Exercise 9.5 answers: Differential Equations

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1. Solve: dy/dx + 2y = sin x.
2. The equation is already dy/dx + Py = Q with P = 2 and Q = sin x.
3. The integrating factor is e^(∫2 dx) = e^2x.
4. Multiply through: d/dx(y·e^2x) = e^2x·sin x.
5. Integrating the right side (by parts twice) gives (2 sin x − cos x)e^2x/5.
6. So y·e^2x = (2 sin x − cos x)e^2x/5 + C.

Answery·e^2x = (2 sin x − cos x)e^2x/5 + C

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1. Solve: dy/dx + 3y = e^(−2x).
2. The equation is dy/dx + Py = Q with P = 3 and Q = e^(−2x).
3. The integrating factor is e^(∫3 dx) = e^3x.
4. Multiply through: d/dx(y·e^3x) = e^3x·e^−2x = e^x.
5. Integrate: y·e^3x = e^x + C.

Answery·e^3x = e^x + C

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1. Solve: dy/dx + y/x = x².
2. The equation is dy/dx + Py = Q with P = 1/x and Q = x².
3. The integrating factor is e^(∫(1/x) dx) = e^(log x) = x.
4. Multiply through: d/dx(y·x) = x·x² = x³.
5. Integrate: y·x = x⁴/4 + C.

Answery·x = x⁴/4 + C

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1. Solve: dy/dx + (sec x) y = tan x.
2. The equation is dy/dx + Py = Q with P = sec x and Q = tan x.
3. The integrating factor is e^(∫sec x dx) = e^(log|sec x + tan x|) = sec x + tan x.
4. Multiply through: d/dx[y(sec x + tan x)] = tan x(sec x + tan x) = sec x tan x + sec²x − 1.
5. Integrate: y(sec x + tan x) = sec x + tan x − x + C.

Answery(sec x + tan x) = sec x + tan x − x + C

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1. Solve: cos²x dy/dx + y = tan x.
2. Divide by cos²x: dy/dx + sec²x·y = sec²x tan x. So P = sec²x and Q = sec²x tan x.
3. The integrating factor is e^(∫sec²x dx) = e^(tan x).
4. Multiply through: d/dx(y·e^{tan x}) = sec²x tan x·e^{tan x}.
5. Writing t = tan x turns the right side into t·e^t, and ∫t e^t dt = (t − 1)e^t.
6. So y·e^{tan x} = (tan x − 1)e^{tan x} + C.

Answery·e^{tan x} = (tan x − 1)e^{tan x} + C

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1. Solve: x dy/dx + 2y = x² log x.
2. Divide by x: dy/dx + (2/x)y = x log x. So P = 2/x and Q = x log x.
3. The integrating factor is e^(∫(2/x) dx) = e^(2 log x) = x².
4. Multiply through: d/dx(y·x²) = x²·x log x = x³ log x.
5. Integrate by parts: ∫x³ log x dx = (x⁴/4) log x − x⁴/16.
6. So y·x² = x⁴(4 log x − 1)/16 + C.

Answery·x² = x⁴(4 log x − 1)/16 + C

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1. Solve: x log x dy/dx + y = (2/x) log x.
2. Divide by x log x: dy/dx + y/(x log x) = 2/x². So P = 1/(x log x) and Q = 2/x².
3. The integrating factor is e^(∫1/(x log x) dx) = e^(log(log x)) = log x.
4. Multiply through: d/dx(y·log x) = (2/x²) log x.
5. Integrate by parts: ∫(2 log x)/x² dx = −2(log x + 1)/x.
6. So y·log x = −2(log x + 1)/x + C.

Answery·log x = −2(log x + 1)/x + C

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1. Divide by dx and by (1 + x²): dy/dx + [2x/(1 + x²)]y = cot x/(1 + x²). So P = 2x/(1 + x²) and Q = cot x/(1 + x²).
2. The integrating factor is e^(∫2x/(1 + x²) dx) = e^(log(1 + x²)) = 1 + x².
3. Multiply through: d/dx[y(1 + x²)] = (1 + x²)·cot x/(1 + x²) = cot x.
4. Integrate: ∫cot x dx = log|sin x|. Keep the modulus: the question allows any x ≠ 0, and sin x is negative for some of those x (for example π < x < 2π).
5. So y(1 + x²) = log|sin x| + C.

Answery(1 + x²) = log|sin x| + C

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1. Solve: x dy/dx + y − x + xy cot x = 0.
2. Rearrange and divide by x: dy/dx + y(1/x + cot x) = 1. So P = 1/x + cot x and Q = 1.
3. The integrating factor is e^(∫(1/x + cot x) dx) = e^(log x + log(sin x)) = x sin x.
4. Multiply through: d/dx(y·x sin x) = x sin x.
5. Integrate by parts: ∫x sin x dx = sin x − x cos x.
6. So y·x·sin x = sin x − x cos x + C.

Answery·x·sin x = sin x − x cos x + C

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1. The equation is awkward for dy/dx but linear if x is treated as the unknown: invert to get dx/dy = x + y, i.e. dx/dy − x = y.
2. Here P = −1 (in y) and Q = y, so the integrating factor is e^(∫(−1) dy) = e^−y.
3. Multiply through: d/dy(x·e^−y) = y·e^−y.
4. Integrate by parts: ∫y e^−y dy = −(y+1)e^−y.
5. So x·e^−y = −(y+1)e^−y + C, i.e. (x+y+1)e^−y = C.

Answer(x + y + 1)e^−y = C, i.e. x + y + 1 = C·e^y

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1. Divide by dy: y·(dx/dy) + x − y² = 0. Dividing by y gives dx/dy + x/y = y, linear in x with P = 1/y and Q = y.
2. The integrating factor is e^(∫(1/y) dy) = e^(log y) = y.
3. Multiply through: d/dy(x·y) = y·y = y².
4. Integrate: x·y = y³/3 + C.

Answerx·y = y³/3 + C

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1. Invert to make x the unknown: dx/dy = (x+3y²)/y = x/y + 3y, i.e. dx/dy − x/y = 3y. P = −1/y, Q = 3y.
2. The integrating factor is e^(∫(−1/y) dy) = e^(−log y) = 1/y.
3. Multiply through: d/dy(x/y) = 3y·(1/y) = 3.
4. Integrate: x/y = 3y + C.

Answerx/y = 3y + C, i.e. x = 3y² + Cy

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1. Solve: dy/dx + 2y tan x = sin x ; y = 0 when x = π/3.
2. The equation is dy/dx + Py = Q with P = 2 tan x and Q = sin x.
3. The integrating factor is e^(∫2 tan x dx) = e^(2 log(sec x)) = sec²x.
4. Multiply through: d/dx(y·sec²x) = sin x·sec²x = sec x tan x.
5. Integrate: y·sec²x = sec x + C.
6. Use y = 0 when x = π/3: 0 = sec(π/3) + C = 2 + C, so C = −2.

Answery·sec²x = sec x − 2

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1. Solve: (1 + x²) dy/dx + 2xy = 1/(1 + x²) ; y = 0 when x = 1.
2. Divide by (1+x²): dy/dx + [2x/(1+x²)]y = 1/(1+x²)². So P = 2x/(1+x²), Q = 1/(1+x²)².
3. The integrating factor is e^(∫2x/(1+x²) dx) = e^(log(1+x²)) = 1+x².
4. Multiply through: d/dx[y(1+x²)] = 1/(1+x²).
5. Integrate: y(1+x²) = arctan x + C.
6. Use y = 0 when x = 1: 0 = π/4 + C, so C = −π/4.

Answery(1 + x²) = arctan x − π/4

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1. Solve: dy/dx − 3y cot x = sin 2x ; y = 2 when x = π/2.
2. The equation is dy/dx + Py = Q with P = −3 cot x and Q = sin 2x.
3. The integrating factor is e^(∫−3 cot x dx) = e^(−3 log(sin x)) = 1/sin³x.
4. Multiply through: d/dx(y/sin³x) = sin 2x/sin³x = 2 cos x/sin²x (using sin 2x = 2 sin x cos x).
5. Integrate: ∫2 cos x/sin²x dx = −2/sin x, so y/sin³x = −2/sin x + C.
6. Use y = 2 when x = π/2: 2 = −2 + C, so C = 4.

Answery/sin³x = −2/sin x + 4, i.e. y = 4 sin³x − 2 sin²x

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1. The slope of the tangent is dy/dx, and it equals the sum of the coordinates: dy/dx = x + y, i.e. dy/dx − y = x.
2. Here P = −1 and Q = x, so the integrating factor is e^(∫(−1) dx) = e^−x.
3. Multiply through: d/dx(y·e^−x) = x·e^−x.
4. Integrate by parts: ∫x e^−x dx = −(x+1)e^−x.
5. So y·e^−x = −(x+1)e^−x + C, i.e. (x + y + 1)e^−x = C.
6. The curve passes through the origin, so y = 0 when x = 0: (0+0+1)·1 = C, giving C = 1.

Answery = eˣ − x − 1

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1. Let (x, y) be any point on the curve. The sum of its coordinates is x + y, and the slope of the tangent there is dy/dx.
2. A word about “magnitude”: at the given point (0, 2) the sum of the coordinates is 2, which is less than 5. A size is never negative, so nothing can be exceeded by 5 there. The question therefore means the slope itself: x + y − dy/dx = 5.
3. So dy/dx = x + y − 5, i.e. dy/dx − y = x − 5. This is linear with P = −1 and Q = x − 5.
4. Integrating factor = e^{∫−1 dx} = e^−x.
5. Multiply through: d/dx(y·e^−x) = (x − 5)e^−x. By parts, ∫(x − 5)e^−x dx = −(x − 5)e^−x − e^−x = −(x − 4)e^−x.
6. So y·e^−x = −(x − 4)e^−x + C, i.e. y = 4 − x + C·e^x.
7. Use the point (0, 2): 2 = 4 − 0 + C·1, so C = −2.

AnswerThe curve is y = 4 − x − 2e^x.

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1. Given: x dy/dx − y = 2x².
2. Divide the whole equation by x (x ≠ 0): dy/dx − y/x = 2x.
3. This is linear with P = −1/x.
4. Integrating factor = e^{∫(−1/x) dx} = e^{−log x} = 1/x.

AnswerThe Integrating Factor is 1/x — option (C).

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1. Given: (1 − y²) dx/dy + yx = ay (−1 < y < 1).
2. Divide by (1 − y²): dx/dy + [y/(1−y²)]x = ay/(1−y²), a linear equation in x with P₁ = y/(1−y²).
3. ∫P₁ dy = ∫ y/(1−y²) dy = −½ log(1−y²) (using 1−y² > 0 since −1 < y < 1).
4. Integrating factor = e^{−½ log(1−y²)} = 1/√(1−y²).

AnswerThe Integrating Factor is 1/√(1−y²) — option (D).

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