Exercise 9.3 answers: Differential Equations

Open NCERT p. 310Matches NCERT’s answer

1. Solve: dy/dx = (1 − cos x)/(1 + cos x).
2. Use the half-angle identities: 1 − cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2).
3. So the right side becomes tan²(x/2), which is sec²(x/2) − 1.
4. The equation is now dy/dx = sec²(x/2) − 1, and x and y are already separated.
5. Integrate both sides: y = 2tan(x/2) − x + C.

Answery = 2 tan(x/2) − x + C

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1. Separate the variables: dy/√(4 − y²) = dx.
2. The left side is a standard integral: ∫dy/√(4−y²) = sin⁻¹(y/2).
3. The right side integrates to x + C.
4. So the general solution is sin⁻¹(y/2) = x + C.

Answersin⁻¹(y/2) = x + C

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1. Rewrite as dy/dx = 1 − y.
2. Separate the variables: dy/(1 − y) = dx.
3. Integrate both sides: −log|1 − y| = x + C₁, so log|1 − y| = −x − C₁.
4. Take exponentials: 1 − y = A e⁻ˣ for a constant A.
5. This can be written as (y − 1)eˣ = C, a new constant.

Answer(y − 1)eˣ = C

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1. Divide throughout by tan x tan y to separate: (sec²x/tan x) dx + (sec²y/tan y) dy = 0.
2. Integrate both sides: log|tan x| + log|tan y| = C₁.
3. Combine the logs: log|tan x tan y| = C₁.
4. Take exponentials: tan x × tan y = C.

Answertan x × tan y = C

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1. Solve: (eˣ + e⁻ˣ) dy − (eˣ − e⁻ˣ) dx = 0.
2. Rewrite as dy/dx = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ).
3. The right side is exactly the derivative of log(eˣ + e⁻ˣ) with respect to x.
4. So integrating both sides gives y = log(eˣ + e⁻ˣ) + C.

Answery = log(eˣ + e⁻ˣ) + C

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1. Separate the variables: dy/(1 + y²) = (1 + x²) dx.
2. Integrate the left side: ∫dy/(1+y²) = tan⁻¹y.
3. Integrate the right side: ∫(1+x²)dx = x + x³/3.
4. So the general solution is tan⁻¹y = x + x³/3 + C.

Answertan⁻¹y = x + x³/3 + C

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1. Rearrange: x dy = y log y dx, so dy/(y log y) = dx/x. (Here y > 0, y ≠ 1 and x ≠ 0.)
2. Left side: put u = log y, so du = dy/y. Then ∫dy/(y log y) = ∫du/u = log|log y|.
3. Right side: ∫dx/x = log|x|.
4. So log|log y| = log|x| + log|C|. Writing the constant as log|C| keeps the next step tidy.
5. Combine the logs: log|log y| = log|Cx|, so |log y| = |Cx|.
6. Let C take either sign: log y = Cx, that is, y = e^(Cx).

Answerlog y = Cx, that is, y = e^(Cx)

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1. Solve: x⁵ dy/dx = −y⁵.
2. Separate the variables: dy/y⁵ = −dx/x⁵.
3. Integrate: −1/(4y⁴) = 1/(4x⁴) + C₁.
4. Multiply through by −4: 1/y⁴ = −1/x⁴ − 4C₁.
5. Move the x term across: 1/x⁴ + 1/y⁴ = C, a new constant.

Answer1/x⁴ + 1/y⁴ = C

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1. Solve: dy/dx = sin⁻¹x.
2. Integrate directly: y = ∫sin⁻¹(x) dx
3. Use integration by parts: u = sin⁻¹(x), dv = dx
4. Then: du = dx/√(1 − x²), v = x
5. By parts: ∫sin⁻¹(x) dx = x·sin⁻¹(x) − ∫x·dx/√(1 − x²)
6. For the remaining integral, let w = 1 − x², dw = −2x dx
7. ∫x·dx/√(1 − x²) = −(1/2)∫dw/√w = −(1/2)·2√w = −√(1 − x²)
8. Therefore: y = x·sin⁻¹(x) + √(1 − x²) + C

Answery = x·sin⁻¹(x) + √(1 − x²) + C

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1. Solve: eˣ tan y dx + (1 − eˣ) sec²y dy = 0.
2. Divide throughout by tan y (1 − eˣ) to separate: sec²y/tan y dy = −eˣ/(1−eˣ) dx.
3. Integrate the left side: ∫sec²y/tan y dy = log|tan y|.
4. Integrate the right side: ∫−eˣ/(1−eˣ) dx = log|1−eˣ| + C₁ (since the numerator is the derivative of the denominator).
5. So log|tan y| = log|1−eˣ| + C₁, which gives tan y = C(1−eˣ).

Answertan y = C(1 − eˣ)

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1. Solve: (x³+x²+x+1) dy/dx = 2x²+x; y = 1 when x = 0.
2. Factor the left bracket: x³+x²+x+1 = (x+1)(x²+1).
3. So dy/dx = (2x²+x) / [(x+1)(x²+1)].
4. Split the right side into partial fractions: (2x²+x)/[(x+1)(x²+1)] = 1/[2(x+1)] + (3x−1)/[2(x²+1)].
5. Integrate each piece: (1/2)log|x+1| + (3/4)log(x²+1) − (1/2)tan⁻¹x + C.
6. Use y = 1 when x = 0: all the log and tan⁻¹ terms are 0, so C = 1.
7. Particular solution: y = (1/2)log|x+1| + (3/4)log(x²+1) − (1/2)tan⁻¹x + 1.

Answery = (1/2)log|x+1| + (3/4)log(x²+1) − (1/2)tan⁻¹x + 1

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1. Solve: x(x²−1) dy/dx = 1; y = 0 when x = 2.
2. Rewrite as dy = dx / [x(x−1)(x+1)] and split into partial fractions.
3. 1/[x(x−1)(x+1)] = −1/x + 1/[2(x−1)] + 1/[2(x+1)].
4. Integrate: y = −log|x| + (1/2)log|x²−1| + C.
5. Use y = 0 when x = 2: 0 = −log2 + (1/2)log3 + C, so C = log2 − (1/2)log3.
6. Combine the logs: y = log[2√(x²−1) / (x√3)].

Answery = log[2√(x²−1) / (x√3)]

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1. Solve: cos(dy/dx) = a (a ∈ R); y = 1 when x = 0.
2. cos(dy/dx) = a says the slope dy/dx is a number whose cosine is a. Such a number exists only when −1 ≤ a ≤ 1.
3. Call that number k, so cos k = a. Then dy/dx = k, a constant.
4. Integrate: y = kx + C.
5. Use y = 1 when x = 0: 1 = 0 + C, so C = 1 and y = 1 + kx.
6. For x ≠ 0 this gives k = (y − 1)/x, and cos k = a becomes cos((y − 1)/x) = a.
7. Taking the principal value k = cos⁻¹a gives one such line: y = 1 + x cos⁻¹a.
8. The answer key at the back of the book prints cos((y − 2)/x) = a, but that curve gives y = 2 when x = 0, and the question says y = 1 when x = 0, so the answer is cos((y − 1)/x) = a.

Answercos((y − 1)/x) = a; with the principal value, y = 1 + x cos⁻¹a

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1. Solve: dy/dx = y tan x; y = 1 when x = 0.
2. Separate the variables: dy/y = tan x dx.
3. Integrate: log|y| = −log|cos x| + C₁ = log|sec x| + C₁.
4. So y = C sec x.
5. Use y = 1 when x = 0: 1 = C × 1, so C = 1.
6. Particular solution: y = sec x.

Answery = sec x

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1. Given: y′ = eˣ sin x.
2. Integrate both sides: y = ∫eˣ sin x dx.
3. This is a standard integral (integrate by parts twice): ∫eˣ sin x dx = eˣ(sin x − cos x)/2 + C.
4. So y = eˣ(sin x − cos x)/2 + C.
5. Use the point (0, 0): 0 = 1×(0−1)/2 + C = −1/2 + C, so C = 1/2.
6. Equation of the curve: y = [eˣ(sin x − cos x) + 1]/2.

Answery = [eˣ(sin x − cos x) + 1]/2

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1. Given: xy dy/dx = (x + 2)(y + 2).
2. Separate the variables: y/(y+2) dy = (x+2)/x dx.
3. Rewrite each side to make it easy to integrate: y/(y+2) = 1 − 2/(y+2), and (x+2)/x = 1 + 2/x.
4. Integrate both sides: y − 2 log|y+2| = x + 2 log|x| + C.
5. Use the point (1, −1): −1 − 2log(1) = 1 + 2log(1) + C, so C = −2.
6. Solution curve: y − x − 2 log|y+2| − 2 log|x| = −2.

Answery − x − 2 log|y+2| − 2 log|x| = −2

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1. Let the curve be y = f(x). At any point (x, y), the slope of the tangent is dy/dx.
2. “Product of the slope and the y coordinate equals the x coordinate” means (dy/dx) × y = x, i.e. y dy = x dx.
3. Integrate both sides: y²/2 = x²/2 + C₁, so y² − x² = C.
4. The curve passes through (0, −2): (−2)² − 0² = C, so C = 4.
5. So the equation of the curve is y² − x² = 4.

Answery² − x² = 4

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1. The line joining (x, y) to (−4, −3) has slope (y + 3)/(x + 4).
2. “Slope of the tangent is twice this” gives dy/dx = 2(y + 3)/(x + 4).
3. Separate the variables: dy/(y + 3) = 2 dx/(x + 4).
4. Integrate: log|y + 3| = 2 log|x + 4| + C₁, so y + 3 = A(x + 4)².
5. The curve passes through (−2, 1): 1 + 3 = A(−2 + 4)², so 4 = 4A, i.e. A = 1.
6. So the curve is y + 3 = (x + 4)², i.e. y = (x + 4)² − 3.

Answery = (x + 4)² − 3

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1. Volume of a sphere: V = (4/3)πr³. “Changes at a constant rate” means dV/dt = k, a constant.
2. So (4/3)π × 3r² × dr/dt = k, i.e. r² dr = m dt, where m = k/(4π).
3. Integrate: r³/3 = mt + C₁, i.e. r³ = 3mt + 3C₁.
4. At t = 0, r = 3: 27 = 3C₁, so the constant term is 27.
5. At t = 3, r = 6: 216 = 3m × 3 + 27, so 3m = 63.
6. So r³ = 63t + 27, i.e. r = ³√(63t + 27).

Answerr = ³√(63t + 27)

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Open NCERT p. 312Matches NCERT’s answer

1. Let P be the principal at time t. “Increases continuously at rate r% per year” means dP/dt = (r/100)P.
2. Separate and integrate: log P = (r/100)t + C, so P = A·e^(r/100)t.
3. Rs 100 doubles in 10 years: 200 = 100 × e^(r/100)×10, so e^r/10 = 2.
4. Take log: r/10 = log 2 = 0.6931 (given).
5. So r = 6.931.

Answerr = 6.931

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1. “Increases continuously at 5% per year” means dP/dt = (5/100)P = P/20.
2. Solving this separable equation gives P = P₀·e^t/20, with P₀ = 1000.
3. After 10 years: P = 1000 × e^10/20 = 1000 × e^0.5.
4. Using e^0.5 = 1.648 (given), P = 1000 × 1.648 = 1648.

Answer₹1648

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1. Let N(t) be the bacteria count. “Rate of growth proportional to N” means dN/dt = kN.
2. Solving gives N = N₀·e^kt, with N₀ = 1,00,000.
3. In 2 hours N grows by 10%, so N(2) = 1,10,000: e^2k = 11/10.
4. We want N(t) = 2,00,000, i.e. e^kt = 2.
5. From e^2k = 11/10, k = (1/2)log(11/10); and kt = log 2 gives t = 2 log 2 / log(11/10).
6. So the count reaches 2,00,000 after t = 2 log 2 / log(11/10) hours (about 14.55 hours).

Answert = 2 log 2 / log(11/10) hours (≈ 14.55 hours)

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1. Write e^x+y = e^x·e^y, so dy/dx = e^xe^y.
2. Separate the variables: e^−y dy = e^x dx.
3. Integrate both sides: −e^−y = e^x + C₁.
4. Rearrange: e^x + e^−y = C.
5. This matches option (A).

Answer(A) e^x + e^−y = C

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