
4. The order of the differential equation is 4.

Teacher's Note:
1. Count only independent constants: any constants that combine into a single one must be counted once.
2. A quick check is to rewrite the solution in its simplest form before counting.

Answer: (a) 1

Explanation:
1. The highest derivative is \( \frac{d^2y}{dx^2} \).
2. It appears with power 1 and the equation is a polynomial in derivatives.
3. So the degree is 1.

Teacher's Note:
1. The cube on \( \frac{dy}{dx} \) belongs to a lower derivative and is ignored while finding the degree.
2. Always identify the order first; the degree depends entirely on which derivative is the highest.

Answer: (c) dy + dx = 0

Explanation:
1. In (a), \( \frac{dy}{dx} \) is squared, so it is first order but second degree.
2. In (b), the highest derivative is \( \frac{d^2y}{dx^2} \), so it is second order.
3. In (c), dy + dx = 0 gives \( \frac{dy}{dx} = -1 \): first order and first degree.
4. So the required equation is dy + dx = 0.

Teacher's Note:
1. An equation in differential form M dx + N dy = 0 is always first order and first degree.
2. Check order first and then degree; both must be 1 for the equation to qualify.

Answer: (c) \( \frac{d^2y}{dx^2} + 9y = 0 \)

Explanation:
1. y = 4 sin 3x gives \( \frac{dy}{dx} = 12\cos 3x \).
2. Differentiating again, \( \frac{d^2y}{dx^2} = -36\sin 3x = -9(4\sin 3x) = -9y \).
3. So \( \frac{d^2y}{dx^2} + 9y = 0 \).

Teacher's Note:
1. For y = A sin kx or A cos kx, the second derivative is always −k²y, so the equation is y″ + k²y = 0.
2. A sine solution can never satisfy a first order equation like y′ ± 8y = 0, whose solutions are exponentials.

Answer: (d) \( \frac{d^2y}{dx^2} - m^2y = 0 \)

Explanation:
1. y = \( ae^{mx} + be^{-mx} \) gives \( \frac{dy}{dx} = m(ae^{mx} - be^{-mx}) \).
2. Differentiating again, \( \frac{d^2y}{dx^2} = m^2(ae^{mx} + be^{-mx}) = m^2y \).
3. So \( \frac{d^2y}{dx^2} - m^2y = 0 \).

Teacher's Note:
1. Two arbitrary constants a and b mean the equation must be of second order, which rules out (a) and (b) at once.
2. Remember the pair: exponentials \( e^{\pm mx} \) give y″ − m²y = 0, while sin mx and cos mx give y″ + m²y = 0.

Answer: (c) \( y = e^x\sin x + c \)

Explanation:
1. Integrate both sides: y = \( \int e^x(\sin x + \cos x)\,dx \).
2. Use \( \int e^x[f(x) + f'(x)]\,dx = e^xf(x) + c \) with f(x) = sin x, since f′(x) = cos x.
3. Hence y = \( e^x\sin x + c \).

Teacher's Note:
1. The pattern \( e^x[f(x) + f'(x)] \) is worth spotting instantly; it saves a full integration by parts.
2. You can check by differentiating: \( \frac{d}{dx}(e^x\sin x) = e^x\sin x + e^x\cos x \).

Answer: (d) \( y = \tan\left(\frac{x^2}{2} + x + c\right) \)

Explanation:
1. Separate the variables: \( \frac{dy}{1 + y^2} = (1 + x)\,dx \).
2. Integrate: \( \tan^{-1}y = x + \frac{x^2}{2} + c \).
3. So y = \( \tan\left(\frac{x^2}{2} + x + c\right) \).

Teacher's Note:
1. Whenever you see \( 1 + y^2 \) on the right, divide by it so that the left side integrates to \( \tan^{-1}y \).
2. Do not forget the factor 1/2 when integrating x; options (a) and (b) are built on that slip.

Answer: (c) \( y = \frac{1}{2}\log_e(1 + x^2) + c \)

Explanation:
1. Separate: \( dy = \frac{x}{1 + x^2}\,dx \).
2. Put t = 1 + x², dt = 2x dx: \( \int \frac{x}{1 + x^2}\,dx = \frac{1}{2}\log_e(1 + x^2) \).
3. So y = \( \frac{1}{2}\log_e(1 + x^2) + c \).

Teacher's Note:
1. When the numerator is a constant multiple of the derivative of the denominator, the integral is a logarithm.
2. \( \tan^{-1}x \) would come only if the numerator were 1, as in \( \int \frac{dx}{1 + x^2} \).

Answer: (b) \( y + \log x + \frac{x^2}{2} + c = 0 \)

Explanation:
1. Rewrite: \( \frac{dy}{dx} = -\frac{1 + x^2}{x} = -\left(\frac{1}{x} + x\right) \).
2. Integrate: y = \( -\log x - \frac{x^2}{2} + C \).
3. Taking all terms to one side, \( y + \log x + \frac{x^2}{2} + c = 0 \) (with c = −C).

Teacher's Note:
1. Split a fraction like \( \frac{1 + x^2}{x} \) term by term before integrating.
2. Since c is arbitrary, a constant on either side of the equation, with any sign, represents the same family.

Answer: (a) \( y = \sec x + \tan x + c \)

Explanation:
1. Expand: \( \frac{dy}{dx} = \sec^2 x + \sec x\tan x \).
2. Integrate: \( \int \sec^2 x\,dx = \tan x \) and \( \int \sec x\tan x\,dx = \sec x \).
3. So y = sec x + tan x + c.

Teacher's Note:
1. Expanding a product of trigonometric functions often turns it into standard integrals.
2. Learn the pairs: the derivative of tan x is sec²x and the derivative of sec x is sec x tan x.

Answer: (a) x = cy

Explanation:
1. y dx = x dy, so \( \frac{dx}{x} = \frac{dy}{y} \).
2. Integrate: log x = log y + log c = log (cy).
3. Hence x = cy.

Teacher's Note:
1. Writing the constant as log c makes it easy to combine logarithms into a single neat answer.
2. The family x = cy is the set of straight lines through the origin, which you can check: y dx − x dy = 0 for y = x/c.

Answer: (a) y = 1 + cxy

Explanation:
1. Write it as \( x\frac{dy}{dx} = y^2 - y = y(y - 1) \), so \( \frac{dy}{y(y - 1)} = \frac{dx}{x} \).
2. Partial fractions: \( \frac{1}{y(y - 1)} = \frac{1}{y - 1} - \frac{1}{y} \), so log (y − 1) − log y = log x + log c.
3. Hence \( \frac{y - 1}{y} = cx \), i.e. y − 1 = cxy.
4. So y = 1 + cxy.

Teacher's Note:
1. Factorising the right side as y(y − 1) is the key step that makes the variables separable.
2. Quick check: y = 1 satisfies the equation (0 + 1 = 1) and is the case c = 0 of y = 1 + cxy, but it never fits y + 1 = cxy.

Answer: (c) x = cot y + c

Explanation:
1. Separate: \( \frac{dy}{\sin^2 y} = -dx \), i.e. \( \text{cosec}^2 y\,dy = -dx \).
2. Integrate: \( -\cot y = -x + c_1 \).
3. So x = cot y + c.

Teacher's Note:
1. Remember \( \int \text{cosec}^2 y\,dy = -\cot y \); the minus sign is easy to lose.
2. Here the answer naturally comes out as x in terms of y, which is perfectly acceptable for a general solution.

Answer: (a) \( e^y = e^x + \frac{x^3}{3} + c \)

Explanation:
1. Factor: \( \frac{dy}{dx} = e^{-y}(e^x + x^2) \).
2. Separate: \( e^y\,dy = (e^x + x^2)\,dx \).
3. Integrate: \( e^y = e^x + \frac{x^3}{3} + c \).

Teacher's Note:
1. Use \( e^{x - y} = e^x \cdot e^{-y} \) to pull out the common factor \( e^{-y} \) before separating.
2. Always divide the power by its new index: \( \int x^2\,dx = \frac{x^3}{3} \), not \( x^3 \).

Answer: (c) \( \sin y = e^x\log x + c \)

Explanation:
1. Divide by x: \( \cos y\,dy = \left(e^x\log x + \frac{e^x}{x}\right)dx \).
2. The right side is \( e^x[f(x) + f'(x)] \) with f(x) = log x, so it integrates to \( e^x\log x \).
3. Hence sin y = \( e^x\log x + c \).

Teacher's Note:
1. The form \( e^x[f(x) + f'(x)] \) appears again here, hidden by the common factor x.
2. Check by differentiating \( e^x\log x \): you get \( e^x\log x + \frac{e^x}{x} \), exactly the right side.

Answer: (d) Both (a) and (c)

Explanation:
1. Separate: \( \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} \), so \( \tan^{-1}y - \tan^{-1}x = C \).
2. Using \( \tan^{-1}y - \tan^{-1}x = \tan^{-1}\frac{y - x}{1 + xy} \), we get \( \frac{y - x}{1 + xy} = c \), i.e. y − x = c(1 + xy), which is (c).
3. Writing the same constant as \( -\frac{1}{c} \) instead gives 1 + xy + c(y − x) = 0, which is (a).
4. (b) and (d) describe \( \tan^{-1}x + \tan^{-1}y \) = constant, which is a different family.
5. So both (a) and (c) are correct.

Teacher's Note:
1. The same family can look different when the arbitrary constant is replaced by its reciprocal or negative.
2. Before rejecting an option, rearrange it to see if it is the same relation with a renamed constant.

Answer: (a) \( y\sin y = x^2\log x + c \)

Explanation:
1. Since log x² = 2 log x, the equation is \( (\sin y + y\cos y)\,dy = (2x\log x + x)\,dx \).
2. Left side: \( \frac{d}{dy}(y\sin y) = \sin y + y\cos y \), so it integrates to y sin y.
3. Right side: \( \int 2x\log x\,dx = x^2\log x - \frac{x^2}{2} \), and adding \( \int x\,dx = \frac{x^2}{2} \) gives \( x^2\log x \).
4. Hence \( y\sin y = x^2\log x + c \).

Teacher's Note:
1. Recognising \( \sin y + y\cos y \) as the derivative of the product y sin y saves integration by parts.
2. Simplify log x² to 2 log x first; the extra x term is exactly what cancels the \( -\frac{x^2}{2} \).

Answer: (a) \( \tan y = c(1 - e^x)^3 \)

Explanation:
1. Divide by \( (1 - e^x)\tan y \): \( \frac{3e^x}{1 - e^x}\,dx + \frac{\sec^2 y}{\tan y}\,dy = 0 \).
2. Integrate: \( -3\log|1 - e^x| + \log|\tan y| = \log c \).
3. So \( \log\frac{\tan y}{(1 - e^x)^3} = \log c \), giving \( \tan y = c(1 - e^x)^3 \).

Teacher's Note:
1. The derivative of \( 1 - e^x \) is \( -e^x \), so \( \int \frac{3e^x}{1 - e^x}\,dx = -3\log|1 - e^x| \); watch that minus sign.
2. Choosing the constant as log c lets you combine all logarithms and remove them cleanly.

Answer: (a) \( \log(1 + y) = x + \frac{x^2}{2} + c \)

Explanation:
1. Factor: 1 + x + y + xy = (1 + x)(1 + y).
2. Separate: \( \frac{dy}{1 + y} = (1 + x)\,dx \).
3. Integrate: \( \log(1 + y) = x + \frac{x^2}{2} + c \).

Teacher's Note:
1. Grouping 1 + x + y + xy as (1 + x)(1 + y) is the standard trick that makes this separable.
2. Try factorising by grouping whenever the right side has four terms like this.

Answer: (c) \( y = Axe^{x - y} \)

Explanation:
1. Factor: \( \frac{dy}{dx} = \frac{y(x + 1)}{x(y + 1)} \), so \( \left(1 + \frac{1}{y}\right)dy = \left(1 + \frac{1}{x}\right)dx \).
2. Integrate: y + log y = x + log x + C.
3. Rearrange: \( \log\frac{y}{x} = x - y + C \), so \( \frac{y}{x} = e^Ce^{x - y} \).
4. Writing \( A = e^C \), y = \( Axe^{x - y} \).

Teacher's Note:
1. Splitting \( \frac{y + 1}{y} \) as \( 1 + \frac{1}{y} \) makes the integration immediate.
2. When the answer mixes x and y in an exponent, collect all logarithms on one side and exponentiate.

Answer: (a) \( y = 2\tan\frac{x}{2} - x + c \)

Explanation:
1. Separate: \( dy = \frac{1 - \cos x}{1 + \cos x}\,dx = \frac{2\sin^2(x/2)}{2\cos^2(x/2)}\,dx = \tan^2\frac{x}{2}\,dx \).
2. Write \( \tan^2\frac{x}{2} = \sec^2\frac{x}{2} - 1 \).
3. Integrate: y = \( 2\tan\frac{x}{2} - x + c \).

Teacher's Note:
1. The half-angle forms 1 − cos x = 2 sin²(x/2) and 1 + cos x = 2 cos²(x/2) simplify such fractions at once.
2. Remember \( \int \sec^2\frac{x}{2}\,dx = 2\tan\frac{x}{2} \), because of the factor 1/2 inside.

Answer: (a) \( e^y + e^{-y} = \log x - \frac{x^2}{2} + c \)

Explanation:
1. Separate: \( \frac{e^{2y} - 1}{e^y}\,dy = -\frac{x^2 - 1}{x}\,dx \).
2. Simplify: \( (e^y - e^{-y})\,dy = \left(\frac{1}{x} - x\right)dx \).
3. Integrate: \( e^y + e^{-y} = \log x - \frac{x^2}{2} + c \).

Teacher's Note:
1. Note that \( \int -e^{-y}\,dy = +e^{-y} \), which is why the left side becomes a sum, not a difference.
2. Dividing each term of the numerator separately by \( e^y \) is cleaner than any substitution here.

Answer: (b) \( (y - 1)^2(1 - x^2) = c^2y^2 \)

Explanation:
1. Rewrite: \( (1 - x^2)\,dy = xy(y - 1)\,dx \), so \( \frac{dy}{y(y - 1)} = \frac{x\,dx}{1 - x^2} \).
2. Integrate: \( \log\frac{y - 1}{y} = -\frac{1}{2}\log(1 - x^2) + \log c \).
3. So \( \frac{y - 1}{y}\sqrt{1 - x^2} = c \).
4. Squaring, \( (y - 1)^2(1 - x^2) = c^2y^2 \).

Teacher's Note:
1. Collect the dx terms first: xy dx − xy² dx = −xy(y − 1) dx, which reveals the separable form.
2. Option (a) has lost the constant; a general solution of a first order equation must contain one arbitrary constant.

Answer: (a) \( y(x^2 + 1) = 4 \)

Explanation:
1. Separate: \( \frac{dy}{y} = -\frac{2x}{x^2 + 1}\,dx \).
2. Integrate: log y = −log (x² + 1) + log C, so y(x² + 1) = C.
3. At (1, 2): 2(1 + 1) = C, so C = 4.
4. The curve is y(x² + 1) = 4.

Teacher's Note:
1. For a curve through a given point, find the general solution first and then substitute the point to fix the constant.
2. Check the answer: at x = 1, y = 4/2 = 2, as required.

Answer: (b) \( \frac{(\log y)^2}{2} + (2 - x^2)\cos x + 2x\sin x = c \)

Explanation:
1. Multiply by sin x and divide by y: \( \frac{\log y}{y}\,dy + x^2\sin x\,dx = 0 \).
2. \( \int \frac{\log y}{y}\,dy = \frac{(\log y)^2}{2} \) (put t = log y).
3. By parts twice, \( \int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x = (2 - x^2)\cos x + 2x\sin x \).
4. So \( \frac{(\log y)^2}{2} + (2 - x^2)\cos x + 2x\sin x = c \).

Teacher's Note:
1. Remember that cosec x = 1/sin x, so multiplying by sin x moves the x part to the dx term.
2. For \( \int x^2\sin x\,dx \), the tabular method (differentiate x², integrate sin x, alternate signs) is quickest.

Answer: (a) \( x^3 - y^3 = C \)

Explanation:
1. Separate: \( y^2\,dy = x^2\,dx \).
2. Integrate: \( \frac{y^3}{3} = \frac{x^3}{3} + c \).
3. Multiply by 3 and rename the constant: \( x^3 - y^3 = C \).

Teacher's Note:
1. A constant can be multiplied, negated or moved to the other side freely, so \( y^3 - x^3 = c \) and \( x^3 - y^3 = C \) are the same family.
2. Option (b) would come from \( \frac{dy}{dx} = -\frac{x^2}{y^2} \); check the sign before you choose.

Answer: (b) sin x = c sec y

Explanation:
1. Separate: \( \tan y\,dy = \cot x\,dx \).
2. Integrate: −log cos y = log sin x − log c.
3. So log (sin x cos y) = log c, i.e. sin x cos y = c.
4. Hence sin x = c sec y.

Teacher's Note:
1. Use \( \int \tan y\,dy = -\log\cos y \) and \( \int \cot x\,dx = \log\sin x \).
2. Dividing by cot y means multiplying by tan y; take care to move each function to the correct side.

Answer: (b) y = log x + c

Explanation:
1. Separate: \( dy = \frac{dx}{x} \).
2. Integrate: y = log x + c.

Teacher's Note:
1. This is the simplest type \( \frac{dy}{dx} = f(x) \): integrate the right side directly.
2. Differentiate your answer to check: the derivative of log x + c is 1/x.

Answer: (a) xy = c

Explanation:
1. Integrate each term: log x + log y = log c.
2. So log (xy) = log c.
3. Hence xy = c.

Teacher's Note:
1. Taking the constant as log c lets you combine logarithms immediately.
2. The equation is the same as \( x\,dy + y\,dx = 0 \), i.e. d(xy) = 0, which also gives xy = c.

Answer: (b) x = c sec y

Explanation:
1. Separate: \( \frac{dx}{x} = \tan y\,dy \).
2. Integrate: log x = log sec y + log c.
3. So x = c sec y.

Teacher's Note:
1. Dividing by cot y is the same as multiplying by tan y.
2. Use \( \int \tan y\,dy = \log\sec y \); the form log sec y keeps the final answer free of negative signs.

Answer: (a) \( e^{\cos x}\tan\frac{y}{2} = C \)

Explanation:
1. Separate: \( \frac{dy}{\sin y} = \sin x\,dx \), i.e. \( \text{cosec}\,y\,dy = \sin x\,dx \).
2. Integrate: \( \log\tan\frac{y}{2} = -\cos x + \log C \).
3. So \( \tan\frac{y}{2} = Ce^{-\cos x} \), i.e. \( e^{\cos x}\tan\frac{y}{2} = C \).

Teacher's Note:
1. Remember \( \int \text{cosec}\,y\,dy = \log\left|\tan\frac{y}{2}\right| \); options with tan y miss the half angle.
2. When a non-logarithmic term like cos x remains, exponentiate both sides to remove the log.