Exercise 9.4 answers: Differential Equations

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1. Write the equation as dy/dx = (x² + y²)/(x² + xy) = F(x, y).
2. F(λx, λy) = λ²(x² + y²)/[λ²(x² + xy)] = λ⁰F(x, y), so F is homogeneous of degree 0 and the equation is homogeneous.
3. Put y = vx, so dy/dx = v + x dv/dx, and the right side becomes (1 + v²)/(1 + v).
4. Then x dv/dx = (1 + v²)/(1 + v) − v = (1 − v)/(1 + v).
5. Separate: (1 + v)/(1 − v) dv = dx/x, and write (1 + v)/(1 − v) = −1 + 2/(1 − v).
6. Integrate: −v − 2 log|1 − v| = log|x| + C₁.
7. Put v = y/x: −y/x − 2 log|x − y| + 2 log|x| = log|x| + C₁, so log|x| − log(x − y)² = C₁ + y/x.
8. So x/(x − y)² = A e^y/x for a constant A; writing C = 1/A gives (x − y)² = Cx e^−y/x.

Answer(x − y)² = Cx e^−y/x

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1. Here dy/dx = (x + y)/x = 1 + y/x, which depends only on the ratio y/x, so it is homogeneous.
2. Put y = vx, so dy/dx = v + x(dv/dx).
3. Substitute: v + x(dv/dx) = 1 + v, so x(dv/dx) = 1.
4. Separate and integrate: dv = dx/x gives v = log|x| + C.
5. Put v = y/x back: y/x − log|x| = C.

Answery/x − log|x| = C (i.e. y = x(log|x| + C))

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1. Write the equation as dy/dx = (x + y)/(x − y) = F(x, y).
2. F(λx, λy) = λ(x + y)/[λ(x − y)] = λ⁰F(x, y), so the equation is homogeneous.
3. Put y = vx, so dy/dx = v + x dv/dx: v + x dv/dx = (1 + v)/(1 − v).
4. So x dv/dx = (1 + v)/(1 − v) − v = (1 + v²)/(1 − v).
5. Separate: (1 − v)/(1 + v²) dv = dx/x, i.e. [1/(1 + v²) − v/(1 + v²)] dv = dx/x.
6. Integrate: tan⁻¹v − (1/2) log(1 + v²) = log|x| + C.
7. Put v = y/x: tan⁻¹(y/x) − (1/2) log[(x² + y²)/x²] = log|x| + C.
8. Since (1/2) log x² = log|x|, this is tan⁻¹(y/x) − (1/2) log(x² + y²) + log|x| = log|x| + C, so tan⁻¹(y/x) − (1/2) log(x² + y²) = C.

Answertan⁻¹(y/x) − (1/2) log(x² + y²) = C

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1. Write dy/dx = (y² − x²)/(2xy), homogeneous of degree 0.
2. Put y = vx, so dy/dx = v + x(dv/dx).
3. Substituting and simplifying: x(dv/dx) = −(1 + v²)/(2v).
4. Separate: 2v/(1 + v²) dv = −dx/x.
5. Integrate: log(1 + v²) = −log|x| + C₁, so (1 + v²)x = A.
6. Put v = y/x back and simplify: x² + y² = Ax, i.e. (x² + y²)/x = C.

Answerx² + y² = Cx

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1. Write dy/dx = (x² − 2y² + xy)/x² = F(x, y).
2. F(λx, λy) = λ²(x² − 2y² + xy)/(λ²x²) = λ⁰F(x, y), so the equation is homogeneous.
3. Put y = vx, so dy/dx = v + x dv/dx: v + x dv/dx = 1 − 2v² + v, so x dv/dx = 1 − 2v².
4. Separate: dv/(1 − 2v²) = dx/x, i.e. (1/2) · dv/[(1/√2)² − v²] = dx/x.
5. Use ∫dv/(a² − v²) = (1/(2a)) log|(a + v)/(a − v)| with a = 1/√2: (1/2) · (√2/2) log|(1/√2 + v)/(1/√2 − v)| = log|x| + C.
6. Multiply top and bottom inside the log by √2: (1/(2√2)) log|(1 + √2v)/(1 − √2v)| = log|x| + C.
7. Put v = y/x: (1/(2√2)) log|(x + √2y)/(x − √2y)| = log|x| + C.

Answer(1/(2√2)) log|(x + √2y)/(x − √2y)| = log|x| + C

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1. Write dy/dx = [y + √(x²+y²)]/x, homogeneous of degree 0.
2. Put y = vx, so dy/dx = v + x(dv/dx).
3. Substituting and simplifying: x(dv/dx) = √(1 + v²).
4. Separate and integrate: dv/√(1+v²) = dx/x gives log[v + √(1+v²)] = log|x| + C.
5. So v + √(1+v²) = Ax. Put v = y/x back and multiply by x: y + √(x²+y²) = Ax².

Answery + √(x² + y²) = Cx²

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1. Write dy/dx = y[x cos(y/x) + y sin(y/x)] / {x[y sin(y/x) − x cos(y/x)]} = F(x, y).
2. Replacing x, y by λx, λy leaves y/x unchanged and multiplies top and bottom by λ², so F(λx, λy) = λ⁰F(x, y): the equation is homogeneous.
3. Put y = vx, so dy/dx = v + x dv/dx: v + x dv/dx = v(cos v + v sin v)/(v sin v − cos v).
4. So x dv/dx = [v cos v + v² sin v − v² sin v + v cos v]/(v sin v − cos v) = 2v cos v/(v sin v − cos v).
5. Separate: (v sin v − cos v)/(v cos v) dv = 2 dx/x, i.e. (tan v − 1/v) dv = 2 dx/x.
6. Integrate: −log|cos v| − log|v| = 2 log|x| + C₁, so log|x² v cos v| = −C₁.
7. So x² v cos v = C. Put v = y/x: x²(y/x) cos(y/x) = C, i.e. xy cos(y/x) = C.

Answerxy cos(y/x) = C

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1. Write dy/dx = y/x − sin(y/x), homogeneous of degree 0.
2. Put y = vx, so dy/dx = v + x(dv/dx).
3. Substituting: v + x(dv/dx) = v − sin(v), so x(dv/dx) = −sin(v).
4. Separate: dv/sin(v) = −dx/x. The standard integral of cosec(v) is log|tan(v/2)|.
5. Integrate: log|tan(v/2)| = −log|x| + C, so x·tan(v/2) = A.
6. Put v = y/x back: x·tan(y/2x) = C.

Answerx·tan(y/(2x)) = C

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1. Collect the dy terms: y dx + x[log(y/x) − 2] dy = 0, so dy/dx = y/{x[2 − log(y/x)]} = F(x, y).
2. F(λx, λy) = λy/{λx[2 − log(y/x)]} = λ⁰F(x, y), so the equation is homogeneous.
3. Put y = vx, so dy/dx = v + x dv/dx: v + x dv/dx = v/(2 − log v).
4. So x dv/dx = v/(2 − log v) − v = v(log v − 1)/(2 − log v).
5. Separate: (2 − log v)/[v(log v − 1)] dv = dx/x.
6. Put t = log v, so dt = dv/v: the left side is (2 − t)/(t − 1) dt = [1/(t − 1) − 1] dt.
7. Integrate: log|t − 1| − t = log|x| + C₁, i.e. log|log v − 1| − log v = log|x| + C₁.
8. So log|(log v − 1)/(vx)| = C₁, i.e. (log v − 1)/(vx) = C.
9. Since vx = y and v = y/x: (log(y/x) − 1)/y = C, i.e. log(y/x) − 1 = Cy.

Answerlog(y/x) − 1 = Cy

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1. Here it is easier to treat x as the unknown and y as the variable: dx/dy = −e^x/y(1 − x/y)/(1 + e^x/y), homogeneous of degree 0.
2. Put x = uy (the mirror substitution), so dx/dy = u + y(du/dy).
3. Substituting and simplifying: y(du/dy) = −(u + e^u)/(1 + e^u).
4. Separate: (1 + e^u)/(u + e^u) du = −dy/y. The left side is exactly d(u + e^u)/(u + e^u).
5. Integrate: log|u + e^u| = −log|y| + C₁, so y(u + e^u) = A.
6. Put u = x/y back and multiply by y: x + y·e^x/y = A.

Answerx + y·e^x/y = C

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1. Write dy/dx = (y − x)/(x + y) = F(x, y). F(λx, λy) = λ(y − x)/[λ(x + y)] = λ⁰F(x, y), so the equation is homogeneous.
2. Put y = vx, so dy/dx = v + x dv/dx: v + x dv/dx = (v − 1)/(1 + v).
3. So x dv/dx = (v − 1)/(1 + v) − v = −(1 + v²)/(1 + v).
4. Separate: (1 + v)/(1 + v²) dv = −dx/x, i.e. [1/(1 + v²) + v/(1 + v²)] dv = −dx/x.
5. Integrate: tan⁻¹v + (1/2) log(1 + v²) = −log|x| + C.
6. Put v = y/x: tan⁻¹(y/x) + (1/2) log(x² + y²) − log|x| = −log|x| + C, so tan⁻¹(y/x) + (1/2) log(x² + y²) = C.
7. Use y = 1 when x = 1: tan⁻¹1 + (1/2) log 2 = C, so C = π/4 + (1/2) log 2.
8. Multiply by 2: log(x² + y²) + 2 tan⁻¹(y/x) = π/2 + log 2.

Answerlog(x² + y²) + 2 tan⁻¹(y/x) = π/2 + log 2

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1. Write dy/dx = −(xy + y²)/x², homogeneous of degree 0.
2. Put y = vx, so dy/dx = v + x(dv/dx).
3. Substituting and simplifying: x(dv/dx) = −v(v + 2).
4. Separate: dv/[v(v+2)] = −dx/x, and split the left side into (1/2)[1/v − 1/(v+2)].
5. Integrate: (1/2)log[v/(v+2)] = −log|x| + C₁, so (v+2)/v = Ax².
6. Put v = y/x back and simplify: (y + 2x)/(x²y) = A.
7. Use y = 1 at x = 1: (1 + 2)/(1) = 3, so A = 3.
8. So the particular solution is (y + 2x)/(x²y) = 3, i.e. y + 2x = 3x²y.

Answery + 2x = 3x²y

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1. Write dy/dx = y/x − sin²(y/x), homogeneous of degree 0.
2. Put y = vx, so dy/dx = v + x(dv/dx).
3. Substituting: v + x(dv/dx) = v − sin²(v), so x(dv/dx) = −sin²(v).
4. Separate: dv/sin²(v) = −dx/x, i.e. cosec²(v) dv = −dx/x.
5. Integrate: −cot(v) = −log|x| + C₁, so cot(v) = log|x| + C.
6. Put v = y/x back: cot(y/x) − log|x| = C.
7. Use y = π/4 at x = 1: cot(π/4) − log 1 = 1 − 0 = 1, so C = 1.
8. So the particular solution is cot(y/x) − log|x| = 1.

Answercot(y/x) − log|x| = 1

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1. Solve: dy/dx − y/x + cosec(y/x) = 0 ; y = 0 when x = 1.
2. Write the equation as dy/dx = y/x − cosec(y/x); the right side depends only on the ratio y/x, so the equation is homogeneous.
3. Substitute y = vx, so dy/dx = v + x·(dv/dx).
4. The equation becomes v + x·(dv/dx) = v − cosec v, which simplifies to x·(dv/dx) = −cosec v.
5. Separate the variables: sin v dv = −dx/x.
6. Integrate both sides: −cos v = −log x + C, i.e. cos v = log x + C₁ (writing C₁ = −C).
7. Replace v by y/x: cos(y/x) = log x + C₁.
8. Use y = 0 when x = 1: cos 0 = log 1 + C₁, so 1 = C₁.

Answercos(y/x) = log x + 1

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1. Solve: 2xy + y² − 2x² dy/dx = 0 ; y = 2 when x = 1.
2. Write the equation as dy/dx = (2xy + y²)/(2x²) = y/x + ½(y/x)²; the right side depends only on y/x, so it is homogeneous.
3. Substitute y = vx, so dy/dx = v + x·(dv/dx).
4. The equation becomes v + x·(dv/dx) = v + v²/2, so x·(dv/dx) = v²/2.
5. Separate the variables: dv/v² = dx/(2x).
6. Integrate both sides: −1/v = ½ log x + C.
7. Replace v by y/x (so 1/v = x/y): −x/y = ½ log x + C.
8. Use y = 2 when x = 1: −½ = 0 + C, so C = −½.
9. Substitute back and multiply by −2: 2x/y = 1 − log x.

Answery(1 − log x) = 2x

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1. Given: dx/dy = h(x/y).
2. The right side is a function of the ratio x/y, so make that ratio the new unknown: put x/y = v, that is, x = vy.
3. Then dx/dy = v + y·(dv/dy), and the equation becomes v + y·(dv/dy) = h(v), that is, dv/(h(v) − v) = dy/y. The variables are now separated.
4. (A) y = vx is the substitution the book uses for dy/dx = F(y/x), where the ratio is y/x. (B) v = yx and (D) x = v do not turn x/y into a single new letter, so h(x/y) would still involve both letters.

Answer(C) x = vy

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1. Scale x and y by the same factor t in each pair of coefficients and see whether both scale by the same power of t.
2. In (A), 4x+6y+5 and 3y+2x+4 carry a stray constant, so they do not scale as a pure power of t.
3. In (B), xy scales as t² but x³+y³ scales as t³ — different powers.
4. In (C), x³+2y² mixes a degree-3 term with a degree-2 term in the same bracket, so it has no single degree.
5. In (D), y² and x²−xy−y² both scale as t² — the same degree — so this equation is homogeneous.

Answer(D) y² dx + (x² − xy − y²) dy = 0

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