Differential Equations: JEE Advanced PYQ, 2007–2026

Answer: (C)

c₁ e^(x + c₂) = (c₁e^(c₂)) eˣ: one constant. sin²x = (1 − cos 2x)/2, so c₃ sin²x + c₄ cos 2x + c₅ = (c₄ − c₃/2) cos 2x + (c₅ + c₃/2): two constants. So y = a eˣ + b cos 2x + c with three independent constants, and the order is 3.

Answer: (A)

Separate the variables: dy/(1 + y^2) = dx/(1 + x^2). Integrating gives arctan y = arctan x + c. At (0, 1), arctan 1 = pi/4 = c. So arctan y = arctan x + pi/4, which gives y = tan(arctan x + pi/4) = (1 + x)/(1 - x) for x < 1. As x tends to -infinity, arctan x tends to -pi/2, so arctan y tends to -pi/4 and y tends to tan(-pi/4) = -1. The rational form agrees: (1 + x)/(1 - x) tends to -1.

Answer: (A)

Write dy/dx = e^x · e^y and separate: e^(−y) dy = e^x dx. Integrating gives −e^(−y) = e^x + C. Using y(0) = 0 gives −1 = 1 + C, so C = −2. Hence e^(−y) = 2 − e^x, and y = −ln(2 − e^x). This is defined only when 2 − e^x > 0, i.e. x < ln 2. As x → ln 2 from the left, y → +∞, so the solution blows up there.

Answer: (D)

Three arbitrary constants mean the order is 3. Differentiating (y − k)² = 4a(x − h) gives (y − k)y' = 2a. Differentiating again gives y'² + (y − k)y'' = 0. Differentiating a third time gives 3y'y'' + (y − k)y''' = 0. From the second relation y − k = −y'²/y''. Substituting: 3y'y'' − y'²y'''/y'' = 0, so 3(y'')² = y'y''' after dividing by y'. The highest-order derivative y''' appears to the first power, so the degree is 1 even though y'' is squared. The pair is (3, 1).

Answer: (A)

The tangent at (x, y) is Y − y = (dy/dx)(X − x). Putting Y = 0 gives the x-intercept X = x − y/(dy/dx). The condition says X = x/2, so y/(dy/dx) = x/2 and dy/dx = 2y/x. Separate: dy/y = 2 dx/x, so ln y = 2 ln x + c and y = Cx². Since y(1) = 3, C = 3 and y = 3x². Therefore y(2) = 12.

Answer: (A)

The equation is homogeneous of degree 0, so put y = vx. Then v + x dv/dx = (1 + v)/(1 − v), which gives x dv/dx = (1 + v²)/(1 − v). Separating variables, ∫(1 − v)/(1 + v²) dv = ∫dx/x, so tan⁻¹v − ½ ln(1 + v²) = ln x + C. Since −½ ln(1 + v²) = −½ ln(x² + y²) + ln x, the ln x terms cancel and tan⁻¹(y/x) = ½ ln(x² + y²) + C. The point (1, 0) gives C = 0. On y = x we have tan⁻¹1 = π/4, so π/4 = ½ ln(2x²). Hence 2x² = e^(π/2) and, for x > 0, x = e^(π/4)/√2.

Answer: (D)

The equation is dx/dy = −e^(x/y)(1 − x/y)/(1 + e^(x/y)), a function of x/y. So substitute x = vy, giving dx/dy = v + y dv/dy. Then v + y dv/dy = −e^v(1−v)/(1+e^v). Hence y dv/dy = −(e^v + v)/(1 + e^v). Separating variables gives ∫(1+e^v)/(e^v+v) dv = −∫dy/y. So ln(e^v + v) = −ln y + c, and y(e^v + v) = C. Replacing v by x/y gives x + y e^(x/y) = C.

Answer: (D)

The family has one parameter c, so the order is 1. Differentiating y² = 2c(x + √c) gives 2yy' = 2c, so c = yy'. Substituting: y² = 2yy'x + 2(yy')^{3/2}, so y² − 2xyy' = 2(yy')^{3/2}. Squaring removes the fractional power: (y² − 2xyy')² = 4y³y'³. The left side has y' to the power 2 at most and the right side has y'³, so the degree is 3. The pair is (1, 3).

Answer: (A)

The degree is defined once the equation is free of fractional powers of derivatives. Raise both sides to the power 6, the LCM of 2 and 3: ((y'')^(1/2))⁶ = ((y' + y)^(1/3))⁶, so (y'')³ = (y' + y)². This is polynomial in y' and y''. The highest-order derivative is y'', with power 3. So the order is 2 and the degree is 3.

Answer: (A)

Put y = vx. Then v + x dv/dx = v + tan v, so x dv/dx = tan v and ∫cot v dv = ∫dx/x. This gives ln|sin v| = ln x + c, so sin(y/x) = Cx. The point (1, π/6) gives C = sin(π/6) = 1/2, so sin(y/x) = x/2. At x = √2, sin(y/x) = √2/2, so y/x = π/4 on the branch in (0, π/2). Therefore y = √2 · π/4 = π√2/4.