CBSE Class 12 Mathematics Differential Equations MCQs Set 15

Answer: (a) \( \log_e(x^2 + y^2) + 2\tan^{-1}\frac{y}{x} + c = 0 \)

Explanation:
1. Put y = vx: \( v + x\frac{dv}{dx} = \frac{v - 1}{v + 1} \), so \( x\frac{dv}{dx} = -\frac{1 + v^2}{1 + v} \).
2. Separate: \( \frac{(1 + v)\,dv}{1 + v^2} = -\frac{dx}{x} \).
3. Integrate: \( \tan^{-1}v + \frac{1}{2}\log(1 + v^2) = -\log x + c_1 \).
4. With v = y/x: \( \tan^{-1}\frac{y}{x} + \frac{1}{2}\log(x^2 + y^2) = c_1 \), i.e. \( \log_e(x^2 + y^2) + 2\tan^{-1}\frac{y}{x} + c = 0 \).

Teacher's Note:
1. The log x terms cancel when you write log(1 + y²/x²) = log(x² + y²) − 2 log x.
2. Splitting (1 + v)/(1 + v²) into 1/(1 + v²) + v/(1 + v²) gives the tan⁻¹ and log parts.

Answer: (a) \( (m + 1)y = x^{m + 1}\cos x + c(m + 1)\cos x \)

Explanation:
1. This is linear with P = tan x, Q = \( x^m\cos x \).
2. \( \text{I.F.} = e^{\int \tan x\,dx} = e^{\log\sec x} = \sec x \).
3. \( y\sec x = \int x^m\cos x\sec x\,dx = \int x^m\,dx = \frac{x^{m + 1}}{m + 1} + c \).
4. Multiply by (m + 1) cos x: \( (m + 1)y = x^{m + 1}\cos x + c(m + 1)\cos x \).

Teacher's Note:
1. The integrating factor sec x is chosen so that cos x in Q cancels neatly.
2. Do not forget to multiply the constant c by cos x when you solve for y.

Answer: (b) \( e^{\tan^{-1}y} \)

Explanation:
1. Rewrite: \( (1 + y^2)\frac{dx}{dy} = \tan^{-1}y - x \), i.e. \( \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2} \).
2. This is linear in x with \( R = \frac{1}{1 + y^2} \).
3. \( \text{I.F.} = e^{\int \frac{dy}{1 + y^2}} = e^{\tan^{-1}y} \).

Teacher's Note:
1. When x appears only to the first power with dx/dy, treat x as the dependent variable.
2. The integrating factor is always an exponential of ∫R dy, so a plain tan⁻¹ y cannot be the I.F.

Answer: (a) \( 3(1 + x^2)y = 4x^3 \)

Explanation:
1. The left side \( (1 + x^2)\frac{dy}{dx} + 2xy \) is exactly \( \frac{d}{dx}[(1 + x^2)y] \).
2. So \( \frac{d}{dx}[(1 + x^2)y] = 4x^2 \), giving \( (1 + x^2)y = \frac{4x^3}{3} + C \).
3. The curve passes through the origin, so 0 = 0 + C, giving C = 0.
4. Hence \( 3(1 + x^2)y = 4x^3 \).

Teacher's Note:
1. Spotting the left side as a product derivative saves finding the I.F. (which is 1 + x² anyway).
2. Check: at x = 0 the answer gives y = 0, so the curve passes through the origin.

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

Explanation:
1. Invert: \( \frac{dx}{dy} = x + y + 1 \), i.e. \( \frac{dx}{dy} - x = y + 1 \), linear in x.
2. \( \text{I.F.} = e^{-y} \), so \( xe^{-y} = \int (y + 1)e^{-y}\,dy \).
3. By parts: \( \int (y + 1)e^{-y}\,dy = -(y + 1)e^{-y} - e^{-y} = -(y + 2)e^{-y} \), so \( xe^{-y} = -(y + 2)e^{-y} + c \).
4. Multiply by \( e^y \): \( x = ce^y - y - 2 \).

Teacher's Note:
1. If dy/dx = 1/(something linear in x), invert it to get a linear equation in x.
2. Option (c) moves x to the same side without changing signs, which is a common slip.

Answer: (d) 2y sin x + cos 2x = c

Explanation:
1. P = cot x, so \( \text{I.F.} = e^{\int \cot x\,dx} = \sin x \).
2. \( y\sin x = \int 2\sin x\cos x\,dx = \int \sin 2x\,dx = -\frac{\cos 2x}{2} + \frac{c}{2} \).
3. Multiply by 2: \( 2y\sin x + \cos 2x = c \).

Teacher's Note:
1. Write 2 sin x cos x as sin 2x before integrating to match the option forms.
2. ∫cot x dx = log sin x, so the I.F. is sin x, not cos x.

Answer: (b) \( y^3 - x = Ay \)

Explanation:
1. Invert: \( \frac{dx}{dy} = \frac{x + 2y^3}{y} \), i.e. \( \frac{dx}{dy} - \frac{x}{y} = 2y^2 \), linear in x.
2. \( \text{I.F.} = e^{-\int \frac{dy}{y}} = \frac{1}{y} \).
3. \( \frac{x}{y} = \int 2y^2 \cdot \frac{1}{y}\,dy = y^2 - A \), so \( x = y^3 - Ay \).
4. Hence \( y^3 - x = Ay \).

Teacher's Note:
1. The term y³ makes the equation non-linear in y, but it is linear in x, so solve for x.
2. Since A is arbitrary, its sign can be chosen to match the option form.

Answer: (d) None of these

Explanation:
1. Write \( y' - y\tan x = -2\sin x \), so P = −tan x and \( \text{I.F.} = e^{-\int \tan x\,dx} = \cos x \).
2. \( y\cos x = \int -2\sin x\cos x\,dx = \int -\sin 2x\,dx = \frac{\cos 2x}{2} + C \).
3. So \( y = \frac{\cos 2x}{2\cos x} + C\sec x \).
4. This does not match (a), (b) or (c), so the answer is None of these.

Teacher's Note:
1. Move y tan x to the left before reading P; here P = −tan x, so the I.F. is cos x.
2. You can test an option quickly: y = tan x + C cos x does not satisfy y′ = y tan x − 2 sin x.

Answer: (a) \( v = ce^{-\frac{k}{m}t} - \frac{mg}{k} \)

Explanation:
1. P = k/m, so \( \text{I.F.} = e^{\frac{k}{m}t} \).
2. \( ve^{\frac{k}{m}t} = \int -ge^{\frac{k}{m}t}\,dt = -\frac{mg}{k}e^{\frac{k}{m}t} + c \).
3. Divide by the I.F.: \( v = ce^{-\frac{k}{m}t} - \frac{mg}{k} \).

Teacher's Note:
1. This is the velocity of a body falling against air resistance; mg/k is the terminal speed.
2. When dividing by the I.F., the constant c also gets divided, giving ce^(−kt/m).

Answer: (c) sec x

Explanation:
1. Divide by cos x: \( \frac{dy}{dx} + y\tan x = \sec x \).
2. P = tan x, so \( \text{I.F.} = e^{\int \tan x\,dx} = e^{\log\sec x} = \sec x \).

Teacher's Note:
1. Always make the coefficient of dy/dx equal to 1 before reading P.
2. ∫tan x dx = log sec x, so the I.F. is sec x, not cos x.

Answer: (d) \( (a + m)y = e^{mx} + Ce^{-ax}(a + m) \)

Explanation:
1. P = a, so \( \text{I.F.} = e^{ax} \).
2. \( ye^{ax} = \int e^{(a + m)x}\,dx = \frac{e^{(a + m)x}}{a + m} + C \).
3. Multiply by \( (a + m)e^{-ax} \): \( (a + m)y = e^{mx} + C(a + m)e^{-ax} \).

Teacher's Note:
1. Combine eᵃˣ · eᵐˣ into e^((a + m)x) before integrating.
2. This form assumes a + m ≠ 0; if a = −m the integral becomes x instead.

Answer: (c) \( y = ce^{-\int P\,dx} \)

Explanation:
1. Separate: \( \frac{dy}{y} = -P(x)\,dx \).
2. Integrate: \( \log y = -\int P\,dx + \log c \).
3. Hence \( y = ce^{-\int P\,dx} \).

Teacher's Note:
1. A homogeneous linear equation (Q = 0) can be solved by separating variables directly.
2. Since P is a function of x, the integral must be with respect to x, which rules out (b) and (d).

Answer: (d) \( y = xe^{-x} \)

Explanation:
1. P = 1, so \( \text{I.F.} = e^x \).
2. \( ye^x = \int e^{-x}e^x\,dx = \int 1\,dx = x + C \).
3. y(0) = 0 gives C = 0, so \( y = xe^{-x} \).

Teacher's Note:
1. When Q is e^(−x) and the I.F. is eˣ, the product is 1, which gives the extra factor x.
2. Check: y = xe^(−x) gives y(0) = 0 and y′ + y = e^(−x).

Answer: (a) \( y = \tan x - 1 + ce^{-\tan x} \)

Explanation:
1. P = \( \sec^2 x \), so \( \text{I.F.} = e^{\tan x} \).
2. \( ye^{\tan x} = \int \tan x\sec^2 x\,e^{\tan x}\,dx \); put t = tan x: \( \int te^t\,dt = (t - 1)e^t \).
3. So \( ye^{\tan x} = (\tan x - 1)e^{\tan x} + c \).
4. Hence \( y = \tan x - 1 + ce^{-\tan x} \).

Teacher's Note:
1. The substitution t = tan x turns the integral into the standard ∫teᵗdt = (t − 1)eᵗ.
2. Option (c) forgets that the right side also carries the factor e^(tan x).

Answer: (b) \( (1 - x^2)^{-1} \)

Explanation:
1. \( P = \frac{2x}{1 - x^2} \), so \( \int P\,dx = -\log(1 - x^2) \).
2. \( \text{I.F.} = e^{-\log(1 - x^2)} = (1 - x^2)^{-1} \).

Teacher's Note:
1. The numerator 2x is minus the derivative of 1 − x², so the log comes with a minus sign.
2. Only P decides the integrating factor; the right side Q plays no role.

Answer: (c) \( \sqrt{(1 - x^2)} \)

Explanation:
1. Divide by \( (1 - x^2) \): \( \frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2} \).
2. \( \int P\,dx = \int \frac{-x}{1 - x^2}\,dx = \frac{1}{2}\log(1 - x^2) \).
3. \( \text{I.F.} = e^{\frac{1}{2}\log(1 - x^2)} = \sqrt{1 - x^2} \).

Teacher's Note:
1. Option (d) is ∫P dx itself; remember the I.F. is e raised to that integral.
2. Make the coefficient of dy/dx equal to 1 before reading P.

Answer: (a) \( -\frac{1}{2} \)

Explanation:
1. Write \( \frac{dy}{dt} - \frac{t}{1 + t}y = \frac{1}{1 + t} \); \( \int -\frac{t}{1 + t}\,dt = -t + \log(1 + t) \).
2. \( \text{I.F.} = (1 + t)e^{-t} \), so \( y(1 + t)e^{-t} = \int e^{-t}\,dt = -e^{-t} + C \).
3. y(0) = −1 gives −1 = −1 + C, so C = 0 and \( y = -\frac{1}{1 + t} \).
4. Hence \( y(1) = -\frac{1}{2} \).

Teacher's Note:
1. Write t/(1 + t) as 1 − 1/(1 + t) to integrate it quickly.
2. Check: y = −1/(1 + t) gives (1 + t)y′ − ty = 1/(1 + t) + t/(1 + t) = 1.

Answer: (a) x

Explanation:
1. \( P = \frac{1}{x} \), so \( \int P\,dx = \log x \).
2. \( \text{I.F.} = e^{\log x} = x \).

Teacher's Note:
1. e^(log x) = x is used again and again in finding integrating factors.
2. Option (b) is ∫P dx itself, not the I.F.

Answer: (b) y sec x = tan x + C

Explanation:
1. Divide by cos x: \( \frac{dy}{dx} + y\tan x = \sec x \); \( \text{I.F.} = \sec x \).
2. \( y\sec x = \int \sec x \cdot \sec x\,dx = \int \sec^2 x\,dx = \tan x + C \).
3. Hence \( y\sec x = \tan x + C \).

Teacher's Note:
1. This continues the earlier question on the same equation, where the I.F. was found to be sec x.
2. Remember to multiply Q = sec x by the I.F. sec x, giving sec² x.

Answer: (d) \( \frac{(2x^2 - 1)}{x(1 - x^2)} \)

Explanation:
1. Divide by \( x(1 - x^2)\,dx \): \( \frac{dy}{dx} + \frac{2x^2 - 1}{x(1 - x^2)}y = \frac{ax^3}{x(1 - x^2)} \).
2. Comparing with \( \frac{dy}{dx} + Py = Q \), \( P = \frac{2x^2 - 1}{x(1 - x^2)} \).

Teacher's Note:
1. Separate the y-terms from the terms without y before dividing; only the y-terms give P.
2. Option (a) wrongly mixes the term ax³, which belongs to Q, into P.

Answer: (d) All of (a), (b) and (c)

Explanation:
1. Divide by x: \( \frac{dy}{dx} + \frac{2}{x}y = x\log x \); \( \text{I.F.} = x^2 \).
2. \( yx^2 = \int x^3\log x\,dx = \frac{x^4}{4}\log x - \frac{x^4}{16} + c \), which is (b).
3. Multiply by 16: \( 16yx^2 = 4x^4\log x - x^4 + c \), which is (c).
4. Also \( 4x^4\log x - x^4 = x^4(\log x^4 - 1) = x^4\log(x^4/e) \), which is (a). So all of (a), (b) and (c) are correct.

Teacher's Note:
1. The same general solution can look very different after renaming the constant or using log rules.
2. To compare such options, bring each one to the form yx² = (expression) + c.

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

Explanation:
1. Divide by x cos x: \( \frac{dy}{dx} + \left(\tan x + \frac{1}{x}\right)y = \frac{\sec x}{x} \).
2. \( \text{I.F.} = e^{\log\sec x + \log x} = x\sec x \).
3. \( xy\sec x = \int \sec^2 x\,dx = \tan x + c \), which is (b).
4. Multiply by cos x: \( xy = \sin x + c\cos x \), which is (a). So both (a) and (b) are correct.

Teacher's Note:
1. The left side is also d/dx(xy sec x) multiplied by cos²x, which is a quick check on the I.F.
2. Option (c) changes the sign of sin x, so it is not equivalent to the others.

Answer: (d) All of (a), (b) and (c)

Explanation:
1. P = −3, \( \text{I.F.} = e^{-3x} \): \( ye^{-3x} = \int e^{-3x}\sin 2x\,dx = -\frac{e^{-3x}}{13}(3\sin 2x + 2\cos 2x) + c \), which is (a).
2. Divide by \( e^{-3x} \): \( y = -\frac{1}{13}(2\cos 2x + 3\sin 2x) + ce^{3x} \), which is (b).
3. \( 2\cos 2x + 3\sin 2x = \sqrt{13}\cos(2x - \alpha) \) with \( \tan\alpha = \frac{3}{2} \), so (c) is the same solution.
4. So all of (a), (b) and (c) are correct.

Teacher's Note:
1. Use ∫e^(ax) sin bx dx = e^(ax)(a sin bx − b cos bx)/(a² + b²) to save two rounds of integration by parts.
2. A sin θ + B cos θ can always be written as one cosine or sine of amplitude √(A² + B²).

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

Explanation:
1. Rewrite: \( \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2} \), linear in x with \( \text{I.F.} = e^{\tan^{-1}y} \).
2. Put t = \( \tan^{-1}y \): \( xe^{t} = \int te^t\,dt = (t - 1)e^t + c \).
3. So \( xe^{\tan^{-1}y} = (\tan^{-1}y - 1)e^{\tan^{-1}y} + c \), which is (b).
4. Divide by \( e^{\tan^{-1}y} \): \( x = \tan^{-1}y - 1 + ce^{-\tan^{-1}y} \), which is (c). So both (b) and (c) are correct.

Teacher's Note:
1. After t = tan⁻¹y, the factor dy/(1 + y²) becomes dt, which makes the integral simple.
2. Option (a) has (1 − t) in place of (t − 1), a sign slip that is easy to make in ∫teᵗdt.

Answer: (c) \( y = x\cot^{-1}(\log_e ex) \)

Explanation:
1. \( \frac{dy}{dx} = \frac{y}{x} - \sin^2\frac{y}{x} \). Put y = vx: \( x\frac{dv}{dx} = -\sin^2 v \).
2. Separate: \( \text{cosec}^2 v\,dv = -\frac{dx}{x} \), so \( -\cot v = -\log x - c \), i.e. \( \cot\frac{y}{x} = \log x + c \).
3. At \( \left(1, \frac{\pi}{4}\right) \): \( \cot\frac{\pi}{4} = 0 + c \), so c = 1 = log e.
4. So \( \cot\frac{y}{x} = \log_e(ex) \), i.e. \( y = x\cot^{-1}(\log_e ex) \).

Teacher's Note:
1. Write the constant 1 as log e so that the two logs combine into log(ex).
2. Options without the factor x in front cannot be right, since the solution of a homogeneous equation gives y/x, not y.

Answer: (c) Parabola

Explanation:
1. Slope of the line joining (x, y) to the origin is \( \frac{y}{x} \), so \( \frac{dy}{dx} = \frac{2y}{x} \).
2. Separate: \( \frac{dy}{y} = \frac{2\,dx}{x} \), so \( \log y = 2\log x + \log c \).
3. Hence \( y = cx^2 \), which is a parabola.

Teacher's Note:
1. Translate the words into an equation first: 'slope of tangent' means dy/dx.
2. If the factor were n instead of 2, the curves would be y = cxⁿ; n = 2 gives a parabola.

Answer: (d) \( N = N_0e^{-kt} \)

Explanation:
1. Rate of decrease proportional to N: \( \frac{dN}{dt} = -kN \).
2. Separate: \( \frac{dN}{N} = -k\,dt \), so \( \log N = -kt + \log N_0 \).
3. Hence \( N = N_0e^{-kt} \), where \( N_0 \) is the value at t = 0.

Teacher's Note:
1. 'Decrement' or 'decay' means the rate is negative, so a minus sign must appear with k.
2. The constant N₀ multiplies the exponential; it is never added to it.

Answer: (c) a = 0, b ≠ 0

Explanation:
1. Separate: \( (by + k)\,dy = (ax + h)\,dx \).
2. Integrate: \( \frac{by^2}{2} + ky = \frac{ax^2}{2} + hx + C \).
3. For a parabola, exactly one of the squared terms must vanish. With a = 0 and b ≠ 0: \( \frac{by^2}{2} + ky = hx + C \), a parabola.
4. With a = b = 0 we get a straight line, and a = 1, b = 2 or a = 2, b = 1 give hyperbolas. So a = 0, b ≠ 0.

Teacher's Note:
1. A second degree curve with only one squared variable (and no xy term) is a parabola.
2. When x² and y² have opposite signs in the same equation, the curve is a hyperbola.

Answer: (b) \( 2e^2 \)

Explanation:
1. \( \phi'(x) = \phi(x) \), so \( \frac{d\phi}{\phi} = dx \) and \( \phi(x) = Ce^x \).
2. φ(1) = 2 gives \( Ce = 2 \), so \( C = \frac{2}{e} \) and \( \phi(x) = 2e^{x - 1} \).
3. Hence \( \phi(3) = 2e^2 \).

Teacher's Note:
1. A function equal to its own derivative is always a multiple of eˣ.
2. A quick shortcut: φ(3)/φ(1) = e^(3 − 1) = e², so φ(3) = 2e².

Answer: (c) A rectangular hyperbola

Explanation:
1. \( \frac{dy}{dx} = \frac{x}{y} \), so \( y\,dy = x\,dx \).
2. Integrate: \( \frac{y^2}{2} = \frac{x^2}{2} + C \), i.e. \( x^2 - y^2 = c \).
3. This is a rectangular hyperbola (equal semi-axes).

Teacher's Note:
1. Abscissa means x and ordinate means y, so the ratio of abscissa to ordinate is x/y.
2. If the slope were −x/y instead, the curves would be circles x² + y² = c.

Answer: (c) \( x = \frac{1}{2} \)

Explanation:
1. Separate: \( \sec^2(\pi x)\,dx = dt \), so \( \frac{\tan \pi x}{\pi} = t + C \).
2. At t = 0, x = 0, so C = 0 and \( \tan(\pi x) = \pi t \), i.e. \( x = \frac{1}{\pi}\tan^{-1}(\pi t) \).
3. As t → ∞, \( \tan^{-1}(\pi t) \to \frac{\pi}{2} \), so x → \( \frac{1}{2} \), but x stays less than \( \frac{1}{2} \) for every finite t.
4. Hence the particle approaches but never reaches \( x = \frac{1}{2} \).

Teacher's Note:
1. The velocity cos²πx becomes zero at x = 1/2, so the particle slows down more and more as it gets near that point.
2. Since it can never pass x = 1/2, points beyond it, such as 3/4 or 1, are also never reached.

Answer: (a) \( 2(x^2 - y^2) = 3x \)

Explanation:
1. \( \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \) is homogeneous. Put y = vx: \( \frac{2v\,dv}{1 - v^2} = \frac{dx}{x} \).
2. Integrate: \( -\log(1 - v^2) = \log x - \log c \), so \( x(1 - v^2) = c \), i.e. \( x^2 - y^2 = cx \).
3. At (2, 1): 4 − 1 = 2c, so \( c = \frac{3}{2} \).
4. Hence \( 2(x^2 - y^2) = 3x \).

Teacher's Note:
1. This is the same differential equation as (x² + y²)dx = 2xy dy, now with a point to fix c.
2. Substitute the point (2, 1) into each option as a quick check: only 2(4 − 1) = 3(2) works.

Answer: (c) 1

Explanation:
1. The slope at any point of an integral curve is given directly by the differential equation.
2. At (1, 0): \( y' = \frac{1^2 + 0^2}{1^2 - 0^2} = \frac{1}{1} = 1 \).
3. Hence the slope is 1.

Teacher's Note:
1. You do not need to solve the equation to find a slope; just substitute the point into y′.
2. The condition y(1) = 2 is extra information here and is not needed for the slope.

Answer: (c) \( \frac{e^6 + 9}{2} \)

Explanation:
1. Separate: \( e^{2y}\,dy = dx \), so \( \frac{e^{2y}}{2} = x + C \).
2. y = 0 when x = 5: \( \frac{1}{2} = 5 + C \), so \( C = -\frac{9}{2} \).
3. For y = 3: \( \frac{e^6}{2} = x - \frac{9}{2} \).
4. Hence \( x = \frac{e^6 + 9}{2} \).

Teacher's Note:
1. Move e^(−2y) to the left as e^(2y) to separate the variables.
2. Find the constant from the given condition first, and only then substitute y = 3.

Answer: (b) \( \sin\frac{(y - 1)}{x} = a \)

Explanation:
1. \( \sin\frac{dy}{dx} = a \) gives \( \frac{dy}{dx} = \sin^{-1}a \), a constant.
2. Integrate: \( y = x\sin^{-1}a + c \); y(0) = 1 gives c = 1.
3. So \( \frac{y - 1}{x} = \sin^{-1}a \).
4. Hence \( \sin\frac{(y - 1)}{x} = a \).

Teacher's Note:
1. Treat sin⁻¹a as a constant slope; the solution is just a straight line through (0, 1).
2. Option (a) applies sin⁻¹ twice; take care to undo the inverse sine correctly.

Answer: (b) \( \log\tan\left(\frac{y}{4}\right) = c - 2\sin\left(\frac{x}{2}\right) \)

Explanation:
1. \( \frac{dy}{dx} = \sin\frac{x - y}{2} - \sin\frac{x + y}{2} = 2\cos\frac{x}{2}\sin\left(-\frac{y}{2}\right) = -2\cos\frac{x}{2}\sin\frac{y}{2} \).
2. Separate: \( \text{cosec}\frac{y}{2}\,dy = -2\cos\frac{x}{2}\,dx \).
3. Integrate: \( 2\log\tan\frac{y}{4} = -4\sin\frac{x}{2} + 2c \).
4. Hence \( \log\tan\left(\frac{y}{4}\right) = c - 2\sin\left(\frac{x}{2}\right) \).

Teacher's Note:
1. Use sin A − sin B = 2 cos((A + B)/2) sin((A − B)/2) to turn the difference into a product.
2. ∫cosec u du = log tan(u/2); with u = y/2 this gives 2 log tan(y/4), not log tan(y/2).

Answer: (c) y = 2x − 4

Explanation:
1. Test y = 2x − 4: \( \frac{dy}{dx} = 2 \).
2. Substitute: \( 2^2 - x(2) + (2x - 4) = 4 - 2x + 2x - 4 = 0 \). ✓
3. y = 2 gives 0 − 0 + 2 ≠ 0; y = 2x gives 4 − 2x + 2x = 4 ≠ 0; \( y = 2x^2 - 4 \) gives \( 16x^2 - 4x^2 + 2x^2 - 4 \ne 0 \).
4. Hence y = 2x − 4 is a solution.

Teacher's Note:
1. For such questions, substituting each option is faster than solving the equation.
2. This is Clairaut's form y = xp − p², whose general solution is y = cx − c²; c = 2 gives y = 2x − 4.

Answer: (c) \( \{3(x - 1)\}^{1/3} \)

Explanation:
1. Differentiate both sides: \( \phi'(x) = \{\phi(x)\}^{-2} \).
2. Separate: \( \phi^2\,d\phi = dx \), so \( \frac{\phi^3}{3} = x + C \).
3. φ(1) = 0 gives 0 = 1 + C, so C = −1 and \( \phi^3 = 3(x - 1) \).
4. Hence \( \phi(x) = \{3(x - 1)\}^{1/3} \).

Teacher's Note:
1. An equation with a function inside its own integral becomes a differential equation on differentiating.
2. Check: with φ = {3(x − 1)}^(1/3), φ′ = {3(x − 1)}^(−2/3) = φ^(−2), and φ(1) = 0.