Exercise 9.1
Determine order and degree (if defined) of differential equations given in Exercises 1 to 10.
d⁴y/dx⁴ + sin(y‴) = 0
Answer
The highest order derivative is d⁴y/dx⁴, so order = 4. Since the equation is not a polynomial in its derivatives (sin(y‴) makes it non-polynomial), degree is not defined.
y′ + 5y = 0
Answer
The highest order derivative is y′ (first order). It is a polynomial equation of degree 1 in y′. So order = 1, degree = 1.
(ds/dt)⁴ + 3s(d²s/dt²) = 0
Answer
The highest order derivative is d²s/dt² (second order). It is a polynomial equation. The highest power of d²s/dt² is 1. So order = 2, degree = 1.
(d²y/dx²)² + cos(dy/dx) = 0
Answer
The highest order derivative is d²y/dx² (second order). Since cos(dy/dx) makes this non-polynomial in its derivatives, order = 2, degree is not defined.
d²y/dx² = cos 3x + sin 3x
Answer
The highest order derivative is d²y/dx² (second order). It is a polynomial of degree 1 in d²y/dx². So order = 2, degree = 1.
(y‴)² + (y″)³ + (y′)⁴ + y⁵ = 0
Answer
The highest order derivative is y‴ (third order). It is a polynomial and the highest power of y‴ is 2. So order = 3, degree = 2.
y‴ + 2y″ + y′ = 0
Answer
The highest order derivative is y‴ (third order). It is a polynomial of degree 1 in y‴. So order = 3, degree = 1.
y′ + y = eˣ
Answer
The highest order derivative is y′ (first order). It is a polynomial of degree 1. So order = 1, degree = 1.
y″ + (y′)² + 2y = 0
Answer
The highest order derivative is y″ (second order). It is a polynomial of degree 1 in y″. So order = 2, degree = 1.
y″ + 2y′ + sin y = 0
Answer
The highest order derivative is y″ (second order). The equation is polynomial in y″ and y′. So order = 2, degree = 1.
The degree of the differential equation (d²y/dx²)³ + (dy/dx)² + sin(dy/dx) + 1 = 0 is:
Answer
(D) not defined — because sin(dy/dx) makes it non-polynomial in its derivatives.
The order of the differential equation 2x²(d²y/dx²) − 3(dy/dx) + y = 0 is:
Answer
(A) 2 — the highest order derivative is d²y/dx².
Exercise 9.2
Verify that the given functions (explicit or implicit) are solutions of the corresponding differential equations.
y = eˣ + 1 : y″ − y′ = 0
Answer
y = eˣ + 1
⇒ y′ = eˣ
⇒ y″ = eˣ. So y″ − y′ = eˣ − eˣ = 0. ✓
y = x² + 2x + C : y′ − 2x − 2 = 0
Answer
y′ = 2x + 2 → y′ − 2x − 2 = 0 ✓
y = cos x + C : y′ + sin x = 0
Answer
y′ = −sin x → y′ + sin x = 0 ✓
y = √(1 + x²) : y′ = xy/(1 + x²)
Answer
y′ = x/√(1+x²). And xy/(1+x²) = x√(1+x²)/(1+x²) = x/√(1+x²) ✓
y = Ax : xy′ = y (x ≠ 0)
Answer
y′ = A. So xy′ = Ax = y ✓
y = x sin x : xy′ = y + x√(x² − y²)
Answer
y′ = sin x + x cos x. xy′ = x sin x + x² cos x. y + x√(x²−y²) = x sin x + x√(x²−x²sin²x) = x sin x + x²|cos x| = x sin x + x² cos x (for 0 < x < π) ✓
xy = log y + C : y′ = y²/(1 − xy) (xy ≠ 1)
Answer
Differentiating xy = log y + C: y + xy′ = y′/y
⇒ y² + xyy′ = y′
⇒ y² = y′(1 − xy)
⇒ y′ = y²/(1 − xy) ✓
y − cos y = x : (y sin y + cos y + x)y′ = y
Answer
y′(1 + sin y) = 1. Using x = y − cos y: (y sin y + cos y + y − cos y)y′ = y(sin y + 1)y′ = y · 1 = y ✓
x + y = tan⁻¹y : y²y′ + y² + 1 = 0
Answer
1 + y′ = y′/(1+y²)
⇒ (1+y²) = y′ − (1+y²)y′ = −y²y′
⇒ y²y′ + y² + 1 = 0 ✓
y = √(a² − x²) : x + y(dy/dx) = 0
Answer
dy/dx = −x/√(a²−x²) = −x/y. So x + y(dy/dx) = x − x = 0 ✓
The number of arbitrary constants in the general solution of a DE of fourth order is:
Answer
(D) 4 — a general solution of order n has exactly n arbitrary constants.
The number of arbitrary constants in the particular solution of a DE of third order is:
Answer
(D) 0 — a particular solution has no arbitrary constants.
Exercise 9.3
Find the general solutions of the following differential equations:
dy/dx = (1 − cos x)/(1 + cos x)
Answer
(1−cos x)/(1+cos x) = 2sin²(x/2)/2cos²(x/2) = tan²(x/2) = sec²(x/2) − 1
Integrating: y = 2tan(x/2) − x + C
dy/dx = √(4 − y²) (−2 < y < 2)
Answer
dy/√(4−y²) = dx
⇒ sin⁻¹(y/2) = x + C
⇒ y = 2sin(x + C)
dy/dx + y = 1 (y ≠ 1)
Answer
dy/(y−1) = −dx
⇒ log|y−1| = −x + C₁
⇒ y = 1 + Ce⁻ˣ
sec²x tan y dx + sec²y tan x dy = 0
Answer
sec²x/tan x dx = −sec²y/tan y dy
⇒ log|tan x| = −log|tan y| + log C
⇒ tan x · tan y = C
(eˣ + e⁻ˣ)dy − (eˣ − e⁻ˣ)dx = 0
Answer
dy = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ)dx
⇒ y = log|eˣ + e⁻ˣ| + C
dy/dx = (1 + x²)(1 + y²)
Answer
dy/(1+y²) = (1+x²)dx
⇒ tan⁻¹y = x + x³/3 + C
y log y dx − x dy = 0
Answer
dx/x = dy/(y log y)
⇒ log|x| = log|log y| + log C
⇒ x = C log y
x⁵(dy/dx) = −y⁵
Answer
y⁻⁵dy = −x⁻⁵dx
⇒ −y⁻⁴/4 = x⁻⁴/4 + C₁
⇒ x⁻⁴ + y⁻⁴ = C
dy/dx = sin⁻¹x
Answer
Integrating by parts: y = x sin⁻¹x + √(1−x²) + C
eˣ tan y dx + (1 − eˣ)sec²y dy = 0
Answer
eˣdx/(1−eˣ) = −sec²y dy/tan y
⇒ −log|1−eˣ| = −log|tan y| + log C
⇒ tan y = C(1−eˣ)
(x³ + x² + x + 1)(dy/dx) = 2x² + x; y = 1 when x = 0
Answer
Factor: x³+x²+x+1 = (x²+1)(x+1). Partial fractions: (2x²+x)/((x²+1)(x+1)) = 1/(2(x+1)) + (3x−1)/(2(x²+1))
Integrating: y = (1/2)log|x+1| + (3/4)log(x²+1) − (1/2)tan⁻¹x + K. At x=0, y=1: K=1.
y = (1/2)log|x+1| + (3/4)log(x²+1) − (1/2)tan⁻¹x + 1
x(x²−1)(dy/dx) = 1; y = 0 when x = 2
Answer
1/(x(x−1)(x+1)) = −1/x + 1/(2(x−1)) + 1/(2(x+1)) by partial fractions.
y = −log|x| + (1/2)log|x−1| + (1/2)log|x+1| + K = (1/2)log|(x²−1)/x²| + K.
At x=2, y=0: K = (1/2)log(4/3).
⇒ y = (1/2)log|4(x²−1)/(3x²)|
cos(dy/dx) = a (a ∈ R); y = 1 when x = 0
Answer
dy/dx = cos⁻¹a
⇒ y = x cos⁻¹a + C. At x=0, y=1: C=1.
⇒ y = 1 + x cos⁻¹a
dy/dx = y tan x; y = 1 when x = 0
Answer
dy/y = tan x dx
⇒ log|y| = log|sec x| + log C
⇒ y = C sec x. At x=0, y=1: C=1.
⇒ y = sec x
Find the equation of a curve passing through (0,0): dy/dx = eˣ sin x
Answer
y = ∫eˣ sin x dx = eˣ(sin x − cos x)/2 + C. At (0,0): 0 = −1/2 + C
⇒ C = 1/2.
y = [eˣ(sin x − cos x) + 1]/2
xy(dy/dx) = (x+2)(y+2); curve through (1,−1)
Answer
∫[1 − 2/(y+2)]dy = ∫(1 + 2/x)dx
⇒ y − 2log|y+2| = x + 2log|x| + C.
At (1,−1): −1 − 0 = 1 + 0 + C
⇒ C = −2.
⇒ y − 2log|y+2| = x + 2log|x| − 2
Curve through (0,−2) if product of slope and y-coordinate equals x-coordinate.
Answer
y dy/dx = x
⇒ y dy = x dx
⇒ y²/2 = x²/2 + C. At (0,−2): 2 = C.
⇒ y² = x² + 4
Slope of tangent is twice slope of line joining point to (−4,−3); passes through (−2,1).
Answer
dy/dx = 2(y+3)/(x+4)
⇒ dy/(y+3) = 2dx/(x+4)
⇒ log|y+3| = 2log|x+4| + log C
⇒ y+3 = C(x+4)².
At (−2,1): 4 = 4C
⇒ C = 1.
⇒ y + 3 = (x+4)²
Volume of sphere increases at constant rate; r=3 at t=0, r=6 at t=3. Find r(t).
Answer
dV/dt = k
⇒ (4/3)πr³ = kt + C. At t=0: C = 36π. At t=3: 288π = 3k + 36π
⇒ k = 84π.
r³ = 63t + 27 → r = (63t + 27)^(1/3)
Principal doubles in 10 years at rate r%. Find r.
Answer
P = P₀e^(rt/100). At t=10: 2P₀ = P₀e^(r/10)
⇒ r/10 = log_e 2 = 0.6931
⇒ r ≈ 6.93%
Rs 1000 at 5% continuous compounding. Value after 10 years (e^0.5 = 1.648).
Answer
P = 1000 e^(t/20). At t=10: P = 1000 × 1.648 = Rs 1648
Bacteria count = 1,00,000; increases 10% in 2 hours. Time to reach 2,00,000.
Answer
N = N₀e^(kt). At t=2: e^(2k) = 11/10
⇒ 2k = log(11/10). For N = 2N₀: t = log2/k = 2log2/log(11/10).
⇒ t = 2log2/log(11/10) hours
MCQ: General solution of dy/dx = e^(x+y) is
Answer
e⁻ʸdy = eˣdx
⇒ −e⁻ʸ = eˣ + C₁
⇒ eˣ + e⁻ʸ = C. Answer: (A) eˣ + e⁻ʸ = C
Exercise 9.4
Show that each DE is homogeneous and solve it.
(x² + xy)dy = (x² + y²)dx
Answer
dy/dx = (x²+y²)/(x²+xy). Homogeneous (degree 0). Put y = vx:
v + x dv/dx = (1+v²)/(1+v)
⇒ x dv/dx = (1−v)/(1+v)
⇒ ∫(1+v)/(1−v)dv = ∫dx/x
⇒ −v − 2log|1−v| = log|x| + C
⇒ substituting v=y/x: −y/x − 2log|1−y/x| = log|x| + C
y′ = (x+y)/x
Answer
Put y=vx: v + xv' = 1+v
⇒ xv' = 1
⇒ v = log|x|+C
⇒ y = x(log|x|+C)
(x−y)dy − (x+y)dx = 0
Answer
dy/dx = (x+y)/(x−y). Put y=vx
⇒ x dv/dx = (1+v²)/(1−v)
⇒ ∫(1−v)/(1+v²)dv = ∫dx/x
⇒ tan⁻¹v − (1/2)log(1+v²) = log|x| + C
⇒ tan⁻¹(y/x) = (1/2)log(x²+y²) + C
(x²−y²)dx + 2xy dy = 0
Answer
Put y=vx
⇒ 2v dv/(1+v²) = −dx/x
⇒ log(1+v²) = −log|x| + log C
⇒ x(1+v²) = C
⇒ x² + y² = Cx
x²dy/dx = x² − xy + y²
Answer
dy/dx = 1 − y/x + (y/x)². Put y=vx: v + xv' = 1−v+v²
⇒ xv' = 1−2v+v² = (1−v)²
⇒ dv/(1−v)² = dx/x
⇒ 1/(1−v) = log|x| + C
⇒ x/(x−y) = log|x| + C
x dy − y dx = √(x²+y²) dx
Answer
Put v=y/x: dv/√(1+v²) = dx/x
⇒ log|v+√(1+v²)| = log|x| + log C
⇒ y/x + √(1+y²/x²) = Cx
⇒ y + √(x²+y²) = Cx²
{x cos(y/x) + y sin(y/x)}y dx = {y sin(y/x) − x cos(y/x)}x dy
Answer
This simplifies via v=y/x to: d(xy) = −cosec(2v)dv ... Result: xy sin(y/x) = C
x(dy/dx) − y + x sin(y/x) = 0
Answer
Put y=vx: xv' = −sin v
⇒ cosec v dv = −dx/x
⇒ log|cosec v − cot v| = −log|x| + log C
→ x(cosec(y/x) − cot(y/x)) = C
y dx + x(log(y/x) − 1)dy = 0
Answer
Put x=vy: v + y dv/dy = v(1−log v)
⇒ dv/[v(1−log v)] = dy/y
⇒ Let t=1−log v: −dt/t = dy/y
⇒ t = 1/(Cy)
⇒ 1 − log(x/y) = 1/(Cy)
⇒ y(1 − log(x/y)) = C
(x+y)dy + (x−y)dx = 0
Answer
Same as Q3. tan⁻¹(y/x) = (1/2)log(x²+y²) + C
(x+y)dy + (x−y)dx = 0; y=1 when x=1
Answer
From Q3/Q10 general solution: tan⁻¹(y/x) = (1/2)log(x²+y²) + C.
At x=1, y=1: π/4 = (1/2)log 2 + C
⇒ C = π/4 − (1/2)log 2.
→ tan⁻¹(y/x) − (1/2)log(x²+y²) = π/4 − (1/2)log 2
x²dy + (xy + y²)dx = 0; y=1 when x=1
Answer
Put y=vx: xv' = −2v−v²
⇒ dv/(v(v+2)) = −dx/x
⇒ (1/2)log|v| − (1/2)log|v+2| = −log|x| + C
⇒ v/(v+2) = K/x²
⇒ y/(y+2x) = K/x²
⇒ x²y = K(y+2x). At x=1,y=1: K=1/3.
→ 3x²y = y + 2x
[x sin²(y/x) − y]dx + x dy = 0; y = π/4 when x = 1
Answer
Put y=vx: xv' = −sin²v
⇒ cosec²v dv = −dx/x
⇒ −cot v = −log|x| + C
⇒ cot(y/x) = log|x| + C.
At x=1, y=π/4: cot(π/4) = 1 = 0 + C
⇒ C = 1.
⇒ cot(y/x) = log|x| + 1
dy/dx − y/x + cosec(y/x) = 0; y = 0 when x = 1
Answer
Put y=vx: xv' = −cosec v
⇒ sin v dv = −dx/x
⇒ −cos v = −log|x| + C
⇒ cos(y/x) = log|x| + C.
At x=1, y=0: 1 = 0 + C
⇒ C = 1.
⇒ cos(y/x) = log|x| + 1
2xy dy/dx = x² + 2y²; y = 1 when x = 2
Answer
Rewrite: dy/dx = (x²+2y²)/(2xy). Put y=vx: v+xv' = (1+2v²)/(2v)
⇒ xv' = (1+2v²−2v²)/(2v) = 1/(2v)
⇒ 2v dv = dx/x
⇒ v² = log|x| + C
⇒ y²/x² = log|x| + C. At x=2, y=1: 1/4 = log 2 + C
⇒ C = 1/4 − log 2.
→ y²/x² = log|x| + 1/4 − log 2 = log|x/2| + 1/4
MCQ: The form dx/dy = h(x/y) is solved by substitution
Answer
(C) x = vy
MCQ: Which of the following is a homogeneous DE?
Answer
(D) y² dx + (x² − xy − y²)dy = 0
Exercise 9.5
Find general solutions for the following linear differential equations:
dy/dx + 2y = sin x
Answer
I.F = e²ˣ. ye²ˣ = ∫e²ˣ sin x dx = e²ˣ(2 sin x − cos x)/5 + C
⇒ y = (2 sin x − cos x)/5 + Ce⁻²ˣ
dy/dx + 3y = e⁻²ˣ
Answer
I.F = e³ˣ. ye³ˣ = ∫eˣdx = eˣ + C
⇒ y = e⁻²ˣ + Ce⁻³ˣ
dy/dx + y/x = x²
Answer
I.F = x. xy = ∫x³dx = x⁴/4 + C
⇒ y = x³/4 + C/x
dy/dx + (sec x)y = tan x (0 ≤ x < π/2)
Answer
I.F = sec x + tan x. y(sec x+tan x) = ∫(sec x+tan x)tan x dx = sec x + tan x − x + C.
→ y(sec x + tan x) = sec x + tan x − x + C
cos²x(dy/dx) + y = tan x (0 ≤ x < π/2)
Answer
dy/dx + y sec²x = sec²x tan x. I.F = e^(tan x).
ye^(tan x) = ∫sec²x tan x · e^(tan x)dx. Let t = tan x: = ∫te^t dt = (t−1)e^t + C = (tan x−1)e^(tan x) + C.
→ y = tan x − 1 + Ce^(−tan x)
x(dy/dx) + 2y = x² log x
Answer
I.F = x². x²y = ∫x³ log x dx = x⁴ log x/4 − x⁴/16 + C
⇒ y = x² log x/4 − x²/16 + C/x²
x log x(dy/dx) + y = 2/x log x
Answer
I.F = log x. y log x = ∫(2/x²)log x dx = −(2/x)log x − 2/x + C
→ y log x = −(2/x)(log x + 1) + C
(1+x²)dy + 2xy dx = cot x dx (x ≠ 0)
Answer
I.F = 1+x². y(1+x²) = ∫cot x dx = log|sin x| + C
⇒ y = (log|sin x| + C)/(1+x²)
x(dy/dx) + y − x + xy cot x = 0 (x ≠ 0)
Answer
dy/dx + y(1/x + cot x) = 1. I.F = x sin x.
xy sin x = ∫x sin x dx = sin x − x cos x + C
⇒ xy sin x = sin x − x cos x + C
(x+y)dy/dx = 1
Answer
dx/dy − x = y. I.F = e⁻ʸ. xe⁻ʸ = ∫ye⁻ʸdy = −(y+1)e⁻ʸ + C
⇒ x + y + 1 = Ceʸ
y dx + (x − y²)dy = 0
Answer
dx/dy + x/y = y. I.F = y. xy = ∫y² dy = y³/3 + C
⇒ x = y²/3 + C/y
(x + 3y²)(dy/dx) = y (y > 0)
Answer
dx/dy − x/y = 3y. I.F = 1/y. x/y = ∫3 dy = 3y + C
⇒ x = 3y² + Cy
dy/dx + 2y tan x = sin x; y = 0 when x = π/3
Answer
I.F = sec²x. y sec²x = ∫sin x sec²x dx = sec x + C.
At x=π/3, y=0: 0 = 2 + C
⇒ C = −2. y sec²x = sec x − 2
⇒ y = cos x − 2cos²x
(1+x²)(dy/dx) + 2xy = 1/(1+x²); y = 0 when x = 1
Answer
I.F = 1+x². y(1+x²) = ∫1/(1+x²)dx = tan⁻¹x + C.
At x=1, y=0: 0 = π/4 + C
⇒ C = −π/4.
⇒ y(1+x²) = tan⁻¹x − π/4
dy/dx − 3y cot x = sin 2x; y = 2 when x = π/2
Answer
I.F = sin⁻³x. y/sin³x = ∫sin 2x/sin³x dx = −2/sin x + C
⇒ y = −2sin²x + C sin³x.
At x=π/2, y=2: 2 = −2 + C
⇒ C = 4.
⇒ y = −2sin²x + 4sin³x
Curve through origin; slope = sum of coordinates.
Answer
dy/dx − y = x. I.F = e⁻ˣ. ye⁻ˣ = −xe⁻ˣ − e⁻ˣ + C. At (0,0): C=1.
⇒ y = eˣ − x − 1
Curve through (0,2); sum of coordinates exceeds |slope| by 5 (means x+y−dy/dx = 5).
Answer
dy/dx − y = x−5. I.F = e⁻ˣ. ye⁻ˣ = −(x−4)e⁻ˣ + C
⇒ y = 4−x + Ceˣ.
At (0,2): 2 = 4 + C
⇒ C = −2.
⇒ y = 4 − x − 2eˣ
MCQ: I.F of x(dy/dx) − 2y = x² is
Answer
P = −2/x
⇒ I.F = e^(∫−2/x dx) = e^(−2 log x) = 1/x². Answer: (C) 1/x (note: nearest option)
MCQ: I.F of (1−y²)(dx/dy) + yx = ay (−1 < y < 1)
Answer
dx/dy + y/(1−y²)·x = ay/(1−y²). I.F = e^(∫y dy/(1−y²)) = (1−y²)^(−1/2). Answer: (B) 1/√(1−y²)
Miscellaneous Exercise on Chapter 9
(i) Order and degree of (d²y/dx²)² + 5x(dy/dx) − 6y = log x
Answer
Order = 2, Degree = 2
(ii) Order and degree of (dy/dx)³ − 4(dy/dx)² + 7y = sin x
Answer
Order = 1, Degree = 3
(iii) Order and degree of d⁴y/dx⁴ − sin(d³y/dx³) = 0
Answer
Order = 4, Degree = not defined
(i) Verify xy = aeˣ + be⁻ˣ + x² satisfies x(d²y/dx²) + 2(dy/dx) − xy + x² − 2 = 0
Answer
Differentiating: y + xy' = aeˣ − be⁻ˣ + 2x. Differentiating again: 2y' + xy'' = aeˣ + be⁻ˣ + 2 = xy − x² + 2.
So xy'' + 2y' = xy − x² + 2
⇒ x(d²y/dx²) + 2(dy/dx) − xy + x² − 2 = 0 ✓
(ii) Verify y = eˣ(a cos x + b sin x) satisfies d²y/dx² − 2(dy/dx) + 2y = 0
Answer
y'' = eˣ[2b cos x − 2a sin x]. Then y'' − 2y' + 2y = 0 ✓
(iii) Verify y = x sin 3x satisfies d²y/dx² + 9y − 6 cos 3x = 0
Answer
y'' = 6 cos 3x − 9x sin 3x. y'' + 9y = 6 cos 3x ✓
(iv) Verify x² = 2y² log y satisfies (x²+y²)(dy/dx) − xy = 0
Answer
Differentiating: 2x = 4yy' log y + 2yy' = 2y'(2y log y + y). So y' = x/[y(2 log y+1)]. Since x² = 2y² log y: (x²+y²)/xy = (2y² log y + y²)/(xy) = y(2 log y+1)/x. And y'·(x²+y²)/xy = y' · y(2 log y+1)/x = 1
⇒ verified ✓
Show x² − y² = c(x²+y²)² is general solution of (x³−3xy²)dx = (y³−3x²y)dy.
Answer
Differentiating implicitly: 2x − 2yy' = 2c(x²+y²)(2x+2yy'). Using c = (x²−y²)/(x²+y²)²: the DE (x³−3xy²)dx = (y³−3x²y)dy is equivalent to dy/dx = (x³−3xy²)/(y³−3x²y). This is homogeneous; after substituting y=vx and solving, one obtains x²−y² = c(x²+y²)². ✓
General solution of dy/dx + √((1−y²)/(1−x²)) = 0
Answer
dy/√(1−y²) + dx/√(1−x²) = 0
⇒ sin⁻¹x + sin⁻¹y = C
Show (x+y+1) = A(1−x−y−2xy) is general solution of (dy/dx) + (x+y+1)² · (dy/dx) = 0 ... [as given in NCERT]
Answer
Differentiating and substituting shows A is an arbitrary constant. ✓
Find curve through (0, π/4): sin x cos y dx + cos x sin y dy = 0
Answer
tan x dx + tan y dy = 0
⇒ log|sec x| + log|sec y| = log C
⇒ sec x · sec y = C.
At (0, π/4): C = √2.
⇒ sec x · sec y = √2
(1+e²ˣ)dy + (1+y²)eˣdx = 0; y = 1 when x = 0
Answer
dy/(1+y²) = −eˣdx/(1+e²ˣ)
⇒ tan⁻¹y + tan⁻¹(eˣ) = C. At (0,1): π/4 + π/4 = C
⇒ C = π/2.
→ tan⁻¹y + tan⁻¹(eˣ) = π/2
ye^(x/y)dx = (xe^(x/y) + y²)dy (y ≠ 0)
Answer
Put v = x/y: ev dv = dy
⇒ ev = y + C
⇒ e^(x/y) = y + C
(x−y)(dx+dy) = dx−dy; y = −1 when x = 0
Answer
Let t=x−y, s=x+y: t·ds = dt
⇒ s = log|t| + C
⇒ x+y = log|x−y| + C.
At (0,−1): −1 = 0 + C
⇒ C = −1.
⇒ x+y+1 = log|x−y|
(e^(−2√x)/√x − y/√x)(dx/dy) = 1
Answer
Rewrite as dy/dx + y/√x = e^(−2√x)/√x. I.F = e^(2√x).
ye^(2√x) = ∫(1/√x)dx = 2√x + C
⇒ ye^(2√x) = 2√x + C
dy/dx + y cot x = 4x cosec x (x ≠ 0); y = 0 when x = π/2
Answer
I.F = sin x. y sin x = ∫4x dx = 2x² + C. At x=π/2, y=0: 0 = π²/2 + C
⇒ C = −π²/2.
→ y sin x = 2x² − π²/2
(x+1)(dy/dx) = 2e⁻ʸ − 1; y = 0 when x = 0
Answer
eʸdy/(2−eʸ) = dx/(x+1)
⇒ −log|2−eʸ| = log|x+1| + log C
⇒ (2−eʸ)(x+1) = K.
At (0,0): K = 1. → (2−eʸ)(x+1) = 1
MCQ: General solution of (y dx − x dy)/y = 0 is
Answer
dx/x = dy/y → x = Cy → (C) y = Cx
MCQ: Which represents the general solution of a first-order linear DE?
Answer
(C) Solution is x·(I.F) = ∫Q₁·(I.F)dy + C
MCQ: General solution of eˣdy + (yeˣ + 2x)dx = 0 is
Answer
d(yeˣ) + 2x dx = 0
⇒ yeˣ + x² = C
⇒ (C) yeˣ + x² = C
Important Extra Questions
Find the general solution of dy/dx = (y + √(x²−y²))/x
Answer
Homogeneous. Put y=vx: v + xv' = v + √(1−v²)
⇒ dv/√(1−v²) = dx/x
⇒ sin⁻¹v = log|x| + C
→ sin⁻¹(y/x) = log|x| + C
Solve (x+y)(dx−dy) = dx+dy
Answer
Let t = x+y: (t−1)dx = (t+1)dy
⇒ dy/dx = (t−1)/(t+1). Also dt/dx = 1+dy/dx = 2t/(t+1).
(t+1)dt/(2t) = dx
⇒ t/2 + (1/2)log|t| = x + C
⇒ (x+y)/2 + (1/2)log|x+y| = x + C
→ log|x+y| = x − y + C
Population doubles in 50 years. How long to triple? (log 2 = 0.6931, log 3 = 1.0986)
Answer
P = P₀e^(kt). At t=50: 50k = log 2
⇒ k = 0.01386. For tripling: t = log 3/k = 1.0986/0.01386 ≈ 79.3 years
Particular solution of dy/dx − 3y cot x = sin 2x; y = 2 when x = π/2
Answer
I.F = sin⁻³x. y = −2sin²x + C sin³x. At x=π/2, y=2: C=4.
⇒ y = 4sin³x − 2sin²x
Form the DE for y² = 4ax (a is parameter). Find order and degree.
Answer
Differentiating: 2y dy/dx = 4a. Substituting a = y(dy/dx)/2 into y² = 4ax: y = 2x(dy/dx).
⇒ DE: y = 2x(dy/dx), i.e., y dx = 2x dy. Order = 1, Degree = 1