CBSE Class 12 Mathematics Differential Equations MCQs Set 12

Answer: (c) \( \frac{2x + 3}{2y - 1} = c \)

Explanation:
1. Separate: \( \frac{dx}{2x + 3} = \frac{dy}{2y - 1} \).
2. Integrate: \( \frac{1}{2}\log(2x + 3) = \frac{1}{2}\log(2y - 1) + \frac{1}{2}\log c \).
3. So log (2x + 3) = log [c(2y − 1)], giving \( \frac{2x + 3}{2y - 1} = c \).

Teacher's Note:
1. \( \int \frac{dx}{ax + b} = \frac{1}{a}\log(ax + b) \); here both sides carry the same factor 1/2, so it cancels.
2. Keep each linear factor with its own variable when separating: x with 2x + 3 and y with 2y − 1.

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

Explanation:
1. Factor: \( x^2(1 - y)\frac{dy}{dx} = -y^2(1 + x) \), so \( \frac{1 - y}{y^2}\,dy = -\frac{1 + x}{x^2}\,dx \).
2. Left: \( \int \left(\frac{1}{y^2} - \frac{1}{y}\right)dy = -\frac{1}{y} - \log y \).
3. Right: \( -\int \left(\frac{1}{x^2} + \frac{1}{x}\right)dx = \frac{1}{x} - \log x \).
4. So \( -\frac{1}{y} - \log y = \frac{1}{x} - \log x + c \), i.e. \( \log\left(\frac{x}{y}\right) = \frac{1}{x} + \frac{1}{y} + c \).

Teacher's Note:
1. Take out the common factors x² and y² first; the variables then separate easily.
2. Split each fraction into simple terms like \( \frac{1}{y^2} - \frac{1}{y} \) before integrating to avoid sign slips.

Answer: (c) \( y = \frac{ae^{bx}}{b} + \frac{c}{m}\sin mx + k \)

Explanation:
1. Integrate directly: y = \( \int (ae^{bx} + c\cos mx)\,dx \).
2. \( \int ae^{bx}\,dx = \frac{a}{b}e^{bx} \) and \( \int c\cos mx\,dx = \frac{c}{m}\sin mx \).
3. So y = \( \frac{ae^{bx}}{b} + \frac{c}{m}\sin mx + k \).

Teacher's Note:
1. For \( e^{bx} \) and \( \cos mx \), always divide by the coefficient of x after integrating.
2. Because c is already used inside the equation, the constant of integration is named k here.

Answer: (d) None of these

Explanation:
1. Integrate by parts with u = log x, dv = x dx: \( \int x\log x\,dx = \frac{x^2}{2}\log x - \int \frac{x^2}{2}\cdot\frac{1}{x}\,dx \).
2. So y = \( \frac{x^2}{2}\log x - \frac{x^2}{4} + c \).
3. None of (a), (b), (c) matches this, so the answer is None of these.

Teacher's Note:
1. In \( \int x\log x\,dx \) take log x as the first function, because it is easy to differentiate but hard to integrate.
2. Before choosing "None of these", differentiate each option quickly to make sure none of them works.

Answer: (c) tan x = c cot y

Explanation:
1. Divide by tan x tan y: \( \frac{\sec^2 x}{\tan x}\,dx + \frac{\sec^2 y}{\tan y}\,dy = 0 \).
2. Integrate: log tan x + log tan y = log c, so tan x tan y = c.
3. Hence tan x = c cot y.

Teacher's Note:
1. Each fraction has the form \( \frac{f'}{f} \), so each integrates to a logarithm.
2. Since 1/tan y = cot y, tan x tan y = c and tan x = c cot y say the same thing.

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

Explanation:
1. Rewrite: \( x^2\,dy + 2xy\,dx = 0 \).
2. The left side is the exact differential \( d(x^2y) \).
3. So \( x^2y = c \).

Teacher's Note:
1. Spotting exact differentials like d(x²y) = x² dy + 2xy dx gives the answer in one line.
2. Separating also works: \( \frac{dy}{y} = -\frac{2\,dx}{x} \) gives log y = −2 log x + log c, the same result.

Answer: (b) cx = sec y + tan y

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

Teacher's Note:
1. Learn \( \int \sec y\,dy = \log|\sec y + \tan y| \); it appears often in such questions.
2. Options (c) and (d) put the trigonometric functions on x, which would require sec on the x side of the equation.

Answer: (a) \( y + \sin^{-1}x = c \)

Explanation:
1. Rewrite: \( dy = -\frac{dx}{\sqrt{1 - x^2}} \).
2. Integrate: \( y = -\sin^{-1}x + c \).
3. So \( y + \sin^{-1}x = c \).

Teacher's Note:
1. \( \int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}x \) is a standard result you should know by heart.
2. A general solution must contain an arbitrary constant, which immediately rules out options equal to fixed numbers.

Answer: (a) sin y = c cos x

Explanation:
1. Divide by cos x sin y: \( \tan x\,dx + \cot y\,dy = 0 \).
2. Integrate: −log cos x + log sin y = log c.
3. So \( \frac{\sin y}{\cos x} = c \), i.e. sin y = c cos x.

Teacher's Note:
1. Divide by the product of the "wrong" factors (cos x from the dy term and sin y from the dx term) to separate.
2. Check: differentiating sin y = c cos x gives cos y y′ = −c sin x, and replacing c by sin y/cos x gives back the equation.

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

Explanation:
1. Separate: \( \frac{dy}{y} + \frac{dx}{(1 + x^2)\tan^{-1}x} = 0 \).
2. Put t = \( \tan^{-1}x \), dt = \( \frac{dx}{1 + x^2} \), so the second integral is log t = log \( \tan^{-1}x \).
3. So log y + log \( \tan^{-1}x \) = log c, i.e. \( y\tan^{-1}x = c \).

Teacher's Note:
1. The factor \( \frac{1}{1 + x^2} \) next to \( \tan^{-1}x \) is a signal to substitute t = \( \tan^{-1}x \).
2. Sum of logarithms becomes the log of a product, which is why the answer is a product equal to c.

Answer: (c) tan y − cot x = c

Explanation:
1. Use 1 + cos 2y = 2 cos²y and 1 − cos 2x = 2 sin²x: \( \frac{dy}{dx} = -\frac{\cos^2 y}{\sin^2 x} \).
2. Separate: \( \sec^2 y\,dy = -\text{cosec}^2 x\,dx \).
3. Integrate: tan y = cot x + c.
4. So tan y − cot x = c.

Teacher's Note:
1. The double angle identities 1 + cos 2θ = 2cos²θ and 1 − cos 2θ = 2sin²θ turn such fractions into squares.
2. \( \int -\text{cosec}^2 x\,dx = +\cot x \); the two minus signs cancel.

Answer: (d) sin y = c cos x

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

Teacher's Note:
1. Both sides are of the form \( \frac{f'}{f} \): cot y integrates to log sin y and −tan x to log cos x.
2. Check by differentiating sin y = c cos x: cos y y′ = −c sin x, and c = sin y/cos x gives back the given equation.

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

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

Teacher's Note:
1. When y appears as a factor on the right, divide by y to get log y on the left.
2. Option (a) has the wrong sign; the correct form gives \( y = e^{e^x + x + c} \), which is always positive.

Answer: (b) \( y = c - \frac{2}{x} \)

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

Teacher's Note:
1. \( \int x^{-2}\,dx = -x^{-1} \); the minus sign is the most common error here.
2. Check by differentiating: \( \frac{d}{dx}\left(c - \frac{2}{x}\right) = \frac{2}{x^2} \), so \( x^2y' = 2 \).

Answer: (d) None of these

Explanation:
1. Integrate: \( y = \int \sec^2 x\,dx = \tan x + c \).
2. None of (a), (b), (c) equals tan x + c.
3. So the answer is None of these.

Teacher's Note:
1. Keep the basic pairs clear: \( \int \sec^2 x\,dx = \tan x \), while \( \int \sec x\tan x\,dx = \sec x \).
2. Option (a), sec x tan x, is the derivative of sec x, not an integral of sec²x.

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

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

Teacher's Note:
1. Numerator 1 over \( 1 + x^2 \) gives \( \tan^{-1}x \); numerator x over \( 1 + x^2 \) would give a logarithm.
2. Check: the derivative of \( \tan^{-1}x \) is \( \frac{1}{1 + x^2} \), so \( (1 + x^2)y' = 1 \).

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

Explanation:
1. Integrate term by term: \( \int e^x = e^x \), \( \int \cos x = \sin x \), \( \int x = \frac{x^2}{2} \).
2. \( \int \tan x\,dx = \log\sec x \).
3. So \( y = e^x + \sin x + \frac{x^2}{2} + \log\sec x + c \).

Teacher's Note:
1. \( \int \tan x\,dx \) can be written as \( -\log\cos x \) or \( \log\sec x \); they are equal.
2. Options (a) and (c) use \( +\log\cos x \), which has the wrong sign.

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

Explanation:
1. Separate: \( \frac{e^y}{e^y + 1}\,dy = -\cot x\,dx \).
2. Integrate: log (eʸ + 1) = −log sin x + log K.
3. So \( (e^y + 1)\sin x = K \).

Teacher's Note:
1. The left side is \( \frac{f'(y)}{f(y)} \) with f(y) = eʸ + 1, so it integrates to log (eʸ + 1).
2. Moving −log sin x to the left turns the difference of logs into the log of a product.

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

Explanation:
1. Integrate: \( y = \int (\sin x + 2x)\,dx \).
2. \( \int \sin x\,dx = -\cos x \) and \( \int 2x\,dx = x^2 \).
3. So \( y = x^2 - \cos x + c \).

Teacher's Note:
1. \( \int \sin x\,dx = -\cos x \); forgetting the minus sign leads to options (b) and (d).
2. A general solution must have an arbitrary constant, so (c) is ruled out at once.

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

Explanation:
1. Separate: \( \frac{dy}{y} = 2x\,dx \).
2. Integrate: log y = x² + log c.
3. So \( y = ce^{x^2} \).

Teacher's Note:
1. Writing the constant as log c gives the neat exponential form directly.
2. Check: if \( y = ce^{x^2} \), then \( y' = 2xce^{x^2} = 2xy \).

Answer: (a) \( y = \frac{1}{5}\log|4 + 5\sin x| + C \)

Explanation:
1. Separate: \( dy = \frac{\cos x}{4 + 5\sin x}\,dx \).
2. Put t = 4 + 5 sin x, dt = 5 cos x dx: \( \int \frac{\cos x}{4 + 5\sin x}\,dx = \frac{1}{5}\int \frac{dt}{t} \).
3. So \( y = \frac{1}{5}\log|4 + 5\sin x| + C \).

Teacher's Note:
1. When the numerator is (almost) the derivative of the denominator, adjust the constant factor and write a logarithm.
2. Here the derivative of 4 + 5 sin x is 5 cos x, which is why the factor 1/5 appears.

Answer: (c) y = x(log x − 1) + C

Explanation:
1. Integrate by parts with u = log x, dv = dx: \( \int \log x\,dx = x\log x - \int x\cdot\frac{1}{x}\,dx \).
2. So y = x log x − x + C.
3. Hence y = x(log x − 1) + C.

Teacher's Note:
1. To integrate log x alone, treat it as 1 · log x and use integration by parts.
2. Check: the derivative of x log x − x is log x + 1 − 1 = log x.

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

Explanation:
1. Separate: \( dy = -\frac{\cos x - \sin x}{\sin x + \cos x}\,dx \).
2. The numerator cos x − sin x is the derivative of sin x + cos x, so y = −log (sin x + cos x) + log c.
3. So log [eʸ] + log (sin x + cos x) = log c, i.e. \( e^y(\sin x + \cos x) = c \).

Teacher's Note:
1. Spot that \( \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \), which turns the integral into a logarithm.
2. To remove a log mixed with y, write y = log eʸ and combine all the logarithms.

Answer: (d) \( x^2 + y^2 = px^3 \)

Explanation:
1. Write \( \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy} \), which is homogeneous. Put y = vx, so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).
2. Then \( v + x\frac{dv}{dx} = \frac{1 + 3v^2}{2v} \), so \( x\frac{dv}{dx} = \frac{1 + v^2}{2v} \).
3. Separate: \( \frac{2v}{1 + v^2}\,dv = \frac{dx}{x} \), so log (1 + v²) = log x + log p.
4. So \( 1 + \frac{y^2}{x^2} = px \), i.e. \( x^2 + y^2 = px^3 \).

Teacher's Note:
1. For a homogeneous equation, every term in the numerator and denominator has the same total degree, here 2.
2. After substituting y = vx, always subtract v from both sides before separating.

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

Explanation:
1. Put y = vx: \( v + x\frac{dv}{dx} = \frac{1 + v}{1 - v} \), so \( x\frac{dv}{dx} = \frac{1 + v^2}{1 - v} \).
2. Separate: \( \frac{1 - v}{1 + v^2}\,dv = \frac{dx}{x} \), so \( \tan^{-1}v - \frac{1}{2}\log(1 + v^2) = \log x + \log c' \).
3. Since \( x\sqrt{1 + v^2} = \sqrt{x^2 + y^2} \), this gives \( \tan^{-1}\frac{y}{x} = \log\left(c'\sqrt{x^2 + y^2}\right) \).
4. So \( c(x^2 + y^2)^{1/2} = e^{\tan^{-1}(y/x)} \).

Teacher's Note:
1. Split \( \frac{1 - v}{1 + v^2} \) into \( \frac{1}{1 + v^2} - \frac{v}{1 + v^2} \) to get an inverse tangent and a logarithm.
2. The answer has \( \sqrt{x^2 + y^2} \), not \( x^2 + y^2 \), because of the factor 1/2 in front of the logarithm.

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

Explanation:
1. Put y = vx: \( v + x\frac{dv}{dx} = \frac{v}{1 + v^2} \), so \( x\frac{dv}{dx} = -\frac{v^3}{1 + v^2} \).
2. Separate: \( \left(\frac{1}{v^3} + \frac{1}{v}\right)dv = -\frac{dx}{x} \), so \( -\frac{1}{2v^2} + \log v = -\log x + C \).
3. Since vx = y, this is \( \log y = \frac{x^2}{2y^2} + C \), so \( y^2 = Ae^{x^2/y^2} \).
4. Writing a = 1/A, \( ay^2 = e^{x^2/y^2} \).

Teacher's Note:
1. Combine log v + log x into log (vx) = log y; this simplification is the key step at the end.
2. Doubling both sides of \( \log y = \frac{x^2}{2y^2} + C \) gives \( \log y^2 \), which is why the answer has \( y^2 \).

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

Explanation:
1. Put y = vx: \( v + x\frac{dv}{dx} = \frac{1}{2v - 1} \), so \( x\frac{dv}{dx} = \frac{1 + v - 2v^2}{2v - 1} = -\frac{(2v + 1)(v - 1)}{2v - 1} \).
2. Partial fractions: \( \frac{2v - 1}{(2v + 1)(v - 1)} = \frac{1}{3(v - 1)} + \frac{4}{3(2v + 1)} \).
3. Integrate: \( \frac{1}{3}\log(v - 1) + \frac{2}{3}\log(2v + 1) = -\log x + C \), i.e. \( (v - 1)(2v + 1)^2x^3 = c' \).
4. Since \( x(v - 1) = y - x \) and \( x(2v + 1) = 2y + x \), this gives \( (x - y)(x + 2y)^2 = c \).

Teacher's Note:
1. Multiply out by powers of x at the end so that each factor in v becomes a linear expression in x and y.
2. Check by implicit differentiation: for \( F = (x - y)(x + 2y)^2 \), \( -\frac{F_x}{F_y} = \frac{3x}{6y - 3x} = \frac{x}{2y - x} \).

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

Explanation:
1. Write \( \frac{dy}{dx} = \frac{2y - x}{y} \). Put y = vx: \( v + x\frac{dv}{dx} = \frac{2v - 1}{v} \).
2. So \( x\frac{dv}{dx} = \frac{-(v - 1)^2}{v} \) and \( \frac{v}{(v - 1)^2}\,dv = -\frac{dx}{x} \).
3. Write \( \frac{v}{(v - 1)^2} = \frac{1}{v - 1} + \frac{1}{(v - 1)^2} \): log (v − 1) − \( \frac{1}{v - 1} \) = −log x + c.
4. Since x(v − 1) = y − x and \( \frac{1}{v - 1} = \frac{x}{y - x} \): \( \log(y - x) = c + \frac{x}{y - x} \).

Teacher's Note:
1. A repeated factor like \( (v - 1)^2 \) in the denominator needs both \( \frac{1}{v - 1} \) and \( \frac{1}{(v - 1)^2} \) terms.
2. Adding log v-type terms to log x lets you return to x and y in a single logarithm.

Answer: (c) \( y^{2/3} - x^{2/3} = C \)

Explanation:
1. Write \( \frac{dy}{dx} = \frac{y^{1/3}}{x^{1/3}} \) and separate: \( y^{-1/3}\,dy = x^{-1/3}\,dx \).
2. Integrate: \( \frac{3}{2}y^{2/3} = \frac{3}{2}x^{2/3} + C' \).
3. So \( y^{2/3} - x^{2/3} = C \).

Teacher's Note:
1. Although it is homogeneous, this equation is directly separable, which is much quicker than putting y = vx.
2. \( \int t^{-1/3}\,dt = \frac{t^{2/3}}{2/3} \): add 1 to the power and divide by the new power.

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

Explanation:
1. Cross multiply: (x + y) dy = (x − y) dx, so x dy + y dx + y dy − x dx = 0.
2. x dy + y dx = d(xy), y dy = \( d\left(\frac{y^2}{2}\right) \), x dx = \( d\left(\frac{x^2}{2}\right) \).
3. Integrate: \( xy + \frac{y^2}{2} - \frac{x^2}{2} = C' \).
4. Multiply by 2: \( y^2 + 2xy - x^2 = C \).

Teacher's Note:
1. Grouping x dy + y dx into d(xy) gives the answer faster than the y = vx substitution.
2. Check by implicit differentiation: 2yy′ + 2y + 2xy′ − 2x = 0 gives \( y' = \frac{x - y}{x + y} \).

Answer: (d) None of these

Explanation:
1. Divide by x: \( \frac{dy}{dx} + \frac{3}{x}y = 1 \), a linear equation with IF = \( e^{\int 3/x\,dx} = x^3 \).
2. Multiply: \( \frac{d}{dx}(x^3y) = x^3 \), so \( x^3y = \frac{x^4}{4} + c \).
3. (a) has the sign of \( \frac{x^4}{4} \) wrong, (b) is not an equation and (c) has no constant, so none matches.
4. The answer is None of these.

Teacher's Note:
1. An expression with no equality sign, like option (b), can never be the solution of a differential equation.
2. Even when the equation appears under a homogeneous heading, it may be easiest to solve as a linear equation.

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

Explanation:
1. This is linear with P = 1, Q = eˣ, so IF = \( e^{\int 1\,dx} = e^x \).
2. Then \( ye^x = \int e^{2x}\,dx = \frac{1}{2}e^{2x} + c \).
3. Divide by eˣ: \( y = \frac{1}{2}e^x + ce^{-x} \).

Teacher's Note:
1. For \( \frac{dy}{dx} + Py = Q \), the solution is \( y \cdot \text{IF} = \int Q \cdot \text{IF}\,dx + c \).
2. Check: \( y = \frac{1}{2}e^x \) gives \( y' + y = e^x \), so the particular part must carry the factor 1/2.

Answer: (b) \( y\frac{dy}{dx} + 4x = 0 \)

Explanation:
1. An equation is linear when y and its derivatives occur only to the first power and are not multiplied together.
2. In (b), the term \( y\frac{dy}{dx} \) is a product of y and its derivative.
3. (a), (c) and (d) are all linear, so (b) is the non-linear equation.

Teacher's Note:
1. Products such as \( y\frac{dy}{dx} \), powers like y², and functions like sin y all make an equation non-linear.
2. Terms in x alone, such as log x or cos x, never affect linearity.

Answer: (d) \( y(1 + x^3) = \frac{x}{2} - \frac{1}{4}\sin 2x + c \)

Explanation:
1. Linear with \( P = \frac{3x^2}{1 + x^3} \), so IF = \( e^{\log(1 + x^3)} = 1 + x^3 \).
2. Then \( y(1 + x^3) = \int \sin^2 x\,dx = \int \frac{1 - \cos 2x}{2}\,dx \).
3. So \( y(1 + x^3) = \frac{x}{2} - \frac{1}{4}\sin 2x + c \).

Teacher's Note:
1. When P is \( \frac{f'(x)}{f(x)} \), the integrating factor is simply f(x).
2. Integrate \( \sin^2 x \) by writing it as \( \frac{1 - \cos 2x}{2} \) first.

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

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

Teacher's Note:
1. Remember \( e^{\log f(x)} = f(x) \), which makes most integrating factors simple.
2. Move all terms except \( \frac{dy}{dx} + Py \) to the right before reading off Q.

Answer: (b) \( x^2\frac{dy}{dx} + y = e^x \)

Explanation:
1. (a) has xy², (c) has xy² and (d) has y², so they are all non-linear in y.
2. (b) is \( x^2\frac{dy}{dx} + y = e^x \): y and \( \frac{dy}{dx} \) appear only to power 1.
3. So (b) is linear (it is \( \frac{dy}{dx} + \frac{1}{x^2}y = \frac{e^x}{x^2} \)).

Teacher's Note:
1. Coefficients that depend only on x, like x², do not spoil linearity.
2. Any power of y higher than 1 makes the equation non-linear, wherever it appears.

Answer: (a) \( 4xy = x^4 + c \)

Explanation:
1. Linear with \( P = \frac{1}{x} \), so IF = \( e^{\log x} = x \).
2. Then \( xy = \int x^3\,dx = \frac{x^4}{4} + c' \).
3. Multiply by 4: \( 4xy = x^4 + c \).

Teacher's Note:
1. For \( P = \frac{k}{x} \), the integrating factor is always \( x^k \).
2. Check: \( y = \frac{x^3}{4} \) gives \( y' + \frac{y}{x} = \frac{3x^2}{4} + \frac{x^2}{4} = x^2 \).

Answer: (d) \( x\frac{dy}{dx} + \frac{3}{\frac{dy}{dx}} = y^2 \)

Explanation:
1. (a), (b) and (c) contain y and its derivatives only to the first power, so they are linear.
2. In (d), clearing the fraction gives \( x\left(\frac{dy}{dx}\right)^2 + 3 = y^2\frac{dy}{dx} \).
3. This has \( \left(\frac{dy}{dx}\right)^2 \) and \( y^2\frac{dy}{dx} \), so (d) is non-linear.

Teacher's Note:
1. A derivative in the denominator, like \( \frac{3}{dy/dx} \), always makes the equation non-linear.
2. A second order equation such as y″ + y = 0 can still be linear; linearity is about powers, not order.

Answer: (a) \( y = \frac{1}{2}(\cos x + \sin x) + ce^{-x} \)

Explanation:
1. Linear with P = 1, so IF = eˣ, and \( ye^x = \int e^x\cos x\,dx \).
2. Using \( \int e^{ax}\cos bx\,dx = \frac{e^{ax}}{a^2 + b^2}(a\cos bx + b\sin bx) \) with a = b = 1: \( \int e^x\cos x\,dx = \frac{e^x}{2}(\cos x + \sin x) \).
3. So \( ye^x = \frac{e^x}{2}(\cos x + \sin x) + c \), i.e. \( y = \frac{1}{2}(\cos x + \sin x) + ce^{-x} \).

Teacher's Note:
1. The standard result for \( \int e^{ax}\cos bx\,dx \) saves applying integration by parts twice.
2. Check: \( y = \frac{1}{2}(\cos x + \sin x) \) gives \( y' + y = \frac{1}{2}(-\sin x + \cos x + \cos x + \sin x) = \cos x \).