CBSE Class 12 Mathematics Differential Equations MCQs Set 14

Answer: (b) 2

Explanation:
1. Isolate the fractional term: \( \left(\frac{d^3y}{dx^3}\right)^{2/3} = 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4 \).
2. Cube both sides: \( \left(\frac{d^3y}{dx^3}\right)^2 = \left(3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4\right)^3 \).
3. The highest derivative \( \frac{d^3y}{dx^3} \) appears with power 2.
4. Hence the degree is 2.

Teacher's Note:
1. After cubing, the second derivative gets power 3, but it is not the highest order derivative.
2. Always identify the highest order derivative first and then read its power.

Answer: (b) \( \sin^{-1}x + \sin^{-1}y = C \)

Explanation:
1. Separate the variables: \( \frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}} \).
2. Integrate: \( \sin^{-1}y = -\sin^{-1}x + C \).
3. Hence \( \sin^{-1}x + \sin^{-1}y = C \).

Teacher's Note:
1. Writing the root of a quotient as a quotient of roots makes the separation obvious.
2. Options (a) and (c) are identities equal to π/2, so they contain no y and cannot be solutions.

Answer: (a) \( y = Ae^{\frac{2}{3}(2a - x)\sqrt{x + a}} \)

Explanation:
1. Separate: \( \frac{dy}{y} = -\frac{x}{\sqrt{a + x}}\,dx \).
2. Put t = a + x: \( \int \frac{t - a}{\sqrt{t}}\,dt = \frac{2}{3}t^{3/2} - 2at^{1/2} = \frac{2}{3}\sqrt{a + x}\,(x - 2a) \).
3. So \( \log y = -\frac{2}{3}(x - 2a)\sqrt{x + a} + \log A = \frac{2}{3}(2a - x)\sqrt{x + a} + \log A \).
4. Hence \( y = Ae^{\frac{2}{3}(2a - x)\sqrt{x + a}} \).

Teacher's Note:
1. For integrals like x/√(a + x), substitute the whole root expression as t and split the fraction.
2. Watch the sign carefully: the minus sign turns (x − 2a) into (2a − x).

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

Explanation:
1. Write it as \( \frac{dy}{dx} = y(1 - 2x) \), so \( \frac{dy}{y} = (1 - 2x)\,dx \).
2. Integrate: \( \log y = x - x^2 + \log c \).
3. Hence \( y = ce^{x - x^2} \).

Teacher's Note:
1. Taking y common on one side often reveals that the variables separate.
2. Check the sign of the exponent: integrating −2x gives −x², not +x².

Answer: (b) \( e^x + e^{-y} = c \)

Explanation:
1. Remove the log: \( \frac{dy}{dx} = e^{x + y} = e^xe^y \).
2. Separate: \( e^{-y}\,dy = e^x\,dx \).
3. Integrate: \( -e^{-y} = e^x - c \).
4. Hence \( e^x + e^{-y} = c \).

Teacher's Note:
1. Whenever the exponent is a sum, split e^(x + y) into e^x · e^y to separate the variables.
2. The integral of e^(−y) is −e^(−y); missing this minus sign leads to option (a).

Answer: (d) None of these

Explanation:
1. Separate: \( \sec x\,\log(\sec x + \tan x)\,dx = \sec y\,\log(\sec y + \tan y)\,dy \).
2. Put \( t = \log(\sec x + \tan x) \), then \( dt = \sec x\,dx \), so the left integral is \( \frac{t^2}{2} \).
3. So \( [\log(\sec x + \tan x)]^2 - [\log(\sec y + \tan y)]^2 = c \).
4. This is not any of (a), (b), (c), so the answer is None of these.

Teacher's Note:
1. The derivative of log(sec x + tan x) is sec x, so the integrand is of the form f · f′.
2. Dividing both sides by cos x cos y is the key first step to separate the variables.

Answer: (c) \( y^2e^{x^2} = cx^2 \)

Explanation:
1. Divide by xy: \( \frac{dx}{x} - \frac{dy}{y} = x\,dx \).
2. Integrate: \( \log x - \log y = \frac{x^2}{2} + C \), so \( \frac{x}{y} = ke^{x^2/2} \).
3. Square both sides: \( \frac{x^2}{y^2} = k^2e^{x^2} \), i.e. \( y^2e^{x^2} = cx^2 \) with \( c = \frac{1}{k^2} \).

Teacher's Note:
1. Dividing by xy is a standard trick when the terms are y dx and x dy.
2. When the options look squared, square your answer and rename the constant.

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

Explanation:
1. Factorise: \( 1 + x + y + xy = (1 + x)(1 + y) \), so \( \frac{dy}{1 + y} = (1 + x)\,dx \).
2. Integrate: \( \log(1 + y) = \frac{(1 + x)^2}{2} + C \).
3. y(−1) = 0 gives \( \log 1 = 0 + C \), so C = 0.
4. Hence \( y = e^{(1 + x)^2/2} - 1 \).

Teacher's Note:
1. Grouping 1 + x + y + xy as (1 + x)(1 + y) is the key factorisation.
2. Check: at x = −1 the answer gives e⁰ − 1 = 0, which matches the condition.

Answer: (b) y = c(x + a)(1 − ay)

Explanation:
1. Rearrange: \( y - ay^2 = (x + a)\frac{dy}{dx} \), so \( \frac{dy}{y(1 - ay)} = \frac{dx}{x + a} \).
2. Partial fractions: \( \frac{1}{y(1 - ay)} = \frac{1}{y} + \frac{a}{1 - ay} \).
3. Integrate: \( \log y - \log(1 - ay) = \log(x + a) + \log c \).
4. So \( \frac{y}{1 - ay} = c(x + a) \), i.e. \( y = c(x + a)(1 - ay) \).

Teacher's Note:
1. Collect all dy/dx terms on one side before trying to separate.
2. Integrating a/(1 − ay) gives −log(1 − ay); the minus sign comes from the chain rule.

Answer: (a) \( cx = e^{x + y^2/2} \)

Explanation:
1. Write \( \frac{dy}{dx} = \frac{1}{xy} - \frac{1}{y} = \frac{1 - x}{xy} \).
2. Separate: \( y\,dy = \left(\frac{1}{x} - 1\right)dx \).
3. Integrate: \( \frac{y^2}{2} = \log x - x + C \), so \( \log x = x + \frac{y^2}{2} - C \).
4. Hence \( x = ke^{x + y^2/2} \), i.e. \( cx = e^{x + y^2/2} \).

Teacher's Note:
1. Taking the LCM xy on the right makes the separation clear.
2. When log x appears with other terms, exponentiate to get the answer in the option form.

Answer: (d) None of these

Explanation:
1. Separate: \( \frac{dy}{1 - y^2} = \frac{dx}{x\sqrt{1 - x^2}} \).
2. Left side: \( \frac{1}{2}\log\frac{1 + y}{1 - y} \).
3. Right side: \( \int \frac{dx}{x\sqrt{1 - x^2}} = -\log\left|\frac{1 + \sqrt{1 - x^2}}{x}\right| \), which has no sec function.
4. So none of the given forms is correct.

Teacher's Note:
1. The sec⁻¹ x result is for 1/(x√(x² − 1)), not for 1/(x√(1 − x²)); notice the order inside the root.
2. The integral of 1/(x√(1 − x²)) is found by putting x = sin θ, which gives −log|cosec θ + cot θ|.

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

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

Teacher's Note:
1. Splitting the numerator 1 + x + x² into (1 + x²) + x avoids long partial fractions.
2. The integral of y/(1 + y²) is ½ log(1 + y²); forgetting the ½ leads to option (d).

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

Explanation:
1. Separate: \( \frac{1 + y}{1 + y^2}\,dy + \frac{1 + x}{1 + x^2}\,dx = 0 \).
2. Split: \( \frac{1 + x}{1 + x^2} = \frac{1}{1 + x^2} + \frac{x}{1 + x^2} \), which integrates to \( \tan^{-1}x + \frac{1}{2}\log(1 + x^2) \).
3. Similarly for y, so \( \tan^{-1}x + \frac{1}{2}\log(1 + x^2) + \tan^{-1}y + \frac{1}{2}\log(1 + y^2) = c \).

Teacher's Note:
1. Divide by (1 + x²)(1 + y²) to put each variable with its own differential.
2. Splitting (1 + x)/(1 + x²) into two standard fractions is a pattern worth remembering.

Answer: (a) \( y^2(\log y) - e^x\sin^2 x + c = 0 \)

Explanation:
1. Separate: \( y(2\log y + 1)\,dy = e^x(\sin^2 x + \sin 2x)\,dx \).
2. Since \( \frac{d}{dy}(y^2\log y) = 2y\log y + y \), the left side integrates to \( y^2\log y \).
3. Since \( \frac{d}{dx}\sin^2 x = \sin 2x \), the right side is \( \int e^x(f + f')\,dx = e^x\sin^2 x \).
4. Hence \( y^2\log y - e^x\sin^2 x + c = 0 \).

Teacher's Note:
1. The pattern ∫eˣ(f(x) + f′(x))dx = eˣf(x) saves a full integration by parts.
2. Recognising 2y log y + y as the derivative of y² log y is quicker than integrating by parts.

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

Explanation:
1. Factorise: \( x(1 - y^2)\,dx = y(1 - x^2)\,dy \).
2. Separate: \( \frac{x\,dx}{1 - x^2} = \frac{y\,dy}{1 - y^2} \).
3. Integrate: \( -\frac{1}{2}\log(1 - x^2) = -\frac{1}{2}\log(1 - y^2) + \log k \).
4. So \( \frac{1 - y^2}{1 - x^2} = c^2 \), i.e. \( (1 - y^2) = c^2(1 - x^2) \).

Teacher's Note:
1. Take x common on the left and y common on the right before separating.
2. Writing the constant as log k (or c²) makes the final form neat.

Answer: (c) \( 4\log\left|\frac{y - 2}{y + 1}\right| = 3\log\left|\frac{x - 1}{x + 3}\right| + c \)

Explanation:
1. Factorise: \( \frac{dy}{(y - 2)(y + 1)} = \frac{dx}{(x + 3)(x - 1)} \).
2. Partial fractions: \( \frac{1}{(y - 2)(y + 1)} = \frac{1}{3}\left(\frac{1}{y - 2} - \frac{1}{y + 1}\right) \) and \( \frac{1}{(x - 1)(x + 3)} = \frac{1}{4}\left(\frac{1}{x - 1} - \frac{1}{x + 3}\right) \).
3. Integrate: \( \frac{1}{3}\log\left|\frac{y - 2}{y + 1}\right| = \frac{1}{4}\log\left|\frac{x - 1}{x + 3}\right| + c_1 \).
4. Multiply by 12: \( 4\log\left|\frac{y - 2}{y + 1}\right| = 3\log\left|\frac{x - 1}{x + 3}\right| + c \).

Teacher's Note:
1. Use ∫dt/((t − a)(t − b)) = (1/(a − b)) log|(t − a)/(t − b)| to save time.
2. Inverting a log fraction changes its sign, so flip both sides together or neither.

Answer: (a) \( y = \log|\tan^2 x + 2\tan x + 5| + c \)

Explanation:
1. Separate: \( dy = \frac{2(1 + \tan x)\sec^2 x}{\tan^2 x + 2\tan x + 5}\,dx \).
2. Put \( t = \tan^2 x + 2\tan x + 5 \); then \( dt = (2\tan x + 2)\sec^2 x\,dx \), which is exactly the numerator.
3. So \( y = \int \frac{dt}{t} = \log|\tan^2 x + 2\tan x + 5| + c \).

Teacher's Note:
1. Before any heavy work, check whether the numerator is the derivative of the denominator.
2. ∫f′(x)/f(x) dx = log|f(x)| + c is one of the most used results in this chapter.

Answer: (a) \( 3y + 2\sqrt{a + x}\,(x - 2a) = 3c \)

Explanation:
1. Separate: \( dy = -\frac{x}{\sqrt{a + x}}\,dx \).
2. With t = a + x: \( \int \frac{x}{\sqrt{a + x}}\,dx = \frac{2}{3}\sqrt{a + x}\,(x - 2a) \).
3. So \( y = -\frac{2}{3}\sqrt{a + x}\,(x - 2a) + c \).
4. Multiply by 3: \( 3y + 2\sqrt{a + x}\,(x - 2a) = 3c \).

Teacher's Note:
1. Writing x = (a + x) − a splits the integrand into √(a + x) − a/√(a + x).
2. A quick check: differentiating the answer should give back dy/dx = −x/√(a + x).

Answer: (a) \( e^{5x} + e^{5y} = c \)

Explanation:
1. Write \( e^{2x}e^{-3y}\,dx + e^{2y}e^{-3x}\,dy = 0 \).
2. Multiply by \( e^{3x}e^{3y} \): \( e^{5x}\,dx + e^{5y}\,dy = 0 \).
3. Integrate: \( \frac{e^{5x}}{5} + \frac{e^{5y}}{5} = c_1 \).
4. Hence \( e^{5x} + e^{5y} = c \).

Teacher's Note:
1. Multiply by e^(3x + 3y) to bring each exponential next to its own variable.
2. Both terms keep a plus sign, so the answer is a sum, not a difference.

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

Explanation:
1. Divide by \( \sqrt{1 + x^2}\sqrt{1 + y^2} \): \( \frac{x\,dx}{\sqrt{1 + x^2}} + \frac{y\,dy}{\sqrt{1 + y^2}} = 0 \).
2. Since \( \int \frac{x\,dx}{\sqrt{1 + x^2}} = \sqrt{1 + x^2} \), integrate both terms.
3. Hence \( \sqrt{1 + x^2} + \sqrt{1 + y^2} = c \).

Teacher's Note:
1. ∫x/√(1 + x²) dx = √(1 + x²) follows from putting t = 1 + x².
2. Always divide by the product of the 'wrong' factors to separate the variables.

Answer: (d) \( c(y + 1)(e^x + 1) = e^y \)

Explanation:
1. Separate: \( \frac{y}{y + 1}\,dy = \frac{e^x}{e^x + 1}\,dx \).
2. Write \( \frac{y}{y + 1} = 1 - \frac{1}{y + 1} \) and integrate: \( y - \log(y + 1) = \log(e^x + 1) + \log c_1 \).
3. So \( y = \log[c_1(y + 1)(e^x + 1)] \), i.e. \( e^y = c_1(y + 1)(e^x + 1) \).
4. Hence \( c(y + 1)(e^x + 1) = e^y \).

Teacher's Note:
1. Write y/(y + 1) as 1 − 1/(y + 1) before integrating; it is a standard step for such fractions.
2. The right side eˣ/(eˣ + 1) is of the form f′/f, so it integrates straight to log(eˣ + 1).

Answer: (d) \( \log[x(1 - y)^2] = \frac{x^2}{2} - \frac{y^2}{2} - 2y + c \)

Explanation:
1. Separate: \( \frac{1 - x^2}{x}\,dx = \frac{y(1 + y)}{1 - y}\,dy \).
2. Divide: \( \frac{y + y^2}{1 - y} = -y - 2 + \frac{2}{1 - y} \).
3. Integrate: \( \log x - \frac{x^2}{2} = -\frac{y^2}{2} - 2y - 2\log(1 - y) + c \).
4. So \( \log x + 2\log(1 - y) = \frac{x^2}{2} - \frac{y^2}{2} - 2y + c \), i.e. \( \log[x(1 - y)^2] = \frac{x^2}{2} - \frac{y^2}{2} - 2y + c \).

Teacher's Note:
1. When the degree of the numerator is not less than the denominator, divide first and then integrate.
2. ∫2/(1 − y) dy = −2 log(1 − y); the minus sign is what moves the log term to the left side.

Answer: (c) \( \frac{1}{2^x} - \frac{1}{2^y} = C \)

Explanation:
1. Write \( \frac{dy}{dx} = \frac{2^y}{2^x} \), so \( 2^{-y}\,dy = 2^{-x}\,dx \).
2. Integrate: \( -\frac{2^{-y}}{\log 2} = -\frac{2^{-x}}{\log 2} + k \).
3. Multiply by log 2: \( 2^{-x} - 2^{-y} = C \), i.e. \( \frac{1}{2^x} - \frac{1}{2^y} = C \).

Teacher's Note:
1. ∫aᵗ dt = aᵗ/log a, and for a⁻ᵗ there is an extra minus sign.
2. The constant log 2 can be absorbed into the arbitrary constant C.

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

Explanation:
1. Separate: \( y\,dy = (x - 1)\,dx \).
2. Integrate: \( \frac{y^2}{2} = \frac{x^2}{2} - x + C \).
3. y(1) = 1 gives \( \frac{1}{2} = \frac{1}{2} - 1 + C \), so C = 1.
4. Hence \( y^2 = x^2 - 2x + 2 \).

Teacher's Note:
1. Find the constant immediately after integrating, before rearranging, to avoid algebra slips.
2. Check: at x = 1 the answer gives y² = 1, which matches y(1) = 1.

Answer: (b) A set of circles, centre on the x-axis

Explanation:
1. Separate: \( y\,dy = (a - x)\,dx \).
2. Integrate: \( \frac{y^2}{2} = ax - \frac{x^2}{2} + C \).
3. So \( x^2 + y^2 - 2ax = 2C \), i.e. \( (x - a)^2 + y^2 = a^2 + 2C \).
4. This is a family of circles with centre (a, 0) on the x-axis.

Teacher's Note:
1. Complete the square to read the centre and radius of the circle.
2. Equal coefficients of x² and y² with no xy term always indicate a circle.

Answer: (b) \( y = \frac{1}{5}\cos^5 x - \frac{1}{3}\cos^3 x + (x - 1)e^x + c \)

Explanation:
1. \( y = \int \sin^3 x\cos^2 x\,dx + \int xe^x\,dx \).
2. Put u = cos x: \( \int (1 - u^2)u^2(-du) = -\frac{u^3}{3} + \frac{u^5}{5} \), i.e. \( \frac{1}{5}\cos^5 x - \frac{1}{3}\cos^3 x \).
3. By parts: \( \int xe^x\,dx = xe^x - e^x = (x - 1)e^x \).
4. Hence \( y = \frac{1}{5}\cos^5 x - \frac{1}{3}\cos^3 x + (x - 1)e^x + c \).

Teacher's Note:
1. For an odd power of sin x, keep one sin x for du and convert the rest to cos x.
2. ∫xeˣdx = (x − 1)eˣ is worth remembering as a ready result.

Answer: (b) \( \frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + c \)

Explanation:
1. Remove the log: \( \frac{dy}{dx} = e^{ax}e^{by} \).
2. Separate: \( e^{-by}\,dy = e^{ax}\,dx \).
3. Integrate: \( \frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + c \).

Teacher's Note:
1. This is the general form of the earlier question log(dy/dx) = x + y, with a = b = 1.
2. Dividing by the coefficient of the variable in the exponent is the usual step while integrating eᵏᵗ.

Answer: (c) \( 2e^{2y + 2} = e^{4x} + e^4 \)

Explanation:
1. Remove the log: \( \frac{dy}{dx} = e^{4x - 2y - 2} \), so \( e^{2y}\,dy = e^{4x - 2}\,dx \).
2. Integrate: \( \frac{e^{2y}}{2} = \frac{e^{4x - 2}}{4} + C \); multiply by \( 4e^2 \): \( 2e^{2y + 2} = e^{4x} + C' \).
3. At x = 1, y = 1: \( 2e^4 = e^4 + C' \), so \( C' = e^4 \).
4. Hence \( 2e^{2y + 2} = e^{4x} + e^4 \).

Teacher's Note:
1. Multiply through by a suitable exponential to clear the constant term in the exponent before using the condition.
2. Substitute the given point into your final form as a check.

Answer: (c) log (xy) + x − y = c

Explanation:
1. Separate: \( \frac{y - 1}{y}\,dy = \frac{1 + x}{x}\,dx \).
2. So \( \left(1 - \frac{1}{y}\right)dy = \left(\frac{1}{x} + 1\right)dx \).
3. Integrate: \( y - \log y = \log x + x + c_1 \).
4. Hence \( \log x + \log y + x - y = c \), i.e. \( \log(xy) + x - y = c \).

Teacher's Note:
1. Split each fraction into simpler terms before integrating.
2. Combine log x + log y into log(xy) to match the option forms.

Answer: (c) \( (e^y + 1)\sin x = c \)

Explanation:
1. Divide by \( (e^y + 1)\sin x \): \( \frac{\cos x}{\sin x}\,dx + \frac{e^y}{e^y + 1}\,dy = 0 \).
2. Both terms are of the form f′/f: \( \log\sin x + \log(e^y + 1) = \log c \).
3. Hence \( (e^y + 1)\sin x = c \).

Teacher's Note:
1. Recognising f′/f on both sides turns this into a one-line integration.
2. Writing the constant as log c lets you combine all the logs at the end.

Answer: (a) \( \tan^{-1}\left(\frac{y}{x}\right) = \log x + c \)

Explanation:
1. The equation is homogeneous: \( \frac{dy}{dx} = 1 + \frac{y}{x} + \frac{y^2}{x^2} \). Put y = vx.
2. \( v + x\frac{dv}{dx} = 1 + v + v^2 \), so \( x\frac{dv}{dx} = 1 + v^2 \).
3. Separate: \( \frac{dv}{1 + v^2} = \frac{dx}{x} \), so \( \tan^{-1}v = \log x + c \).
4. Hence \( \tan^{-1}\left(\frac{y}{x}\right) = \log x + c \).

Teacher's Note:
1. Dividing the whole equation by x² shows clearly that the right side depends only on y/x.
2. After y = vx, the v on both sides cancels here, which makes separation easy.

Answer: (a) \( \tan^{-1}\left(\frac{x}{y}\right) + \log y + c = 0 \)

Explanation:
1. Write as \( \frac{dx}{dy} = -\frac{x^2 - xy + y^2}{y^2} \) and put x = uy, so \( \frac{dx}{dy} = u + y\frac{du}{dy} \).
2. \( u + y\frac{du}{dy} = -(u^2 - u + 1) \), so \( y\frac{du}{dy} = -(1 + u^2) \).
3. Separate: \( \frac{du}{1 + u^2} = -\frac{dy}{y} \), so \( \tan^{-1}u + \log y + c = 0 \).
4. Hence \( \tan^{-1}\left(\frac{x}{y}\right) + \log y + c = 0 \).

Teacher's Note:
1. When dx/dy is the simpler form, use x = uy instead of y = vx.
2. The options contain x/y, which is a hint that the substitution x = uy is intended.

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

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

Teacher's Note:
1. ∫2v/(1 − v²) dv = −log(1 − v²); keep track of this minus sign.
2. A quick check with the curve x = c(x² − y²): differentiating gives back the same slope (x² + y²)/(2xy).

Answer: (c) \( x\sin\left(\frac{y}{x}\right) = c \)

Explanation:
1. Divide by x: \( \frac{dy}{dx} = \frac{y}{x} - \tan\frac{y}{x} \). Put y = vx.
2. \( v + x\frac{dv}{dx} = v - \tan v \), so \( \cot v\,dv = -\frac{dx}{x} \).
3. Integrate: \( \log\sin v = -\log x + \log c \), so \( x\sin v = c \).
4. Hence \( x\sin\left(\frac{y}{x}\right) = c \).

Teacher's Note:
1. ∫cot v dv = log sin v is the key standard integral here.
2. Be careful to keep y/x inside the sine; option (a) has x/y, which is a trap.

Answer: (b) \( y + \sqrt{x^2 + y^2} = cx^2 \)

Explanation:
1. \( \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} = v + \sqrt{1 + v^2} \) with y = vx (x > 0).
2. So \( x\frac{dv}{dx} = \sqrt{1 + v^2} \), i.e. \( \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} \).
3. Integrate: \( \log(v + \sqrt{1 + v^2}) = \log x + \log c \), so \( v + \sqrt{1 + v^2} = cx \).
4. Multiply by x: \( y + \sqrt{x^2 + y^2} = cx^2 \).

Teacher's Note:
1. ∫dv/√(1 + v²) = log(v + √(1 + v²)) is the standard integral that makes this work.
2. Multiplying v + √(1 + v²) = cx by x raises the power of x, giving cx² and not cx.

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

Explanation:
1. \( \frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy} \). Put y = vx: \( v + x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v} \).
2. So \( x\frac{dv}{dx} = -\frac{2v(v + 2)}{1 + v} \), i.e. \( \frac{(1 + v)\,dv}{v(v + 2)} = -\frac{2\,dx}{x} \).
3. Partial fractions: \( \frac{1 + v}{v(v + 2)} = \frac{1}{2}\left(\frac{1}{v} + \frac{1}{v + 2}\right) \), so \( \frac{1}{2}\log[v(v + 2)] = -2\log x + \log k \).
4. So \( v(v + 2)x^4 = c^2 \), i.e. \( x^2(y^2 + 2xy) = c^2 \).

Teacher's Note:
1. After y = vx, move v to the right and simplify the numerator carefully; most errors happen here.
2. Notice that (1 + v) is half the derivative of v(v + 2), which gives the same result faster.

Answer: (a) \( \phi\left(\frac{y}{x}\right) = kx \)

Explanation:
1. Put y = vx: \( v + x\frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)} \).
2. So \( \frac{\phi'(v)}{\phi(v)}\,dv = \frac{dx}{x} \).
3. Integrate: \( \log\phi(v) = \log x + \log k \).
4. Hence \( \phi\left(\frac{y}{x}\right) = kx \).

Teacher's Note:
1. The v terms cancel at once, leaving φ′/φ, which integrates to log φ.
2. This result works for any function φ, so try φ(v) = v to check: y/x = kx gives a valid family.

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

Explanation:
1. \( \frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} \). Put y = vx: \( v + x\frac{dv}{dx} = \frac{1 - 3v^2}{v^3 - 3v} \).
2. So \( x\frac{dv}{dx} = \frac{1 - v^4}{v^3 - 3v} \), i.e. \( \frac{(v^3 - 3v)\,dv}{1 - v^4} = \frac{dx}{x} \).
3. Partial fractions: \( \frac{v^3 - 3v}{(1 - v^2)(1 + v^2)} = -\frac{v}{1 - v^2} - \frac{2v}{1 + v^2} \), which integrates to \( \frac{1}{2}\log(1 - v^2) - \log(1 + v^2) \).
4. So \( \frac{\sqrt{1 - v^2}}{1 + v^2} = cx \); squaring and putting v = y/x: \( \frac{x^2(x^2 - y^2)}{(x^2 + y^2)^2} = c^2x^2 \).
5. Hence \( x^2 - y^2 = c^2(x^2 + y^2)^2 \).

Teacher's Note:
1. Factorise 1 − v⁴ as (1 − v²)(1 + v²) before splitting into partial fractions.
2. Check: differentiating (x² − y²)/(x² + y²)² along the given slope gives zero, so the family fits the equation.

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

Explanation:
1. \( \frac{dy}{dx} = \frac{x^3 + y^3}{3xy^2} \). Put y = vx: \( v + x\frac{dv}{dx} = \frac{1 + v^3}{3v^2} \).
2. So \( x\frac{dv}{dx} = \frac{1 - 2v^3}{3v^2} \), i.e. \( \frac{3v^2\,dv}{1 - 2v^3} = \frac{dx}{x} \).
3. Integrate: \( -\frac{1}{2}\log(1 - 2v^3) = \log x + \log k \), so \( 1 - 2v^3 = \frac{c}{x^2} \).
4. Multiply by x³: \( x^3 - 2y^3 = cx \).

Teacher's Note:
1. The numerator 3v² is −½ times the derivative of 1 − 2v³, so use the f′/f idea.
2. Check the answer at a simple point: it gives back slope (x³ + y³)/(3xy²) on differentiation.