Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
An equation that contains the derivatives of an unknown function (the dependent variable) with respect to one or more independent variables. We study equations with one independent variable, called ordinary differential equations.
The order is the order of the highest order derivative present.
| Equation | Highest derivative | Order |
|---|---|---|
| dy/dx − cos x = 0 | dy/dx | 1 |
| xy d²y/dx² + x(dy/dx)² − y dy/dx = 0 | d²y/dx² | 2 |
| y‴ + y² + e^(y′) = 0 | y‴ | 3 |
The degree is defined only when the equation is a polynomial in its derivatives. It is the power of the highest order derivative, after the equation is free of radicals and fractions in the derivatives.
Functions of y or x alone, such as sin y or eˣ, do not affect the degree. Functions of a derivative, such as sin(y′), do.
A function y = φ(x) is a solution if the equation becomes an identity when φ and its derivatives are substituted. Its graph is a solution curve.
Differentiate the given function as many times as needed, substitute in the equation and check that the left side equals the right side.
Given a family with n arbitrary constants, differentiate n times and eliminate the constants.
Solve dy/dx = −x/y.
F(x, y) is homogeneous of degree n if F(λx, λy) = λⁿF(x, y). The equation dy/dx = F(x, y) is homogeneous if F is homogeneous of degree 0, that is F can be written as a function of y/x. For M dx + N dy = 0, M and N must be homogeneous of the same degree.
Solve (x² − y²) dx + 2xy dy = 0.
Solve dy/dx + y/x = x² (x > 0).
| Type | Test | Method |
|---|---|---|
| Variables separable | dy/dx = f(x)g(y) | dy/g(y) = f(x) dx, integrate |
| Homogeneous | dy/dx = F(y/x) | y = vx (or x = vy), then separate |
| Linear in y | dy/dx + Py = Q | I.F. = e^(∫P dx) |
| Linear in x | dx/dy + P₁x = Q₁ | I.F. = e^(∫P₁ dy) |
| Order, degree | highest derivative; its power | degree needs a polynomial equation |
Try each problem first, then open the solution to check your working.
Find the order and degree of (d²y/dx²)³ + 2(dy/dx)⁴ + y = x.
The highest derivative is d²y/dx², so the order is 2. Its power is 3, so the degree is 3. (The power 4 of dy/dx does not matter.)
Answer: order 2, degree 3
Find the order and degree (if defined) of y‴ + (y″)² + cos(y′) = 0.
The highest derivative is y‴, so the order is 3.
cos(y′) is not a polynomial in y′, so the degree is not defined.
Answer: order 3, degree not defined
Show that y = Ae^(−x) + Be^(−2x) is a solution of y″ + 3y′ + 2y = 0.
y′ = −Ae^(−x) − 2Be^(−2x) and y″ = Ae^(−x) + 4Be^(−2x).
y″ + 3y′ + 2y = (A − 3A + 2A)e^(−x) + (4B − 6B + 2B)e^(−2x) = 0. ✓
Answer: Verified
Form the differential equation of the family y = Ae^(3x).
One constant, so differentiate once: y′ = 3Ae^(3x) = 3y.
Answer: dy/dx = 3y
Form the differential equation of the family y = A cos 2x + B sin 2x.
Two constants, so differentiate twice: y′ = −2A sin 2x + 2B cos 2x, y″ = −4A cos 2x − 4B sin 2x = −4y.
Answer: y″ + 4y = 0
Find the general solution of dy/dx = 2xy².
dy/y² = 2x dx. Integrate: −1/y = x² + C₁, so 1/y = C − x² where C = −C₁.
Answer: 1/y = C − x²
Find the general solution of dy/dx = (1 + y²)eˣ.
dy/(1 + y²) = eˣ dx. Integrate: tan⁻¹y = eˣ + C.
Answer: tan⁻¹ y = eˣ + C
Find the general solution of y dy/dx = x + 1.
y dy = (x + 1) dx. Integrate: y²/2 = (x + 1)²/2 + C₁, so y² = (x + 1)² + C.
Answer: y² = (x + 1)² + C
Find the particular solution of dy/dx = y cos x given y = 1 when x = 0.
dy/y = cos x dx, so log|y| = sin x + c and y = Ce^(sin x). At x = 0, y = 1: C = 1.
Answer: y = e^(sin x)
Find the particular solution of dy/dx = 2x(1 + y²) given y = 1 when x = 0.
dy/(1 + y²) = 2x dx, so tan⁻¹y = x² + C. At (0, 1): π/4 = C.
So tan⁻¹y = x² + π/4, that is y = tan(x² + π/4).
Answer: y = tan(x² + π/4)
Solve (x² + y²) dx = xy dy.
Use: y = vx
dy/dx = (x² + y²)/(xy) = (1 + v²)/v, so v + x dv/dx = (1 + v²)/v and x dv/dx = 1/v.
v dv = dx/x, so v²/2 = log|x| + c, that is y²/x² = 2 log|x| + C.
Hence y² = 2x² log|x| + Cx².
Answer: y² = 2x² log|x| + Cx²
Solve (x + y) dy = (x − y) dx.
dy/dx = (1 − v)/(1 + v) with y = vx, so x dv/dx = (1 − 2v − v²)/(1 + v).
(1 + v)/(1 − 2v − v²) dv = dx/x. The numerator is −(1/2) times the derivative of the denominator.
−(1/2) log|1 − 2v − v²| = log|x| + c, so x²(1 − 2v − v²) = C, that is x² − 2xy − y² = C.
Answer: x² − 2xy − y² = C
Solve x dy/dx = y + x e^(y/x).
dy/dx = y/x + e^(y/x). With y = vx: x dv/dx = e^v, so e^(−v) dv = dx/x.
−e^(−v) = log|x| + c, that is e^(−y/x) + log|x| = C.
Answer: e^(−y/x) + log|x| = C
Find the particular solution of (x² − y²) dx + 2xy dy = 0 given y = 1 when x = 1.
From the general solution x² + y² = Cx, put x = 1, y = 1: 2 = C.
So x² + y² = 2x.
Answer: x² + y² = 2x
Solve dy/dx + 3y/x = x (x > 0).
Use: I.F. = e^(∫P dx)
I.F. = e^(3 log x) = x³. x³y = ∫x⁴ dx = x⁵/5 + C.
So y = x²/5 + C/x³.
Answer: y = x²/5 + C/x³
Solve dy/dx + y cot x = 2 cos x (0 < x < π).
I.F. = e^(∫cot x dx) = sin x. y sin x = ∫2 sin x cos x dx = sin²x + C.
So y = sin x + C/sin x.
Answer: y = sin x + C/sin x
Solve y dx + (x − y³) dy = 0 (y > 0).
Write dx/dy + x/y = y². P₁ = 1/y, so I.F. = e^(log y) = y.
xy = ∫y³ dy = y⁴/4 + C.
Answer: xy = y⁴/4 + C
Find the particular solution of dy/dx − 2y = e^(3x) given y = 0 when x = 0.
I.F. = e^(−2x). y e^(−2x) = ∫eˣ dx = eˣ + C. At (0, 0): 0 = 1 + C, so C = −1.
y = e^(3x) − e^(2x).
Answer: y = e^(3x) − e^(2x)
A population grows at a rate proportional to its size. It doubles in 5 years. In how many years does it triple?
Use: dP/dt = kP
dP/P = k dt gives P = P₀e^(kt). Doubling in 5 years: e^(5k) = 2, so k = (log 2)/5.
Tripling: e^(kt) = 3, so t = (log 3)/k = 5 log 3/log 2 ≈ 7.92 years.
Answer: 5 log 3 / log 2 ≈ 7.92 years
A radioactive substance decays at a rate proportional to the amount present. If 20% decays in 10 days, find its half-life.
Use: dN/dt = −kN
N = N₀e^(−kt). After 10 days 80% remains: e^(−10k) = 0.8, so k = (log 1.25)/10.
Half-life: e^(−kt) = 1/2, so t = (log 2)/k = 10 log 2/log 1.25 ≈ 31.06 days.
Answer: 10 log 2 / log 1.25 ≈ 31.06 days
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
The highest derivative is d⁴y/dx⁴, so the order is 4.
The equation contains sin(y‴), which is not a polynomial in y‴, so it is not a polynomial equation in the derivatives.
Answer: order 4, degree not defined
The highest derivative is y′, so the order is 1. It is a polynomial in y′ with y′ to the power 1, so the degree is 1.
Answer: order 1, degree 1
The highest derivative is d²s/dt², so the order is 2.
The power of d²s/dt² is 1 (the (ds/dt)⁴ term does not matter), so the degree is 1.
Answer: order 2, degree 1
The order is 2.
The term cos(dy/dx) is not a polynomial in dy/dx, so the degree is not defined.
Answer: order 2, degree not defined
The order is 2 and d²y/dx² appears to the first power, so the degree is 1.
Answer: order 2, degree 1
The highest derivative is y‴, so the order is 3.
The power of y‴ is 2, so the degree is 2.
Answer: order 3, degree 2
The order is 3 and the power of y‴ is 1, so the degree is 1.
Answer: order 3, degree 1
The order is 1. The equation is a polynomial in y′ of power 1, so the degree is 1.
Answer: order 1, degree 1
The order is 2. The highest derivative y″ has power 1, so the degree is 1.
Answer: order 2, degree 1
The order is 2. sin y is a function of y, not of a derivative, so the equation is still a polynomial in the derivatives and the degree is 1.
Answer: order 2, degree 1
The term sin(dy/dx) is not a polynomial in dy/dx, so the equation is not a polynomial in its derivatives.
Therefore the degree is not defined.
Answer: (D) not defined
The highest derivative is d²y/dx², so the order is 2.
Answer: (A) 2
y′ = eˣ and y″ = eˣ.
L.H.S. = y″ − y′ = eˣ − eˣ = 0 = R.H.S.
Answer: Verified: the left side equals the right side, so the given function is a solution.
y′ = 2x + 2.
L.H.S. = (2x + 2) − 2x − 2 = 0 = R.H.S.
Answer: Verified: the left side equals the right side, so the given function is a solution.
y′ = −sin x.
L.H.S. = −sin x + sin x = 0 = R.H.S.
Answer: Verified: the left side equals the right side, so the given function is a solution.
y′ = 2x/(2√(1 + x²)) = x/√(1 + x²).
R.H.S. = x√(1 + x²)/(1 + x²) = x/√(1 + x²) = L.H.S.
Answer: Verified: the left side equals the right side, so the given function is a solution.
y′ = A, so xy′ = Ax = y.
Answer: Verified: the left side equals the right side, so the given function is a solution.
y′ = sin x + x cos x, so xy′ = x sin x + x² cos x = y + x² cos x.
x² − y² = x²(1 − sin²x) = x² cos²x, so x√(x² − y²) = x² cos x (taking cos x > 0 in the stated region).
Hence xy′ = y + x√(x² − y²).
Answer: Verified: the left side equals the right side, so the given function is a solution.
Differentiate xy = log y + C: y + xy′ = y′/y.
So y′(1/y − x) = y, that is y′(1 − xy)/y = y, giving y′ = y²/(1 − xy).
Answer: Verified: the left side equals the right side, so the given function is a solution.
Differentiate: y′ + sin y · y′ = 1, so y′ = 1/(1 + sin y).
From the relation, x = y − cos y, so y sin y + cos y + x = y sin y + cos y + y − cos y = y(1 + sin y).
L.H.S. = y(1 + sin y) × 1/(1 + sin y) = y = R.H.S.
Answer: Verified: the left side equals the right side, so the given function is a solution.
Differentiate: 1 + y′ = y′/(1 + y²), so y′(1/(1 + y²) − 1) = 1.
That is y′ · (−y²/(1 + y²)) = 1, so y′ = −(1 + y²)/y².
Hence y²y′ + y² + 1 = 0.
Answer: Verified: the left side equals the right side, so the given function is a solution.
y′ = −x/√(a² − x²) = −x/y.
So y dy/dx = −x, giving x + y dy/dx = 0.
Answer: Verified: the left side equals the right side, so the given function is a solution.
The general solution of an nth-order equation has n independent arbitrary constants. Here n = 4.
Answer: (D) 4
A particular solution has all its constants fixed by the given conditions, so it has no arbitrary constants.
Answer: (D) 0
(1 − cos x)/(1 + cos x) = 2 sin²(x/2) / 2 cos²(x/2) = tan²(x/2) = sec²(x/2) − 1.
So dy = (sec²(x/2) − 1) dx, and integrating, y = 2 tan(x/2) − x + C.
Answer: y = 2 tan(x/2) − x + C
Separate the variables: dy/√(4 − y²) = dx.
Integrate: sin⁻¹(y/2) = x + C, so y = 2 sin(x + C).
Answer: sin⁻¹(y/2) = x + C, that is y = 2 sin(x + C)
dy/dx = 1 − y, so dy/(1 − y) = dx.
Integrate: −log|1 − y| = x + C₁, so 1 − y = e^(−x − C₁) = A e^(−x), where A = ±e^(−C₁).
Hence y = 1 − A e^(−x). Writing C for −A gives y = 1 + C e^(−x).
Answer: y = 1 + C e^(−x)
Divide by tan x tan y: (sec²x/tan x) dx + (sec²y/tan y) dy = 0.
Integrate: log|tan x| + log|tan y| = log C, so tan x tan y = C.
Answer: tan x tan y = C
dy = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ) dx. The numerator is the derivative of the denominator.
Integrate: y = log(eˣ + e⁻ˣ) + C.
Answer: y = log(eˣ + e^(−x)) + C
dy/(1 + y²) = (1 + x²) dx.
Integrate: tan⁻¹y = x + x³/3 + C.
Answer: tan⁻¹ y = x + x³/3 + C
Separate: dy/(y log y) = dx/x.
Integrate: log|log y| = log|x| + log C, so log y = Cx.
Hence y = e^(Cx).
Answer: log y = Cx, that is y = e^(Cx)
Separate: dy/y⁵ = −dx/x⁵.
Integrate: −1/(4y⁴) = 1/(4x⁴) + C₁.
Multiply by −4: 1/x⁴ + 1/y⁴ = C, where C = −4C₁.
Answer: 1/x⁴ + 1/y⁴ = C
y = ∫ sin⁻¹x dx. Integrate by parts with u = sin⁻¹x, dv = dx: x sin⁻¹x − ∫ x/√(1 − x²) dx.
∫ x/√(1 − x²) dx = −√(1 − x²), so y = x sin⁻¹x + √(1 − x²) + C.
Answer: y = x sin⁻¹ x + √(1 − x²) + C
Separate: sec²y/tan y dy = −eˣ/(1 − eˣ) dx.
Integrate: log|tan y| = log|1 − eˣ| + log C, so tan y = C(1 − eˣ).
Answer: tan y = C(1 − eˣ)
x³ + x² + x + 1 = (x + 1)(x² + 1). Split into partial fractions: (2x² + x)/((x + 1)(x² + 1)) = (1/2)/(x + 1) + ((3/2)x − 1/2)/(x² + 1).
Integrate: y = (1/2) log|x + 1| + (3/4) log(x² + 1) − (1/2) tan⁻¹x + C.
Put x = 0, y = 1: 1 = 0 + 0 − 0 + C, so C = 1.
Answer: y = (1/2) log|x + 1| + (3/4) log(x² + 1) − (1/2) tan⁻¹x + 1
dy = dx/(x(x² − 1)) = [−1/x + (1/2)/(x − 1) + (1/2)/(x + 1)] dx.
y = −log|x| + (1/2) log|x − 1| + (1/2) log|x + 1| + C = (1/2) log|(x² − 1)/x²| + C.
At x = 2, y = 0: 0 = (1/2) log(3/4) + C, so C = −(1/2) log(3/4).
Hence y = (1/2) log[(x² − 1)/x²] − (1/2) log(3/4) = (1/2) log[4(x² − 1)/(3x²)].
Answer: y = (1/2) log[4(x² − 1)/(3x²)]
cos(dy/dx) = a, so dy/dx = cos⁻¹a (a constant).
Integrate: y = x cos⁻¹a + C. Put x = 0, y = 1: C = 1.
So y = x cos⁻¹a + 1, that is (y − 1)/x = cos⁻¹a.
Answer: y = x cos⁻¹ a + 1
dy/y = tan x dx, so log|y| = log|sec x| + log C, giving y = C sec x.
At x = 0, y = 1: 1 = C · 1, so C = 1. Hence y = sec x.
Answer: y = sec x
y = ∫ eˣ sin x dx = (eˣ/2)(sin x − cos x) + C.
At (0, 0): 0 = (1/2)(0 − 1) + C, so C = 1/2.
Hence y = (1/2)[eˣ(sin x − cos x) + 1].
Answer: y = (1/2)[eˣ(sin x − cos x) + 1]
Separate: y/(y + 2) dy = (x + 2)/x dx, i.e. (1 − 2/(y + 2)) dy = (1 + 2/x) dx.
Integrate: y − 2 log|y + 2| = x + 2 log|x| + C.
At (1, −1): −1 − 0 = 1 + 0 + C, so C = −2.
So y − 2 log|y + 2| = x + 2 log|x| − 2, that is y − x + 2 = log[x²(y + 2)²].
Answer: y − x + 2 = log[x²(y + 2)²]
The condition is y dy/dx = x, so y dy = x dx.
Integrate: y²/2 = x²/2 + C₁, that is y² − x² = C.
At (0, −2): 4 = C. So y² − x² = 4.
Answer: y² − x² = 4
The slope of the segment joining (x, y) and (−4, −3) is (y + 3)/(x + 4). So dy/dx = 2(y + 3)/(x + 4).
Separate: dy/(y + 3) = 2 dx/(x + 4). Integrate: log|y + 3| = 2 log|x + 4| + log C, so y + 3 = C(x + 4)².
At (−2, 1): 4 = C · 4, so C = 1. Hence y + 3 = (x + 4)².
Answer: y + 3 = (x + 4)²
V = (4/3)πr³ and dV/dt = k (a constant), so (4π/3) d(r³)/dt = k, that is d(r³)/dt = m, a constant. Integrate: r³ = mt + C.
At t = 0, r = 3: 27 = C. At t = 3, r = 6: 216 = 3m + 27, so m = 63.
Hence r³ = 63t + 27.
Answer: r = (63t + 27)^(1/3) units
Both coefficients are homogeneous of degree 2, so the equation is homogeneous.
dy/dx = (x² + y²)/(x² + xy); put y = vx, so dy/dx = v + x dv/dx:
v + x dv/dx = (1 + v²)/(1 + v), so x dv/dx = (1 − v)/(1 + v).
Separate: (1 + v)/(1 − v) dv = dx/x, where (1 + v)/(1 − v) = −1 + 2/(1 − v).
Integrate: −v − 2 log|1 − v| = log|x| + C₁. Put v = y/x: −y/x − 2 log|(x − y)/x| = log|x| + C₁.
Simplify: 2 log|x − y| = log|x| − y/x + C, where C = −C₁.
Answer: 2 log|x − y| = log|x| − y/x + C
Both coefficients are homogeneous of degree 0, so the equation is homogeneous.
The right side (x + y)/x = 1 + y/x depends only on y/x. Put y = vx:
v + x dv/dx = 1 + v, so x dv/dx = 1 and dv = dx/x.
Integrate: v = log|x| + C, so y = x log|x| + Cx.
Answer: y = x log|x| + Cx
Both coefficients are homogeneous of degree 1, so the equation is homogeneous.
dy/dx = (x + y)/(x − y). Put y = vx: v + x dv/dx = (1 + v)/(1 − v), so x dv/dx = (1 + v²)/(1 − v).
Separate: (1 − v)/(1 + v²) dv = dx/x.
Integrate: tan⁻¹v − (1/2) log(1 + v²) = log|x| + C.
Since (1/2) log(1 + v²) + log|x| = (1/2) log(x² + y²), we get tan⁻¹(y/x) = (1/2) log(x² + y²) + C.
Answer: tan⁻¹(y/x) = (1/2) log(x² + y²) + C
Both coefficients are homogeneous of degree 2, so the equation is homogeneous.
dy/dx = −(x² − y²)/(2xy). Put y = vx: v + x dv/dx = (v² − 1)/(2v), so x dv/dx = −(1 + v²)/(2v).
Separate: 2v/(1 + v²) dv = −dx/x. Integrate: log(1 + v²) = −log|x| + log C.
So x(1 + v²) = C, that is x² + y² = Cx.
Answer: x² + y² = Cx
Both coefficients are homogeneous of degree 2, so the equation is homogeneous.
dy/dx = 1 − 2(y/x)² + y/x. Put y = vx: v + x dv/dx = 1 − 2v² + v, so x dv/dx = 1 − 2v².
Separate: dv/(1 − 2v²) = dx/x. Since ∫dv/(1 − 2v²) = (1/(2√2)) log|(1 + √2 v)/(1 − √2 v)|,
(1/(2√2)) log|(1 + √2 v)/(1 − √2 v)| = log|x| + C. Put v = y/x.
Answer: (1/(2√2)) log|(x + √2y)/(x − √2y)| = log|x| + C
Both coefficients are homogeneous of degree 1, so the equation is homogeneous.
dy/dx = (y + √(x² + y²))/x = v + √(1 + v²) for x > 0, with y = vx.
So x dv/dx = √(1 + v²), and dv/√(1 + v²) = dx/x.
Integrate: log|v + √(1 + v²)| = log|x| + log C, so v + √(1 + v²) = Cx.
Put v = y/x and multiply by x: y + √(x² + y²) = Cx².
Answer: y + √(x² + y²) = Cx²
Both coefficients are homogeneous of degree 2, so the equation is homogeneous.
Put y = vx, dy = v dx + x dv. The equation becomes v(cos v + v sin v) dx = (v sin v − cos v)(v dx + x dv).
Simplify: 2v cos v dx = x(v sin v − cos v) dv, so (v sin v − cos v)/(v cos v) dv = 2 dx/x.
That is (tan v − 1/v) dv = 2 dx/x. Integrate: −log|cos v| − log|v| = 2 log|x| + c.
So log|v cos v| + log x² = −c, giving x² v cos v = C, that is xy cos(y/x) = C.
Answer: xy cos(y/x) = C
Both coefficients are homogeneous of degree 1, so the equation is homogeneous.
dy/dx = y/x − sin(y/x). Put y = vx: v + x dv/dx = v − sin v, so x dv/dx = −sin v.
Separate: dv/sin v = −dx/x, that is cosec v dv = −dx/x.
Integrate: log|tan(v/2)| = −log|x| + log C, so x tan(v/2) = C.
Hence x tan(y/(2x)) = C.
Answer: x tan(y/(2x)) = C
Both coefficients are homogeneous of degree 1, so the equation is homogeneous.
dy/dx = y/(2x − x log(y/x)) = v/(2 − log v) with y = vx. So x dv/dx = v/(2 − log v) − v = v(log v − 1)/(2 − log v).
Separate: (2 − log v)/(v(log v − 1)) dv = dx/x. Put u = log v: ∫(2 − u)/(u − 1) du = ∫(−1 + 1/(u − 1)) du = −u + log|u − 1|.
So −log v + log|log v − 1| = log|x| + c, giving (log v − 1)/(vx) = C.
Since vx = y: log(y/x) − 1 = Cy.
Answer: log(y/x) − 1 = Cy
Both coefficients are homogeneous of degree 0, so the equation is homogeneous.
dx/dy = −e^(x/y)(1 − x/y)/(1 + e^(x/y)) depends only on x/y. Put x = vy: v + y dv/dy = −e^v(1 − v)/(1 + e^v).
So y dv/dy = −(e^v + v)/(1 + e^v). Separate: (1 + e^v)/(v + e^v) dv = −dy/y.
Integrate: log|v + e^v| = −log|y| + log C, so y(v + e^v) = C.
Since vy = x: x + y e^(x/y) = C.
Answer: x + y e^(x/y) = C
Both coefficients are homogeneous of degree 1, so the equation is homogeneous.
dy/dx = (y − x)/(x + y). Put y = vx: v + x dv/dx = (v − 1)/(v + 1), so x dv/dx = −(1 + v²)/(1 + v).
Separate: (1 + v)/(1 + v²) dv = −dx/x. Integrate: tan⁻¹v + (1/2) log(1 + v²) = −log|x| + c.
That is tan⁻¹(y/x) + (1/2) log(x² + y²) = c. At (1, 1): c = π/4 + (1/2) log 2.
Answer: tan⁻¹(y/x) + (1/2) log(x² + y²) = π/4 + (1/2) log 2
Both coefficients are homogeneous of degree 2, so the equation is homogeneous.
dy/dx = −(v + v²) with y = vx, so x dv/dx = −v(v + 2).
Separate: dv/(v(v + 2)) = −dx/x. Since 1/(v(v + 2)) = (1/2)(1/v − 1/(v + 2)), (1/2) log|v/(v + 2)| = −log|x| + c.
So v/(v + 2) = C/x², that is y/(y + 2x) = C/x², or x²y = C(y + 2x). At (1, 1): 1 = 3C, so C = 1/3.
Hence 3x²y = y + 2x.
Answer: 3x²y = y + 2x
Both coefficients are homogeneous of degree 1, so the equation is homogeneous.
dy/dx = y/x − sin²(y/x). Put y = vx: v + x dv/dx = v − sin²v, so x dv/dx = −sin²v.
Separate: cosec²v dv = −dx/x. Integrate: −cot v = −log|x| + c, so cot(y/x) = log|x| + C.
At x = 1, y = π/4: cot(π/4) = 1 = 0 + C, so C = 1.
Answer: cot(y/x) = log x + 1
Both coefficients are homogeneous of degree 0, so the equation is homogeneous.
dy/dx = y/x − cosec(y/x). Put y = vx: v + x dv/dx = v − cosec v, so x dv/dx = −cosec v.
Separate: sin v dv = −dx/x. Integrate: −cos v = −log|x| + c, so cos(y/x) = log|x| + C.
At x = 1, y = 0: 1 = 0 + C, so C = 1.
Answer: cos(y/x) = log x + 1
Both coefficients are homogeneous of degree 2, so the equation is homogeneous.
dy/dx = (2xy + y²)/(2x²) = v + v²/2 with y = vx. So x dv/dx = v²/2.
Separate: 2 dv/v² = dx/x. Integrate: −2/v = log|x| + c, that is −2x/y = log|x| + c.
At (1, 2): −1 = 0 + c, so c = −1. Then 2x/y = 1 − log|x|, giving y = 2x/(1 − log|x|).
Answer: y = 2x/(1 − log|x|)
The right side depends on x/y, so put x/y = v, that is x = vy. Then dx/dy = v + y dv/dy and the variables separate.
Answer: (C) x = vy
In (D) both coefficients, y² and x² − xy − y², are homogeneous of degree 2.
(A) has constant terms, (B) has coefficients of degrees 2 and 3, and (C) has coefficients of degrees 3 and 2, so none of these is homogeneous.
Answer: (D) y² dx + (x² − xy − y²) dy = 0
P = 2, so I.F. = e^(∫2 dx) = e^(2x).
y e^(2x) = ∫ e^(2x) sin x dx = (e^(2x)/5)(2 sin x − cos x) + C.
Divide by e^(2x): y = (1/5)(2 sin x − cos x) + C e^(−2x).
Answer: y = (1/5)(2 sin x − cos x) + C e^(−2x)
I.F. = e^(3x). y e^(3x) = ∫ e^(3x) e^(−2x) dx = ∫ eˣ dx = eˣ + C.
Hence y = e^(−2x) + C e^(−3x).
Answer: y = e^(−2x) + C e^(−3x)
I.F. = e^(∫dx/x) = x. xy = ∫ x · x² dx = x⁴/4 + C.
Hence y = x³/4 + C/x.
Answer: y = x³/4 + C/x
I.F. = e^(∫sec x dx) = e^(log(sec x + tan x)) = sec x + tan x.
y(sec x + tan x) = ∫ tan x (sec x + tan x) dx = ∫ (sec x tan x + sec²x − 1) dx = sec x + tan x − x + C.
Answer: y(sec x + tan x) = sec x + tan x − x + C
Divide by cos²x: dy/dx + (sec²x) y = sec²x tan x. I.F. = e^(tan x).
y e^(tan x) = ∫ tan x sec²x e^(tan x) dx. Put t = tan x: ∫ t eᵗ dt = eᵗ(t − 1) + C.
So y e^(tan x) = e^(tan x)(tan x − 1) + C, that is y = tan x − 1 + C e^(−tan x).
Answer: y = tan x − 1 + C e^(−tan x)
Divide by x: dy/dx + (2/x) y = x log x. I.F. = e^(2 log x) = x².
x²y = ∫ x³ log x dx = (x⁴/4) log x − ∫ x³/4 dx = (x⁴/4) log x − x⁴/16 + C.
Hence y = (x²/4) log x − x²/16 + C x⁻².
Answer: y = (1/4)x² log x − x²/16 + C/x²
Divide by x log x: dy/dx + y/(x log x) = 2/x². I.F. = e^(∫dx/(x log x)) = e^(log log x) = log x.
y log x = ∫ (2/x²) log x dx = 2[−(log x)/x − 1/x] + C.
So y log x = −(2/x)(1 + log x) + C.
Answer: y log x = −(2/x)(1 + log x) + C
dy/dx + (2x/(1 + x²)) y = cot x/(1 + x²). I.F. = e^(log(1 + x²)) = 1 + x².
y(1 + x²) = ∫ cot x dx = log|sin x| + C.
Answer: y(1 + x²) = log|sin x| + C
dy/dx + (1/x + cot x) y = 1. I.F. = e^(log x + log sin x) = x sin x.
y x sin x = ∫ x sin x dx = −x cos x + sin x + C.
Answer: xy sin x = sin x − x cos x + C
Write it as dx/dy = x + y, that is dx/dy − x = y. Here P₁ = −1 and Q₁ = y, so I.F. = e^(−y).
x e^(−y) = ∫ y e^(−y) dy = −y e^(−y) − e^(−y) + C.
Hence x = −y − 1 + C e^y.
Answer: x = −y − 1 + C e^y
dx/dy + x/y = y. Here P₁ = 1/y, so I.F. = e^(log y) = y.
xy = ∫ y · y dy = y³/3 + C.
Answer: xy = y³/3 + C
dx/dy = (x + 3y²)/y, that is dx/dy − x/y = 3y. Here P₁ = −1/y, so I.F. = e^(−log y) = 1/y.
x/y = ∫ (1/y)(3y) dy = 3y + C.
Hence x = 3y² + Cy.
Answer: x = 3y² + Cy
I.F. = e^(2 log sec x) = sec²x. y sec²x = ∫ sin x sec²x dx = ∫ sec x tan x dx = sec x + C.
So y = cos x + C cos²x. At x = π/3, y = 0: 0 = 1/2 + C/4, so C = −2.
Hence y = cos x − 2cos²x.
Answer: y = cos x − 2cos²x
Divide by (1 + x²): dy/dx + (2x/(1 + x²)) y = 1/(1 + x²)². I.F. = 1 + x².
y(1 + x²) = ∫ dx/(1 + x²) = tan⁻¹x + C. At (1, 0): 0 = π/4 + C, so C = −π/4.
Hence y = (tan⁻¹x − π/4)/(1 + x²).
Answer: y(1 + x²) = tan⁻¹x − π/4
P = −3 cot x, so I.F. = e^(−3 log sin x) = 1/sin³x.
y/sin³x = ∫ (2 sin x cos x)/sin³x dx = 2∫ cos x/sin²x dx = −2/sin x + C.
So y = −2 sin²x + C sin³x. At x = π/2, y = 2: 2 = −2 + C, so C = 4.
Hence y = 4 sin³x − 2 sin²x.
Answer: y = 4 sin³x − 2 sin²x
dy/dx = x + y, so dy/dx − y = x. I.F. = e^(−x).
y e^(−x) = ∫ x e^(−x) dx = −x e^(−x) − e^(−x) + C, so y = −x − 1 + C eˣ.
Through (0, 0): 0 = −1 + C, so C = 1. Hence y = eˣ − x − 1.
Answer: y = eˣ − x − 1
The condition is x + y = dy/dx + 5, so dy/dx − y = x − 5. I.F. = e^(−x).
y e^(−x) = ∫ (x − 5) e^(−x) dx = −(x − 5)e^(−x) − e^(−x) + C = (4 − x)e^(−x) + C.
So y = 4 − x + C eˣ. Through (0, 2): 2 = 4 + C, so C = −2.
Hence y = 4 − x − 2eˣ, that is x + y = 4 − 2eˣ.
Answer: y = 4 − x − 2eˣ
Divide by x: dy/dx − (1/x) y = 2x, so P = −1/x.
I.F. = e^(∫−dx/x) = e^(−log x) = 1/x.
Answer: (C) 1/x
Divide by (1 − y²): dx/dy + (y/(1 − y²)) x = ay/(1 − y²), so P₁ = y/(1 − y²).
∫ y/(1 − y²) dy = −(1/2) log(1 − y²), so I.F. = (1 − y²)^(−1/2) = 1/√(1 − y²).
Answer: (D) 1/√(1 − y²)
(i) The highest derivative is d²y/dx², so the order is 2, and its power is 1, so the degree is 1.
(ii) The only derivative is dy/dx, so the order is 1. Its highest power is 3, so the degree is 3.
(iii) The highest derivative is d⁴y/dx⁴, so the order is 4. The term sin(d³y/dx³) is not a polynomial in the derivatives, so the degree is not defined.
Answer: (i) order 2, degree 1; (ii) order 1, degree 3; (iii) order 4, degree not defined
(i) Differentiate xy = a eˣ + b e⁻ˣ + x² twice: y + xy′ = a eˣ − b e⁻ˣ + 2x and 2y′ + xy″ = a eˣ + b e⁻ˣ + 2 = xy − x² + 2. So xy″ + 2y′ − xy + x² − 2 = 0.
(ii) y′ = eˣ[(a + b) cos x + (b − a) sin x] and y″ = eˣ[2b cos x − 2a sin x].
Then y″ − 2y′ + 2y = eˣ[(2b − 2a − 2b + 2a) cos x + (−2a − 2b + 2a + 2b) sin x] = 0.
(iii) y′ = sin 3x + 3x cos 3x and y″ = 6 cos 3x − 9x sin 3x. So y″ + 9y − 6 cos 3x = 6 cos 3x − 9x sin 3x + 9x sin 3x − 6 cos 3x = 0.
(iv) Differentiate x² = 2y² log y: 2x = (4y log y + 2y) y′, so y′ = x/(y(2 log y + 1)). From the relation, 2 log y = x²/y², so 2 log y + 1 = (x² + y²)/y², and y′ = xy/(x² + y²). Hence (x² + y²)y′ − xy = 0.
Answer: All four are verified.
Differentiate x² − y² = c(x² + y²)²: 2x − 2y y′ = 2c(x² + y²)(2x + 2y y′), so (x − y y′) = 2c(x² + y²)(x + y y′).
From the relation, c = (x² − y²)/(x² + y²)². Substitute: (x − y y′)(x² + y²) = 2(x² − y²)(x + y y′).
Expand: x³ + xy² − (x²y + y³)y′ = 2x³ − 2xy² + (2x²y − 2y³)y′.
Collect terms: 0 = x³ − 3xy² + (3x²y − y³)y′, so (x³ − 3xy²) = (y³ − 3x²y)y′, that is (x³ − 3xy²) dx = (y³ − 3x²y) dy.
Answer: Proved: the relation satisfies the equation for every value of c.
Separate: dy/√(1 − y²) = −dx/√(1 − x²).
Integrate: sin⁻¹y = −sin⁻¹x + C, so sin⁻¹x + sin⁻¹y = C.
Answer: sin⁻¹ x + sin⁻¹ y = C
Separate: dy/(y² + y + 1) + dx/(x² + x + 1) = 0. Since ∫dx/(x² + x + 1) = (2/√3) tan⁻¹((2x + 1)/√3):
tan⁻¹((2x + 1)/√3) + tan⁻¹((2y + 1)/√3) = c.
Use tan⁻¹a + tan⁻¹b = tan⁻¹[(a + b)/(1 − ab)]: [(2x + 2y + 2)/√3] / [1 − (2x + 1)(2y + 1)/3] = tan c, that is 2√3(x + y + 1)/(2 − 2x − 2y − 4xy) = tan c.
So x + y + 1 = A(1 − x − y − 2xy), where A = tan c/√3.
Answer: x + y + 1 = A(1 − x − y − 2xy)
Divide by cos x cos y: tan x dx + tan y dy = 0.
Integrate: −log|cos x| − log|cos y| = −log C, so cos x cos y = C.
At (0, π/4): C = 1 · cos(π/4) = 1/√2.
Hence cos x cos y = 1/√2.
Answer: cos x cos y = 1/√2
Separate: dy/(1 + y²) = −eˣ dx/(1 + e^(2x)). Integrate (with t = eˣ on the right): tan⁻¹y = −tan⁻¹(eˣ) + c.
At (0, 1): π/4 = −π/4 + c, so c = π/2.
Hence tan⁻¹y + tan⁻¹(eˣ) = π/2. Since tan⁻¹a + tan⁻¹(1/a) = π/2 for a > 0, this means y = 1/eˣ = e⁻ˣ.
Answer: tan⁻¹y + tan⁻¹(eˣ) = π/2, that is y = e⁻ˣ
dx/dy = (x e^(x/y) + y²)/(y e^(x/y)). Put x = vy: v + y dv/dy = v + y e^(−v), so dv/dy e^v = 1.
Separate: e^v dv = dy. Integrate: e^v = y + C.
Hence e^(x/y) = y + C.
Answer: e^(x/y) = y + C
Put t = x − y, so dx − dy = dt. The equation is t(dx + dy) = dt, so dx + dy = dt/t.
Integrate: x + y = log|t| + C, that is x + y = log|x − y| + C.
At (0, −1): −1 = 0 + C, so C = −1.
Hence log|x − y| = x + y + 1.
Answer: log|x − y| = x + y + 1
Invert: dy/dx = e^(−2√x)/√x − y/√x, so dy/dx + y/√x = e^(−2√x)/√x. I.F. = e^(∫x^(−1/2) dx) = e^(2√x).
y e^(2√x) = ∫ (e^(−2√x)/√x) e^(2√x) dx = ∫ x^(−1/2) dx = 2√x + C.
Answer: y e^(2√x) = 2√x + C
I.F. = e^(∫cot x dx) = sin x. y sin x = ∫ 4x cosec x sin x dx = ∫ 4x dx = 2x² + C.
At (π/2, 0): 0 = π²/2 + C, so C = −π²/2.
Hence y sin x = 2x² − π²/2.
Answer: y sin x = 2x² − π²/2
Separate: dy/(2e^(−y) − 1) = dx/(x + 1), that is e^y dy/(2 − e^y) = dx/(x + 1).
Integrate: −log|2 − e^y| = log|x + 1| + c, so (x + 1)(2 − e^y) = C.
At (0, 0): 1 · 1 = C, so C = 1.
Hence (x + 1)(2 − e^y) = 1, that is e^y = (2x + 1)/(x + 1) and y = log[(2x + 1)/(x + 1)].
Answer: (x + 1)(2 − e^y) = 1, that is y = log[(2x + 1)/(x + 1)]
Divide: dx/x − dy/y = 0. Integrate: log|x| − log|y| = log|C₁|, so x/y = C₁, that is y = Cx.
Answer: (C) y = Cx
Here x is the dependent variable and y the independent one, so the integrating factor is e^(∫P₁ dy).
Multiplying by it gives d/dy(x · I.F.) = Q₁ · I.F., hence x e^(∫P₁ dy) = ∫(Q₁ e^(∫P₁ dy)) dy + C.
Answer: (C) x e^(∫P₁ dy) = ∫(Q₁ e^(∫P₁ dy)) dy + C
The equation is eˣ dy + y eˣ dx + 2x dx = 0, that is d(y eˣ) + d(x²) = 0.
Integrate: y eˣ + x² = C.
Answer: (C) y eˣ + x² = C