Q5. Solve dy/dx = sin−1 x (NCERT Exercise 9.3, Q9, page 311).
y = ∫sin−1 x dx. By parts, with sin−1 x as the first function, this is x sin−1 x − ∫x/√(1 − x2) dx. So y = x sin−1 x + √(1 − x2) + C.
Q6. In a bank, principal increases continuously at r% per year. Find r if Rs 100 doubles in 10 years, taking loge 2 = 0.6931 (NCERT Exercise 9.3, Q20, page 312).
dP/dt = rP/100 gives P = 100ert/100. Doubling in 10 years means er/10 = 2, so r = 10 loge 2 = 6.931. The rate is about 6.93%.
Q7. Which of these is a homogeneous differential equation? (A) (4x + 6y + 5) dy − (3y + 2x + 4) dx = 0; (B) (xy) dx − (x3 + y3) dy = 0; (C) (x3 + 2y2) dx + 2xy dy = 0; (D) y2 dx + (x2 − xy − y2) dy = 0 (NCERT Exercise 9.4, Q17, page 322).
Only in (D) do all the terms have the same degree, 2. (A) has constants, and (B) and (C) mix degrees 2 and 3: (D).
Q8. Solve (x + y) dy + (x − y) dx = 0, given y = 1 when x = 1 (NCERT Exercise 9.4, Q11, page 321).
dy/dx = (y − x)/(x + y). With y = vx, x dv/dx = −(1 + v2)/(1 + v). Separating, (1 + v) dv/(1 + v2) = −dx/x, so tan−1 v + ½ log (1 + v2) = −log |x| + C. Doubling and putting v = y/x gives log (x2 + y2) + 2 tan−1 (y/x) = K. At (1, 1), K = log 2 + π/2. So log (x2 + y2) + 2 tan−1 (y/x) = π/2 + log 2.
Q9. Solve cos2 x dy/dx + y = tan x, for 0 ≤ x < π/2 (NCERT Exercise 9.5, Q5, page 328).
Divide by cos2 x: dy/dx + y sec2 x = tan x sec2 x. So I.F. = etan x, and yetan x = ∫tan x sec2 x etan x dx. With t = tan x this is ∫tet dt = (t − 1)et + C. So y = tan x − 1 + Ce−tan x.
Q10. Solve dy/dx + 2y tan x = sin x, given y = 0 when x = π/3 (NCERT Exercise 9.5, Q13, page 329).
I.F. = e2 log sec x = sec2 x. Then y sec2 x = ∫sin x sec2 x dx = ∫sec x tan x dx, which is sec x + C. So y = cos x + C cos2 x. At x = π/3: 0 = ½ + C/4, so C = −2, and y = cos x − 2 cos2 x.
Q11. Find the curve through (0, 2) for which the sum of the coordinates of any point exceeds the slope of the tangent there by 5 (NCERT Exercise 9.5, Q17, page 329, without its word “magnitude”: see the answer).
x + y = dy/dx + 5, so dy/dx − y = x − 5. I.F. = e−x, and ye−x = ∫(x − 5)e−x dx, which is (4 − x)e−x + C. So y = 4 − x + Cex, and (0, 2) gives C = −2: y = 4 − x − 2ex. NCERT slip (page 329): NCERT’s question says the sum exceeds “the magnitude of the slope”. Then no curve passes through (0, 2): there the magnitude would be 0 + 2 − 5 = −3. NCERT’s answer, y = 4 − x − 2ex, solves the equation without the magnitude, as above. In an exam, write dy/dx = x + y − 5 and give that answer.
Q12. The integrating factor of (1 − y2) dx/dy + yx = ay, for −1 < y < 1, is (A) 1/(y2 − 1); (B) 1/√(y2 − 1); (C) 1/(1 − y2); (D) 1/√(1 − y2) (NCERT Exercise 9.5, Q19, page 329).
Divide by 1 − y2: it is linear in x with P1 = y/(1 − y2). ∫P1 dy = −½ log (1 − y2), so I.F. = 1/√(1 − y2): (D).
Q13. Beyond the NCERT text (JEE): solve dy/dx = (x + y)2.
It is not separable, homogeneous or linear, but the substitution u = x + y helps: du/dx = 1 + dy/dx = 1 + u2. That separates: du/(1 + u2) = dx, so locally tan−1 (x + y) = x + C. Equivalently y = tan (x + C) − x on any interval between consecutive poles; the principal arctangent form uses the corresponding local branch.