JEE Mathematics Differential Equations MCQs Set 22

Answer: (d) p < q

Explanation:
1. The highest derivative is y'', so the order is p = 2.
2. Isolate the fractional power: 3x(y')^1/3 = sin x − x²y² − y²(y'')². Cubing gives 27x³y' = (sin x − x²y² − y²(y'')²)³, a polynomial in derivatives.
3. The highest power of y'' is 2 × 3 = 6, so the degree is q = 6 and p < q.

Teacher's Note:
1. Remove fractional powers of derivatives before reading the degree.

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

Explanation:
1. The integrating factor is e^kt/m, so (ve^kt/m)' = −g e^kt/m.
2. Integrating, ve^kt/m = −(mg/k)e^kt/m + c, so v = ce^−kt/m − mg/k.

Teacher's Note:
1. A linear equation y' + Py = Q is solved with the integrating factor e^{∫P dx}.

Answer: (a) \( \dfrac{1}{y_1(x)}\displaystyle\int r(x)y_1(x)\,dx \)

Explanation:
1. y₁ satisfies y₁' = f y₁, so y₁ = e^{∫f dx} (up to a constant).
2. For y' + f y = r the integrating factor is e^{∫f dx} = y₁, so y₁y = ∫ r y₁ dx.
3. Hence y = (1/y₁)∫ r y₁ dx.

Teacher's Note:
1. The integrating factor equals y₁ here.

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

Explanation:
1. Invert: dx/dy = x + y + 1, a linear equation in x: dx/dy − x = y + 1.
2. The integrating factor is e^−y: d(xe^−y)/dy = (y + 1)e^−y, so xe^−y = −(y + 2)e^−y + c.
3. x = −(y + 2) + ce^y.

Teacher's Note:
1. Treat x as the dependent variable when y appears awkwardly.

Answer: (c) 2x

Explanation:
1. Circles touching the x-axis at the origin: x² + y² + 2ky = 0, i.e. k = −(x² + y²)/(2y).
2. Differentiating: 2x + 2yy' + 2ky' = 0, so x + y' (y + k) = 0. Substituting k: y + k = (y² − x²)/(2y), giving x = y'(x² − y²)/(2y).
3. So (x² − y²)y' = 2xy and g(x) = 2x.

Teacher's Note:
1. Write the family with one parameter and eliminate it.

Answer: (c) yy'' = (y')²

Explanation:
1. ln y = ln c₁ + c₂x, so y'/y = c₂ (a constant).
2. Differentiating again: (y''y − y'²)/y² = 0, i.e. yy'' = (y')².

Teacher's Note:
1. Take logarithms to isolate the constant c₂.

Answer: (a) \( y^2=x^2+2xy\dfrac{dy}{dx} \)

Explanation:
1. The family is x² + y² − 2gx = 0.
2. Differentiating: 2x + 2yy' − 2g = 0, so g = x + yy'.
3. Substituting: x² + y² = 2x(x + yy'), i.e. y² = x² + 2xyy'.

Teacher's Note:
1. One arbitrary constant needs one differentiation.

Answer: (c) order 1, degree 3

Explanation:
1. Differentiating: 2yy' = 2c, so c = yy'.
2. Substitute: y² = 2yy'(x + √(yy')), i.e. y² − 2xyy' = 2yy'√(yy').
3. Squaring to remove the root: (y² − 2xyy')² = 4y³y'³. The highest power of y' is 3 and only y' appears, so the order is 1 and the degree is 3.

Teacher's Note:
1. Eliminate the radical by squaring before reading off the degree.

Answer: (b) 1/3

Explanation:
1. The equation says d/dx[(y + 1)(2 + sin x)] = 0, so (y + 1)(2 + sin x) = constant.
2. At x = 0: (1 + 1)(2) = 4. At x = π/2: (y + 1)(3) = 4, so y = 1/3.

Teacher's Note:
1. Recognise the left side as an exact derivative of a product.

Answer: (b) 4/5

Explanation:
1. Rewrite as x dy − y dx = xy² dx and divide by y²: −d(x/y) = x dx.
2. So x/y = −x²/2 + c. At (1, −1): −1 = −1/2 + c, so c = −1/2 and y = −2x/(x² + 1).
3. f(−1/2) = 1/(1/4 + 1) = 4/5.

Teacher's Note:
1. The combination x dy − y dx suggests d(x/y) or d(y/x).

Answer: (d) 31/18

Explanation:
1. The limit is 0/0; differentiating the numerator in t at t = x gives 2xf(x) − x²f'(x) = 1.
2. So f' − 2f/x = −1/x², and (f/x²)' = −1/x⁴, giving f/x² = 1/(3x³) + C.
3. f(1) = 1 gives C = 2/3, so f(x) = 1/(3x) + 2x²/3 and f(3/2) = 2/9 + 3/2 = 31/18.

Teacher's Note:
1. A difference quotient in t at t = x gives a derivative.

Answer: (b) 3

Explanation:
1. dx/dy = x/y + 2y, so d(x/y)/dy = 2, giving x = 2y² + cy.
2. f(−1) = 1: 2 − c = 1, so c = 1 and f(y) = 2y² + y.
3. f(1) = 3.

Teacher's Note:
1. Divide by y to see the linear equation in x.

Answer: (c) 400 − 300e^t/2

Explanation:
1. p' − p/2 = −200 has the constant solution p = 400, so p = 400 + Ce^t/2.
2. p(0) = 100 gives C = −300.

Teacher's Note:
1. Find the equilibrium value and add the homogeneous solution.

Answer: (a) (1 + eˣ)tan y = 2

Explanation:
1. Separate: eˣ/(1 + eˣ) dx + sec²y/tan y dy = 0.
2. Integrating: ln(1 + eˣ) + ln tan y = ln C, so (1 + eˣ)tan y = C.
3. At (0, π/4): 2 · 1 = 2, so C = 2.

Teacher's Note:
1. Separate variables and integrate both sides.

Answer: (c) 3500

Explanation:
1. P = 2000 + ∫₀²⁵(100 − 12√x)dx.
2. ∫ = [100x − 8x^3/2]₀²⁵ = 2500 − 8 · 125 = 1500.
3. P = 3500.

Teacher's Note:
1. Integrate the rate from x = 0 to x = 25 and add the initial production.

Answer: (a) −3/2

Explanation:
1. y = x + 1/x + c. At (2, 7/2): 2 + 1/2 + c = 7/2, so c = 1.
2. At x = −2: y = −2 − 1/2 + 1 = −3/2.

Teacher's Note:
1. Integrate the slope and use the given point for the constant.

Answer: (a) 2 ln 18

Explanation:
1. Equilibrium at p = 900, so p = 900 − 50e^t/2 (using p(0) = 850).
2. p = 0 gives e^t/2 = 18, so t = 2 ln 18.

Teacher's Note:
1. Write p = equilibrium + Ce^t/2.

Answer: (c) 7/2

Explanation:
1. Separate: dy/(1 + y) = cos x/(2 + sin x) dx, so ln(1 + y) = ln(2 + sin x) + C.
2. At x = 0: ln 3 = ln 2 + C, so 1 + y = (3/2)(2 + sin x).
3. At x = π/2: 1 + y = 9/2, so y = 7/2.

Teacher's Note:
1. Integrate both separated sides; the right side is a logarithmic derivative.

Answer: (d) y = x ln x + x

Explanation:
1. Put y = vx: v + xv' = 1 + v, so xv' = 1 and v = ln x + c.
2. y = x ln x + cx. At (1, 1): c = 1, so y = x ln x + x.

Teacher's Note:
1. The equation is homogeneous; substitute y = vx.

Answer: (c) \( \log\left(\frac yx\right)=cx \)

Explanation:
1. Put y = vx: v + xv' = v(ln v + 1), so xv' = v ln v.
2. dv/(v ln v) = dx/x gives ln(ln v) = ln x + ln c, so ln v = cx.
3. Hence log(y/x) = cx.

Teacher's Note:
1. The substitution y = vx turns it into a separable equation.

Answer: (b) (−1/3, 1/3)

Explanation:
1. dx/dy = x/y + 3y, so d(x/y) = 3 dy and x = 3y² + cy.
2. Through (1, 1): 1 = 3 + c, so c = −2 and x = 3y² − 2y.
3. Check: y = 1/3 gives x = 1/3 − 2/3 = −1/3, so (−1/3, 1/3) lies on the curve.

Teacher's Note:
1. Substitute each option into x = 3y² − 2y.

Answer: (d) \( y^2=1-\dfrac{x}{\sec x+\tan x} \)

Explanation:
1. Multiply by 2y and put u = y²: u' + u sec x = tan x.
2. The integrating factor is sec x + tan x, and ∫ tan x(sec x + tan x)dx = sec x + tan x − x + C.
3. So u(sec x + tan x) = sec x + tan x − x + C; at x = 0, u = 1 gives C = 0, so y² = 1 − x/(sec x + tan x).

Teacher's Note:
1. A Bernoulli equation in y with power −1 becomes linear in y².

Answer: (d) \( y\sqrt{\cot x}=x+c \)

Explanation:
1. y' − y/sin 2x = √tan x. The integrating factor is exp(−∫ csc 2x dx) = exp(−½ ln tan x) = 1/√tan x.
2. d/dx(y/√tan x) = √tan x/√tan x = 1, so y/√tan x = x + c.
3. That is y√cot x = x + c.

Teacher's Note:
1. ∫ csc 2x dx = ½ ln tan x.

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

Explanation:
1. dy/dx = y tan x − y² sec x is a Bernoulli equation; put u = 1/y.
2. u' = −y'/y² = −u tan x + sec x, so u' + u tan x = sec x with integrating factor sec x.
3. u sec x = tan x + c, i.e. sec x/y = tan x + c, so sec x = (tan x + c)y.

Teacher's Note:
1. Divide by y² and substitute u = 1/y.

Answer: (a) x = y³ + cy

Explanation:
1. dx/dy = x/y + 2y², so d(x/y)/dy = 2y.
2. x/y = y² + c, i.e. x = y³ + cy.

Teacher's Note:
1. Invert the equation to make x the dependent variable.

Answer: (a) x³sin³y = 3y² sin x + C

Explanation:
1. M = x² sin³y − y² cos x and N = x³ cos y sin²y − 2y sin x satisfy ∂M/∂y = 3x² sin²y cos y − 2y cos x = ∂N/∂x, so the equation is exact.
2. ∫M dx = x³ sin³y/3 − y² sin x, so the solution is x³ sin³y/3 − y² sin x = c.
3. That is x³ sin³y = 3y² sin x + C.

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

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

Explanation:
1. ∂M/∂y = 1 + 1/x + cos y = ∂N/∂x, so the equation is exact.
2. ∫M dx = y(x + log x) + x sin y.
3. So xy + y log x + x sin y = C.

Teacher's Note:
1. Verify exactness; the potential function is y(x + ln x) + x sin y.

Answer: (a) y = C₁e^3x + C₂e^4x

Explanation:
1. The auxiliary equation m² − 7m + 12 = 0 has roots m = 3 and 4 (distinct).
2. So y = C₁e^3x + C₂e^4x.

Teacher's Note:
1. Distinct real roots give a sum of pure exponentials.