JEE Mathematics Differential Equations MCQs Set 17

Answer: (b) two

Explanation:
1. The highest derivative present is d²y/dx², a second derivative.
2. Hence the order is 2.

Teacher's Note:
1. Order is the order of the highest derivative; degree is its power once the equation is a polynomial in derivatives.
2. Do not confuse the square on dy/dx with the order.

Answer: (b) two

Explanation:
1. Square both sides to remove the radical: \( 1+\left(\frac{dy}{dx}\right)^2 = x^4 \).
2. The highest derivative dy/dx now appears with power 2, so the degree is 2.

Teacher's Note:
1. Degree is defined only after the equation is free from radicals and fractions in the derivatives.
2. A radical alone does not make the degree fractional.

Answer: (c) n = 1, m = 2

Explanation:
1. The family has two independent arbitrary constants, so the order is m = 2.
2. Eliminating them gives y″ − 2y′ + 2y = 0, which is linear in the highest derivative, so the degree is n = 1.

Teacher's Note:
1. The order equals the number of independent arbitrary constants.
2. Check: y = e^x(A cos x + B sin x) corresponds to roots 1 ± i of r² − 2r + 2 = 0.

Answer: (b) 2

Explanation:
1. Write y = a e^c · e^{bx} = A e^{bx} with A = a e^c.
2. Only two independent constants (A and b) remain, so the order is 2.

Teacher's Note:
1. Constants that combine into one are not independent.
2. Count independent constants only.

Answer: (a) 1, 2

Explanation:
1. One arbitrary constant, so the order is 1.
2. Differentiating: 2(y + c)y′ = c, so y + c = c/(2y′). Substituting back gives c = 4xy′².
3. Then y = c(1/(2y′) − 1) = 2xy′(1 − 2y′), which is a polynomial of degree 2 in y′.
4. Order 1, degree 2.

Teacher's Note:
1. After elimination, expand and read the power of the highest derivative.
2. Here the highest derivative is y′ and its highest power is 2.

Answer: (a) 1, 2

Explanation:
1. Parabolas with axis along the x-axis: y² = 4a(x − h), with two constants a and h, so the order is 2.
2. Differentiate twice: 2yy′ = 4a and y y″ + y′² = 0.
3. The highest derivative y″ has power 1, so the degree is 1.
4. Degree 1, order 2.

Teacher's Note:
1. Read carefully: the question asks degree first and order second.
2. Two constants give a second-order equation.

Answer: (a) 1, 1

Explanation:
1. Such circles are x² + y² − 2ky = 0 with one constant k, so the order is 1.
2. Differentiating: x + yy′ = ky′, so k = (x + yy′)/y′.
3. Substituting gives (x² − y²)y′ = 2xy, which is of degree 1.
4. Order 1, degree 1.

Teacher's Note:
1. Fix the geometric condition first to count the free parameters correctly.
2. Touching at the origin fixes the centre on the y-axis.

Answer: (d) 1, 2

Explanation:
1. With foci (±c, 0) fixed, the family is \( \frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1 \) with one parameter a, so the order is 1.
2. Eliminating a gives xy(y′)² + (x² − y² − c²)y′ − xy = 0, which is quadratic in y′.
3. Order 1, degree 2.

Teacher's Note:
1. Fixed foci leave only the major-axis length free.
2. The elimination produces a quadratic in y′, hence degree 2.

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

Explanation:
1. Write it as \( \frac{dy}{dt}-\frac{t}{1+t}y=\frac{1}{1+t} \); the integrating factor is \( e^{-t}(1+t) \).
2. Then \( \frac{d}{dt}\left[y(1+t)e^{-t}\right]=e^{-t} \), so y(1 + t)e^{−t} = −e^{−t} + C.
3. y(0) = −1 gives C = 0, so y = −1/(1 + t).
4. y(1) = −1/2.

Teacher's Note:
1. Always bring a linear equation to the form y′ + Py = Q before finding the integrating factor.
2. Use the initial condition to fix the constant at the end.

Answer: (b) y(x − 1 + log y) + 1 = 0

Explanation:
1. Treat x as a function of y: \( \frac{dx}{dy}+\frac{x}{y}=-\frac{\log y}{y} \), with integrating factor y.
2. Then xy = −∫log y dy = −(y log y − y) + C.
3. At y = 1, x = 0 gives 0 = 1 + C, so C = −1.
4. Hence xy + y log y − y + 1 = 0, that is, y(x − 1 + log y) + 1 = 0.

Teacher's Note:
1. When the equation is linear in x rather than y, swap the roles of the variables.
2. Integration by parts gives ∫log y dy = y log y − y.

Answer: (a) \( 2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+k \)

Explanation:
1. Rewrite as \( \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{\tan^{-1}y}}{1+y^2} \), which is linear in x.
2. The integrating factor is \( e^{\tan^{-1}y} \).
3. With t = tan⁻¹y: x eᵗ = ∫e^{2t}dt = e^{2t}/2 + C.
4. So 2x e^{tan⁻¹y} = e^{2tan⁻¹y} + k.

Teacher's Note:
1. Recognise that ∫dy/(1+y²) = tan⁻¹y to get the integrating factor quickly.
2. Substitute t = tan⁻¹y to simplify the right side.

Answer: (c) 64

Explanation:
1. Put u = x³: dx/x = du/(3u), and the limits become 1 and 64.
2. The integral becomes \( \int_1^{64}\frac{e^{\sin u}}{u}du = f(64)-f(1) \).
3. So k = 64.

Teacher's Note:
1. Substitution should turn the integrand into the given derivative f′(u).
2. Remember to change the limits along with the variable.

Answer: (a) \( ke^{x^2/2} \)

Explanation:
1. The left side is d(xy)/dx. Put u = xy: \( \frac{du}{dx}=x\frac{f(u)}{f'(u)} \).
2. Separating, \( \frac{f'(u)}{f(u)}du = x\,dx \), so ln f(u) = x²/2 + C.
3. Hence f(xy) = k e^{x²/2}.

Teacher's Note:
1. Spot the product-rule pattern x dy/dx + y = d(xy)/dx.
2. Separate variables treating u = xy as the dependent variable.

Answer: (b) \( \frac{1}{y}\phi(x)-x \)

Explanation:
1. Rearrange: φ dy − yφ′ dx + y² dx = 0.
2. Divide by y²: (φ dy − yφ′ dx)/y² + dx = 0.
3. Since d(φ/y) = φ′dx/y − φ dy/y², the first group equals −d(φ/y).
4. So d(x − φ/y) = 0, and equivalently d(φ(x)/y − x) = 0, giving f = φ(x)/y − x.

Teacher's Note:
1. Look for the quotient rule pattern in the combination φ dy − yφ′ dx.
2. Dividing by the square of the denominator makes the expression an exact differential.

Answer: (c) \( \sqrt{3}\,e \)

Explanation:
1. The equation is homogeneous; put y = vx: x dv/dx = v/(1 + v²) − v = −v³/(1 + v²).
2. Separating: ∫(1 + v²)/v³ dv = −∫dx/x, so −1/(2v²) + ln v = −ln x + C, i.e. ln y = x²/(2y²) + C.
3. y(1) = 1 gives C = −1/2.
4. For y = e: 1 = x₀²/(2e²) − 1/2, so x₀² = 3e² and x₀ = √3 e.

Teacher's Note:
1. For homogeneous equations always substitute y = vx and separate variables.
2. Use the initial condition to fix the constant, then substitute the second point.