CBSE Class 12 Mathematics Differential Equations MCQs Set 13

Answer: (b) \( x^3y = \frac{x^4}{4} + c \)

Explanation:
1. Divide by x: \( \frac{dy}{dx} + \frac{3}{x}y = 1 \), so IF = \( e^{3\log x} = x^3 \).
2. Then \( \frac{d}{dx}(x^3y) = x^3 \).
3. Integrate: \( x^3y = \frac{x^4}{4} + c \).

Teacher's Note:
1. Always divide by the coefficient of \( \frac{dy}{dx} \) first, so that the equation is in the form y′ + Py = Q.
2. Check: \( y = \frac{x}{4} \) gives \( xy' + 3y = \frac{x}{4} + \frac{3x}{4} = x \).

Answer: (c) \( \frac{1}{\cos x} \)

Explanation:
1. Write \( \frac{dy}{dx} + (\tan x)y = \sec x \), so P = tan x.
2. IF = \( e^{\int \tan x\,dx} = e^{\log\sec x} = \sec x \).
3. sec x = \( \frac{1}{\cos x} \).

Teacher's Note:
1. The integrating factor depends only on P, never on Q.
2. Options may write the same factor in a different form, so convert sec x to 1/cos x before matching.

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

Explanation:
1. Divide by x log x: \( \frac{dy}{dx} + \frac{1}{x\log x}y = \frac{2}{x} \).
2. \( \int \frac{dx}{x\log x} = \log(\log x) \), so IF = log x.
3. Then \( y\log x = \int \frac{2\log x}{x}\,dx = (\log x)^2 + c \).

Teacher's Note:
1. Put t = log x to evaluate both \( \int \frac{dx}{x\log x} \) and \( \int \frac{\log x}{x}\,dx \) quickly.
2. Note that \( x\log x\frac{dy}{dx} + y = 2\log x \) is also \( \frac{d}{dx}(y\log x) = \frac{2\log x}{x} \) after dividing by x, which confirms the IF.

Answer: (a) \( y\sin^2 x = x^3 + c \)

Explanation:
1. P = 2 cot x, so IF = \( e^{2\log\sin x} = \sin^2 x \).
2. Then \( y\sin^2 x = \int 3x^2\,\text{cosec}^2 x\sin^2 x\,dx = \int 3x^2\,dx \).
3. So \( y\sin^2 x = x^3 + c \).

Teacher's Note:
1. The integrating factor is chosen to cancel awkward functions in Q; here \( \sin^2 x \) cancels \( \text{cosec}^2 x \).
2. Use \( e^{k\log f} = f^k \), so \( e^{2\log\sin x} = \sin^2 x \).

Answer: (b) \( y = 3 + Ce^{-x/3} \)

Explanation:
1. Linear with \( P = \frac{1}{3} \), so IF = \( e^{x/3} \).
2. Then \( ye^{x/3} = \int e^{x/3}\,dx = 3e^{x/3} + C \).
3. So \( y = 3 + Ce^{-x/3} \).

Teacher's Note:
1. For constant P, the integrating factor is \( e^{Px} \) and the solution decays to the steady value Q/P.
2. Check: y = 3 is a constant solution, since \( 0 + \frac{3}{3} = 1 \).

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

Explanation:
1. Rewrite: \( \frac{dy}{dx} - y = x^2 \), so IF = \( e^{-x} \).
2. Then \( ye^{-x} = \int x^2e^{-x}\,dx = -e^{-x}(x^2 + 2x + 2) + C \).
3. Multiply by eˣ: \( y = -(x^2 + 2x + 2) + Ce^x \), i.e. \( y + x^2 + 2x + 2 = Ce^x \).

Teacher's Note:
1. Bring the y term to the left with the correct sign before finding P; here P = −1.
2. \( \int x^2e^{-x}\,dx \) is quickly done by the tabular method: \( -e^{-x}(x^2 + 2x + 2) \).

Answer: (c) \( y = x^2 + ax \)

Explanation:
1. Rewrite: \( \frac{dy}{dx} - \frac{1}{x}y = x \), so IF = \( e^{-\log x} = \frac{1}{x} \).
2. Then \( \frac{y}{x} = \int x\cdot\frac{1}{x}\,dx = x + a \).
3. So \( y = x^2 + ax \).

Teacher's Note:
1. For \( P = -\frac{1}{x} \), the integrating factor is \( \frac{1}{x} \), not x; the sign of P matters.
2. Check: y = x² gives \( xy' = 2x^2 = x^2 + x^2 = y + x^2 \).

Answer: (c) \( \frac{1}{x}\frac{d^2y}{dx^2} = e^x \)

Explanation:
1. (a) gives \( \frac{dy}{dx} = -\frac{\sqrt{1 - x^2}}{\sqrt{1 - y^2}} \), which contains \( \sqrt{1 - y^2} \), so it is non-linear.
2. (b) has \( \left(\frac{ds}{dt}\right)^4 \) and a product \( s\frac{d^2s}{dt^2} \), so it is non-linear.
3. (d) gives \( \frac{dy}{dx} = -\frac{x(y^2 + 1)}{y(1 - x^2)} \), with y² terms, so it is non-linear.
4. (c) is \( \frac{d^2y}{dx^2} = xe^x \): y″ appears to power 1 with no products, so (c) is linear.

Teacher's Note:
1. Linearity can hold for any order; a second order equation is linear if y, y′, y″ appear only to the first power.
2. Rewrite differential forms such as M dx + N dy = 0 as \( \frac{dy}{dx} = -\frac{M}{N} \) to test them.

Answer: (a) \( xy = \frac{x^3}{3} + \frac{3}{2}x^2 + 2x + c \)

Explanation:
1. The left side is \( x\frac{dy}{dx} + y = \frac{d}{dx}(xy) \).
2. So \( \frac{d}{dx}(xy) = x^2 + 3x + 2 \).
3. Integrate: \( xy = \frac{x^3}{3} + \frac{3}{2}x^2 + 2x + c \).

Teacher's Note:
1. Recognising \( x\,y' + y \) as the derivative of xy means the equation is already in integrated form.
2. This is the same as dividing by x and using IF = x.

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

Explanation:
1. Treat x as the dependent variable: \( \frac{dx}{dy} = \frac{x + 2y^3}{y} \), i.e. \( \frac{dx}{dy} - \frac{1}{y}x = 2y^2 \).
2. IF = \( e^{-\log y} = \frac{1}{y} \).
3. Then \( \frac{x}{y} = \int 2y^2\cdot\frac{1}{y}\,dy = y^2 + c \).

Teacher's Note:
1. If the equation is not linear in y but becomes linear in x when written as dx/dy, solve it as a linear equation in x.
2. Here the term \( 2y^3 \) multiplying \( \frac{dy}{dx} \) is the clue to flip the derivative.

Answer: (a) y = cx − (1 + log x)

Explanation:
1. Rewrite: x dy = (y + log x) dx, so \( \frac{dy}{dx} - \frac{1}{x}y = \frac{\log x}{x} \) with IF = \( \frac{1}{x} \).
2. Then \( \frac{y}{x} = \int \frac{\log x}{x^2}\,dx = -\frac{\log x}{x} - \frac{1}{x} + c \) (by parts).
3. Multiply by x: y = cx − (1 + log x).

Teacher's Note:
1. For \( \int \frac{\log x}{x^2}\,dx \), take log x as the first function and \( \frac{1}{x^2} \) as the second.
2. Check: y = −(1 + log x) gives \( y' = -\frac{1}{x} \) and \( xy' - y = -1 + 1 + \log x = \log x \).

Answer: (b) xy + y log y − y = c

Explanation:
1. Treat x as dependent: \( \frac{dx}{dy} + \frac{1}{y}x = -\frac{\log y}{y} \), so IF = y.
2. Then \( xy = -\int \log y\,dy = -(y\log y - y) + c \).
3. So xy + y log y − y = c.

Teacher's Note:
1. When log y sits next to dy, the equation is usually linear in x; write it as dx/dy.
2. Check: the left side (x + log y) dy + y dx is the exact differential of xy + y log y − y.

Answer: (a) \( y = x^2 \)

Explanation:
1. Slope \( \frac{dy}{dx} = \frac{2y}{x} \), so \( \frac{dy}{y} = \frac{2\,dx}{x} \).
2. Integrate: log y = 2 log x + log C, so y = Cx².
3. At (1, 1): C = 1, so y = x².

Teacher's Note:
1. "Slope at (x, y)" always means \( \frac{dy}{dx} \) as a function of x and y.
2. Check: for y = x², \( \frac{dy}{dx} = 2x = \frac{2x^2}{x} = \frac{2y}{x} \).

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

Explanation:
1. Separate: \( \frac{dx}{x} = \frac{y\,dy}{1 + y^2} \).
2. Integrate: log x = ½ log (1 + y²) + log c, so x² = k(1 + y²).
3. At (1, 0): 1 = k, so x² = 1 + y², i.e. x² − y² = 1.

Teacher's Note:
1. Doubling the logarithm equation removes the 1/2 and gives a neat algebraic curve.
2. Substitute the given point into each option as a quick first filter, then check the slope condition.

Answer: (c) \( 6xy = 3x^3 + 29x - 6 \)

Explanation:
1. Integrate: \( y = \frac{x^2}{2} - \frac{1}{x} + C \).
2. At (3, 9): \( 9 = \frac{9}{2} - \frac{1}{3} + C \), so \( C = \frac{29}{6} \).
3. So \( y = \frac{x^2}{2} - \frac{1}{x} + \frac{29}{6} \); multiply by 6x: \( 6xy = 3x^3 + 29x - 6 \).

Teacher's Note:
1. Multiply through by the LCM of the denominators (here 6x) to match the form of the options.
2. Check with the point: 6(3)(9) = 162 and 3(27) + 29(3) − 6 = 81 + 87 − 6 = 162.

Answer: (b) \( xy = x^2 + x + 1 \)

Explanation:
1. \( \frac{dy}{dx} = 1 - \frac{1}{x^2} \), so \( y = x + \frac{1}{x} + C \).
2. At \( \left(2, \frac{7}{2}\right) \): \( \frac{7}{2} = 2 + \frac{1}{2} + C \), so C = 1.
3. So \( y = x + \frac{1}{x} + 1 \); multiply by x: \( xy = x^2 + x + 1 \).

Teacher's Note:
1. "Gradient" means the same as slope, i.e. \( \frac{dy}{dx} \).
2. Check with the point: \( 2 \times \frac{7}{2} = 7 \) and \( 4 + 2 + 1 = 7 \).

Answer: (a) (y − 1)(x + 1) + 2x = 0

Explanation:
1. Separate: \( \frac{dy}{y - 1} = \frac{dx}{x(x + 1)} = \left(\frac{1}{x} - \frac{1}{x + 1}\right)dx \).
2. Integrate: \( \log(y - 1) = \log\frac{x}{x + 1} + \log C \), so \( y - 1 = \frac{Cx}{x + 1} \).
3. At (1, 0): \( -1 = \frac{C}{2} \), so C = −2.
4. So (y − 1)(x + 1) = −2x, i.e. (y − 1)(x + 1) + 2x = 0.

Teacher's Note:
1. Use \( \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1} \); this partial fraction appears very often.
2. Check with the point: (0 − 1)(1 + 1) + 2(1) = −2 + 2 = 0.

Answer: (c) \( y^2 = x + 5 \)

Explanation:
1. Slope = reciprocal of twice the ordinate: \( \frac{dy}{dx} = \frac{1}{2y} \).
2. Separate: 2y dy = dx, so y² = x + C.
3. At (4, 3): 9 = 4 + C, so C = 5, and the curve is y² = x + 5.

Teacher's Note:
1. "Ordinate" means the y coordinate, and "reciprocal of twice the ordinate" is \( \frac{1}{2y} \).
2. Translate the words into \( \frac{dy}{dx} \) carefully before solving; most errors in such questions are in this step.

Answer: (b) Straight line passing through origin

Explanation:
1. x dy = y dx, so \( \frac{dy}{y} = \frac{dx}{x} \).
2. Integrate: log y = log x + log c, so y = cx.
3. y = cx is a straight line passing through the origin.

Teacher's Note:
1. The slope condition \( \frac{dy}{dx} = \frac{y}{x} \) says the tangent always points along the line to the origin.
2. Every line through the origin satisfies this, which is why the solution is the whole family y = cx.

Answer: (c) −1

Explanation:
1. y′ = 1 + y, so \( \frac{dy}{1 + y} = dx \) and log (1 + y) = x + C, i.e. 1 + y = Aeˣ.
2. y(0) = −1 gives 1 + (−1) = A, so A = 0.
3. So 1 + y = 0, i.e. y(x) = −1.

Teacher's Note:
1. A constant solution, where y′ = 0, is worth checking first: here y = −1 makes y′ − y = 0 + 1 = 1.
2. Option (d) also gives y(0) = −1, but it fails the equation, so always test both conditions.

Answer: (a) None

Explanation:
1. Separate: \( \frac{dy}{y + 1} = \frac{dx}{x - 1} \), so log (y + 1) = log (x − 1) + log C, i.e. y + 1 = C(x − 1).
2. At x = 1, every member of this family gives y + 1 = 0, i.e. y = −1.
3. So no solution can satisfy y(1) = 2; the number of solutions is none.

Teacher's Note:
1. All solution curves y + 1 = C(x − 1) pass through the single point (1, −1).
2. When the right side is undefined at the initial point, test whether any solution can reach the given value at all.

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

Explanation:
1. Factor: \( y' = (1 + x) + y^2(1 + x) = (1 + x)(1 + y^2) \).
2. Separate: \( \frac{dy}{1 + y^2} = (1 + x)\,dx \), so \( \tan^{-1}y = x + \frac{x^2}{2} + C \).
3. y(0) = 0 gives C = 0, so \( y = \tan\left(x + \frac{x^2}{2}\right) \).

Teacher's Note:
1. Group the terms as 1 + x and y²(1 + x) to see the common factor (1 + x).
2. With an initial condition, the final answer has no arbitrary constant, which rules out (b) and (c).

Answer: (d) 2 and 4

Explanation:
1. The highest derivative is \( \frac{d^2y}{dx^2} \), so the order is 2.
2. Raise both sides to the power 4 to remove the fractional power: \( \left(\frac{d^2y}{dx^2}\right)^4 = y + \left(\frac{dy}{dx}\right)^2 \).
3. Now the equation is a polynomial in derivatives and the power of \( \frac{d^2y}{dx^2} \) is 4.
4. Hence order = 2 and degree = 4.

Teacher's Note:
1. Always clear fractional powers of the derivatives first; the degree is read only after that.
2. Order never changes when you raise both sides to a power, only the degree can change.

Answer: (d) 4

Explanation:
1. Take the radical term alone: \( \left(\frac{d^2y}{dx^2}\right)^2 - y^3 = \left(\frac{dy}{dx}\right)^{1/2} \).
2. Square both sides: \( \left[\left(\frac{d^2y}{dx^2}\right)^2 - y^3\right]^2 = \frac{dy}{dx} \).
3. The highest power of \( \frac{d^2y}{dx^2} \) now is \( (2)(2) = 4 \).
4. Hence the degree is 4.

Teacher's Note:
1. Isolate the radical term on one side before squaring, otherwise you may leave a radical behind.
2. A fractional degree like 1/2 is never possible; degree is always a positive integer when it is defined.

Answer: (d) 6

Explanation:
1. Square both sides: \( \left(\frac{d^2y}{dx^2}\right)^6 = 1 + \frac{dy}{dx} \).
2. The equation is now a polynomial in derivatives with highest order derivative \( \frac{d^2y}{dx^2} \).
3. Its power is 6, so the degree is 6.

Teacher's Note:
1. Do not read the degree as 3 straight away; the right side still has a square root of a derivative.
2. When you square to remove a half power, every power on the other side doubles.

Answer: (a) Degree 3 and order 2

Explanation:
1. The highest order derivative present is \( \frac{d^2y}{dx^2} \), so the order is 2.
2. The equation is already a polynomial in the derivatives (no radicals or fractions).
3. The power of \( \frac{d^2y}{dx^2} \) is 3, so the degree is 3.
4. Hence degree 3 and order 2.

Teacher's Note:
1. Degree is the power of the highest order derivative, not the largest power seen anywhere in the equation.
2. Here the power 4 belongs to dy/dx, which is a lower order derivative, so it does not decide the degree.

Answer: (b) \( y'y'' + y = \sin x \)

Explanation:
1. \( y'^2 + x = y^2 \) contains only y′, so it is of order 1.
2. \( y'y'' + y = \sin x \) contains y″ as the highest derivative, so it is of order 2.
3. \( y''' + y'' + y = 0 \) is of order 3 and \( y' = y \) is of order 1.
4. Hence \( y'y'' + y = \sin x \) is the second order equation.

Teacher's Note:
1. The number of primes on y tells you the order of that derivative directly.
2. A product like y′y″ does not change the order; it only makes the equation non-linear.

Answer: (c) 3, 3

Explanation:
1. The highest derivative is \( \frac{d^3y}{dx^3} \), so the order is 3.
2. Cube both sides: \( \left(1 + 3\frac{dy}{dx}\right)^2 = 64\left(\frac{d^3y}{dx^3}\right)^3 \).
3. The power of \( \frac{d^3y}{dx^3} \) is 3, so the degree is 3.
4. Hence order 3, degree 3.

Teacher's Note:
1. The fractional power 2/3 is removed by cubing, not squaring.
2. Remember to cube the coefficient 4 as well, although it does not affect the degree.

Answer: (a) 3

Explanation:
1. The solution \( y = a\cos x + b\sin x + ce^{-x} \) has three arbitrary constants a, b and c.
2. None of them can be combined into one, since \( \cos x \), \( \sin x \) and \( e^{-x} \) are different functions.
3. Eliminating three independent constants needs three differentiations.
4. Hence the order is 3.

Teacher's Note:
1. The order of the differential equation equals the number of independent arbitrary constants in its general solution.
2. Before counting, always check whether two constants can be merged into one.

Answer: (d) m = 3 and n = 2

Explanation:
1. The highest derivative is \( \frac{d^3y}{dx^3} \), so m = 3.
2. Multiply the equation by \( \frac{d^3y}{dx^3} \) to remove it from the denominator: \( \left(\frac{d^2y}{dx^2}\right)^5\frac{d^3y}{dx^3} + 4\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{d^3y}{dx^3}\right)^2 = (x^2 - 1)\frac{d^3y}{dx^3} \).
3. The highest power of \( \frac{d^3y}{dx^3} \) is now 2, so n = 2.
4. Hence m = 3 and n = 2.

Teacher's Note:
1. A derivative in the denominator must be cleared before reading the degree.
2. After clearing, look only at the power of the highest order derivative, not the power 5 of the second derivative.

Answer: (a) 2, 2

Explanation:
1. Cross-multiply: \( \rho\,\frac{d^2y}{dx^2} = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} \).
2. Square both sides: \( \rho^2\left(\frac{d^2y}{dx^2}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^3 \).
3. Highest derivative \( \frac{d^2y}{dx^2} \) gives order 2 and its power 2 gives degree 2.
4. Hence order 2, degree 2.

Teacher's Note:
1. This is the radius of curvature formula, a favourite in competitive papers.
2. Clear the denominator first and then remove the half power by squaring.

Answer: (a) 1, 1

Explanation:
1. Only the first derivative \( \frac{dy}{dx} \) appears, so the order is 1.
2. The right side contains only x and y, and fractions in x and y do not affect the degree.
3. \( \frac{dy}{dx} \) appears to the power 1, so the degree is 1.
4. Hence order 1, degree 1.

Teacher's Note:
1. Radicals or fractions in x and y do not matter for degree; only those in the derivatives do.
2. This same equation can be solved by separating variables, which is done in a later question.

Answer: (d) Order two and degree one

Explanation:
1. The highest derivative is \( \frac{d^2y}{dx^2} \), so the order is two.
2. \( \frac{d^2y}{dx^2} \) appears to the first power, so the degree is one.
3. The term sin y is not of first degree in y, so the equation is not linear.
4. So it is of order two and degree one.

Teacher's Note:
1. A term like sin y, y², or y·y′ makes an equation non-linear even if its degree is one.
2. Linear and first degree are different ideas: every linear equation has degree one, but not the other way round.

Answer: (c) 1, 1

Explanation:
1. Write the equation as \( \frac{dy}{dx} = -\frac{xy^2 + x}{y - x^2y} \).
2. Only \( \frac{dy}{dx} \) appears, and to the first power.
3. Hence the order is 1 and the degree is 1.

Teacher's Note:
1. Any equation of the form M dx + N dy = 0 is always of first order and first degree.
2. Dividing by dx turns the differential form into dy/dx form immediately.

Answer: (c) \( \left[1 + \left(\frac{dy}{dx}\right)^3\right]^{2/3} = 4\frac{d^3y}{dx^3} \)

Explanation:
1. (a): order 4, degree 1. (b): order 3, degree 4.
2. (c): cube both sides: \( \left[1 + \left(\frac{dy}{dx}\right)^3\right]^2 = 64\left(\frac{d^3y}{dx^3}\right)^3 \), so order 3, degree 3.
3. (d): \( \left(y - x^2\frac{dy}{dx}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2 \), so order 1, degree 2.
4. Only (c) has the same order and degree.

Teacher's Note:
1. Work out each option fully, removing radicals before you decide its degree.
2. In (d) squaring is needed because of the square root of a derivative, which raises the degree to 2.

Answer: (c) 3

Explanation:
1. \( C_1 + C_2 \) is a single constant, say A.
2. \( C_4e^{x + C_5} = C_4e^{C_5}e^x = Be^x \), which is one constant B.
3. So \( y = A\cos(x + C_3) - Be^x \) has only three independent constants A, \( C_3 \) and B.
4. Hence the order is 3.

Teacher's Note:
1. Constants that only add or multiply together must be merged before you count them.
2. A phase constant inside cos(x + C) is genuinely independent and must be counted.

Answer: (b) 2

Explanation:
1. Square both sides: \( 9\left(\frac{d^2y}{dx^2}\right)^2 = \left\{1 + \left(\frac{dy}{dx}\right)^2\right\}^3 \).
2. The highest derivative \( \frac{d^2y}{dx^2} \) now appears with power 2.
3. Hence the degree is 2.

Teacher's Note:
1. The cube on the right is on dy/dx, a lower derivative, so it does not decide the degree.
2. Do not forget to square the coefficient 3, although only the power of y″ matters here.

Answer: (a) 1

Explanation:
1. Clear the fraction: \( y\frac{dy}{dx}\left[\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^3\right] = x \).
2. Only the first derivative \( \frac{dy}{dx} \) appears.
3. Hence the order is 1.

Teacher's Note:
1. The order only depends on the highest derivative present, so fractions can be ignored when finding order.
2. If degree were asked here, it would be 4 after clearing the fraction.

Answer: (c) 2, 3

Explanation:
1. The highest derivative is \( \frac{d^2y}{dx^2} \), so the order is 2.
2. Cube both sides: \( \left[4 + \left(\frac{dy}{dx}\right)^2\right]^2 = \left(\frac{d^2y}{dx^2}\right)^3 \).
3. The power of \( \frac{d^2y}{dx^2} \) is 3, so the degree is 3.
4. Hence order 2, degree 3.

Teacher's Note:
1. The power 2/3 is removed by cubing, which moves the cube onto the side with y″.
2. Option (d) swaps order and degree, a very common slip.