JEE Mathematics Differential Equations MCQs Set 23

Answer: (a) \( \log\left(\dfrac{\sqrt3+6x-8y}{\sqrt3-6x+8y}\right)=4\sqrt3\,x \)

Explanation:
1. Put v = 3x − 4y, so v' = 3 − 4y' = 3 − 4v².
2. Separating, dv/(3 − 4v²) = dx, and ∫ dv/((√3 − 2v)(√3 + 2v)) = (1/(4√3)) ln((√3 + 2v)/(√3 − 2v)).
3. So ln((√3 + 2v)/(√3 − 2v)) = 4√3 x + c, and with 2v = 6x − 8y the constant is 0 because y(0) = 0.
4. Hence log((√3 + 6x − 8y)/(√3 − 6x + 8y)) = 4√3 x.

Teacher's Note:
1. The substitution v = 3x − 4y separates the variables.

Answer: (d) \( \dfrac{3e^2}{4} \)

Explanation:
1. The left side is (xy)', so (xy)' = x ln x and xy = (x²/2) ln x − x²/4 + C.
2. At x = 2: 2y = 2 ln 2 − 1, while the right side is 2 ln 2 − 1 + C, so C = 0.
3. At x = e²: xy = (e⁴/2)(2) − e⁴/4 = 3e⁴/4, so y = 3e²/4.

Teacher's Note:
1. Recognise x dy/dx + y as the derivative of xy.

Answer: (b) \( (1,5) \)

Explanation:
1. dy/dx + 2y/x = x. The integrating factor is x², so (x²y)' = x³ and x²y = x⁴/4 + C.
2. Through (−1, 5): 5 = 1/4 + C, so C = 19/4 and y = x²/4 + 19/(4x²).
3. At x = 1: y = 1/4 + 19/4 = 5, so (1, 5) lies on the curve. For the other options: at (−1, 2) the formula gives 5, at (3, 0) it gives 9/4 + 19/36 and at (√2, 1) it gives 1/2 + 19/8, none of which matches.

Teacher's Note:
1. Solve the linear equation and test the options.

Answer: (b) 3

Explanation:
1. Putting x = 0: f(0) = f(0)f(y), and f(0) ≠ 0, so f(y) = 1 for all y.
2. Then dy/dx = 1 and y = x + 1.
3. y(2/3) + y(1/3) = 5/3 + 4/3 = 3.

Teacher's Note:
1. The condition f(0) ≠ 0 forces f to be the constant 1.

Answer: (d) a circle with centre on the x-axis

Explanation:
1. Put y = vx: v + x v' = (v² − 1)/(2v), so x v' = −(1 + v²)/(2v).
2. Separating, 2v dv/(1 + v²) = −dx/x, so (1 + v²)x = c, i.e. x² + y² = cx.
3. This is a circle with centre (c/2, 0) on the x-axis (through (2, 1): c = 5/2).

Teacher's Note:
1. The equation is homogeneous; the substitution y = vx gives x² + y² = cx.

Answer: (a) exists and equals 8

Explanation:
1. Put t = 1/x → ∞: the limit is L = lim f(t)/t.
2. By L'Hospital, L = lim f'(t) = lim (14 − (3/4)f(t)/t) = 14 − (3/4)L, so (7/4)L = 14 and L = 8.
3. (Check: the general solution is f = 8x + Cx^−3/4, and xf(1/x) = 8 + Cx^7/4 → 8.)

Teacher's Note:
1. Rewrite the limit as lim f(t)/t and apply L'Hospital with the given f'.

Answer: (c) \( y=x\tan(c-x) \)

Explanation:
1. Multiply out: x dy = (y − x² − y²) dx, so x dy − y dx = −(x² + y²) dx.
2. Dividing by x² + y²: (x dy − y dx)/(x² + y²) = d(tan⁻¹(y/x)) = −dx.
3. So tan⁻¹(y/x) = −x + c, i.e. y = x tan(c − x).

Teacher's Note:
1. x dy − y dx over x² + y² is the differential of tan⁻¹(y/x).

Answer: (a) \( x\log_e|y|=2(x-1) \)

Explanation:
1. The centre of the circle is (1, 1).
2. dy/y = 2 dx/x², so ln|y| = −2/x + C. At (1, 1): 0 = −2 + C, so C = 2.
3. ln|y| = 2 − 2/x = 2(x − 1)/x, i.e. x ln|y| = 2(x − 1).

Teacher's Note:
1. Separate the variables and use the centre of the circle as the given point.

Answer: (d) \( y=\dfrac{x^2}{4}+\dfrac3{4x^2} \)

Explanation:
1. Multiply by x: x²y' + 2xy = x³, so (x²y)' = x³ and x²y = x⁴/4 + C.
2. At x = 1, y = 1: 1 = 1/4 + C, so C = 3/4.
3. y = x²/4 + 3/(4x²).

Teacher's Note:
1. The integrating factor of a linear equation with coefficient 2/x is x².

Answer: (c) \( -\dfrac{\pi^2}{2\sqrt3} \)

Explanation:
1. The left side is (y cos x)', so (y cos x)' = 6x and y cos x = 3x² + C.
2. At x = π/3, y = 0: 0 = 3π²/9 + C, so C = −π²/3.
3. At x = π/6: y(√3/2) = 3π²/36 − π²/3 = −π²/4, so y = −π²/(2√3).

Teacher's Note:
1. cos x y' − y sin x is the derivative of y cos x.

Answer: (b) \( y'\left(\dfrac\pi4\right)-y'\left(-\dfrac\pi4\right)=\pi-\sqrt2 \)

Explanation:
1. The integrating factor is sec x: (y sec x)' = 2x sec x + x² sec x tan x = (x² sec x)', so y sec x = x² sec x + C.
2. y = x² + C cos x, and y(0) = 1 gives C = 1, so y = x² + cos x and y' = 2x − sin x.
3. y' is odd, so y'(π/4) − y'(−π/4) = 2y'(π/4) = 2(π/2 − √2/2) = π − √2. The sum of the y' values is 0, y is even (so the difference of the y values is 0) and the sum of the y values is π²/8 + √2, which rules out the others.

Teacher's Note:
1. Solve the linear equation completely, then use the parity of y and y'.

Answer: (d) \( \dfrac32-\dfrac1{\sqrt e} \)

Explanation:
1. Treat x as a function of y: dx/dy + x/y² = 1/y³. The integrating factor is e^−1/y.
2. With u = −1/y: ∫ e^−1/y/y³ dy = ∫ (−u)e^u du = (1 − u)e^u = (1 + 1/y)e^−1/y. So x = 1 + 1/y + Ce^1/y.
3. At (1, 1): 1 = 2 + Ce, so C = −1/e. At y = 2: x = 3/2 − e^1/2/e = 3/2 − 1/√e.

Teacher's Note:
1. Treat x as the dependent variable to obtain a linear equation.

Answer: (d) \( \dfrac{41}{12} \)

Explanation:
1. Multiply by x³: (x⁴y)' = x⁵, so x⁴y = x⁶/6 + C. At x = 1: 1 = 1/6 + C, so C = 5/6.
2. y = x²/6 + 5/(6x⁴). At x = 1/√2, x² = 1/2 and x⁴ = 1/4: y = 1/12 + 5/(6/4) = 1/12 + 10/3 = 41/12.

Teacher's Note:
1. The integrating factor is x⁴/x = x³.

Answer: (b) \( \dfrac1{16} \)

Explanation:
1. Divide by (x² + 1): (x² + 1)y' + 2xy = 1/(x² + 1), i.e. ((x² + 1)y)' = 1/(x² + 1).
2. So (x² + 1)y = tan⁻¹x + C, and y(0) = 0 gives C = 0. Then y(1) = (π/4)/2 = π/8.
3. √a · π/8 = π/32 gives √a = 1/4 and a = 1/16.

Teacher's Note:
1. The left side is the derivative of (x² + 1)y after dividing by x² + 1.

Answer: (c) \( e-2 \)

Explanation:
1. Put t = tan x: dy/dt = t − y, so y' + y = t with solution y = t − 1 + Ce^−t.
2. y(0) = 0 (t = 0) gives −1 + C = 0, so C = 1.
3. At x = −π/4, t = −1: y = −1 − 1 + e = e − 2.

Teacher's Note:
1. Substituting t = tan x turns it into a standard linear equation.

Answer: (a) \( y^2+2x^3+cx^2=0 \)

Explanation:
1. With u = y²: y dy/dx = (y² − x³)/x gives u'/2 = (u − x³)/x, i.e. u' − 2u/x = −2x².
2. Integrating factor x⁻²: (u/x²)' = −2, so u/x² = −2x + c'.
3. u = −2x³ + c'x², i.e. y² + 2x³ + cx² = 0 (with c = −c').

Teacher's Note:
1. Substitute u = y² to get a linear equation in u.

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

Explanation:
1. Write (2x + 3y + 1)dx + (3x + 2y − 1)dy = 0. With M = 2x + 3y + 1 and N = 3x + 2y − 1, ∂M/∂y = 3 = ∂N/∂x: the equation is exact.
2. Integrating M with respect to x gives x² + 3xy + x + g(y). Matching ∂/∂y with N: 3x + g'(y) = 3x + 2y − 1, so g = y² − y.
3. The solution is x² + 3xy + y² + x − y = c.

Teacher's Note:
1. Check ∂M/∂y = ∂N/∂x to confirm exactness, then integrate.

Answer: (c) \( \sin^{-1}\left(\dfrac{\sqrt{x^2+y^2}}a\right)=\tan^{-1}\dfrac yx+c \)

Explanation:
1. In polar form x dx + y dy = r dr and x dy − y dx = r² dθ.
2. The equation becomes (r dr)/(r² dθ) = √(a² − r²)/r, i.e. dr/√(a² − r²) = dθ.
3. So sin⁻¹(r/a) = θ + c, i.e. sin⁻¹(√(x² + y²)/a) = tan⁻¹(y/x) + c.

Teacher's Note:
1. Polar substitution turns the equation into separable form.

Answer: (a) TTT

Explanation:
1. S1: y² = 4ax gives 2yy' = 4a, so y² = 2xyy', i.e. y = 2xy' and dy/y = dx/(2x): separable, true.
2. S2: The lines are y = mx + p√(1 + m²), so (y − xy')² = p²(1 + y'²): order 1 and degree 2, true.
3. S3: The conics are ax² + by² = 1, with two arbitrary constants, so the differential equation has order 2: true.
4. All three are true.

Teacher's Note:
1. Count the arbitrary constants for the order and clear radicals for the degree.

Answer: (c) \( xy+y\log x+x\sin y=C \)

Explanation:
1. M = y + y/x + sin y and N = x + log x + x cos y. ∂M/∂y = 1 + 1/x + cos y = ∂N/∂x, so the equation is exact.
2. ∫M dx = xy + y log x + x sin y, and its y-derivative x + log x + x cos y equals N.
3. So the solution is xy + y log x + x sin y = C.

Teacher's Note:
1. Check exactness, then integrate M with respect to x.

Answer: (b) −16

Explanation:
1. At x = 0: 1 + y(0) = 5, so y(0) = 4.
2. Differentiating: e^xy²(y² + 2xyy') + y' cos(x²) − 2xy sin(x²) = 0.
3. At x = 0: 1 · y(0)² + y'(0) = 0, i.e. 16 + y'(0) = 0, so y'(0) = −16.

Teacher's Note:
1. Differentiate implicitly and substitute x = 0 with y(0) = 4.