Miscellaneous Exercise answers: Differential Equations

Open NCERT p. 333Matches NCERT’s answer

(i)

1. Given: d²y/dx² + 5x(dy/dx)² − 6y = log x.
2. The highest derivative present is d²y/dx² (a second derivative), so the order is 2.
3. The equation is a polynomial in its derivatives, and the highest derivative d²y/dx² appears to the power 1, so the degree is 1.

AnswerOrder 2, degree 1.

(ii)

1. Given: (dy/dx)³ − 4(dy/dx)² + 7y = sin x.
2. The only derivative present is dy/dx (a first derivative), so the order is 1.
3. The equation is a polynomial in dy/dx. It appears cubed and squared; the highest power is 3, so the degree is 3.

AnswerOrder 1, degree 3.

(iii)

1. Given: d⁴y/dx⁴ − sin(d³y/dx³) = 0.
2. The highest derivative present is d⁴y/dx⁴, so the order is 4.
3. The equation contains sin(d³y/dx³) — a derivative inside a sine — so it is not a polynomial in its derivatives, and the degree is not defined.

AnswerOrder 4, degree not defined.

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Open NCERT p. 334One way to think about it

(i) xy = a eˣ + b e⁻ˣ + x²

1. Differentiate xy = a eˣ + b e⁻ˣ + x² once: y + x·dy/dx = a eˣ − b e⁻ˣ + 2x.
2. Differentiate again: 2·dy/dx + x·d²y/dx² = a eˣ + b e⁻ˣ + 2.
3. From the given function, a eˣ + b e⁻ˣ = xy − x². Substitute: x·d²y/dx² + 2·dy/dx = xy − x² + 2.
4. Rearranged: x·d²y/dx² + 2·dy/dx − xy + x² − 2 = 0, exactly the given equation.

In shortVerified: the function satisfies the equation.

(ii)

1. Given: y = eˣ (a cos x + b sin x).
2. Differentiate y = eˣ(a cos x + b sin x) by the product rule: dy/dx = eˣ(a cos x + b sin x) + eˣ(b cos x − a sin x) = y + eˣ(b cos x − a sin x).
3. So eˣ(b cos x − a sin x) = dy/dx − y.
4. Differentiate again: d²y/dx² = dy/dx + eˣ(b cos x − a sin x) + eˣ(−b sin x − a cos x) = dy/dx + (dy/dx − y) − y.
5. So d²y/dx² = 2·dy/dx − 2y, i.e. d²y/dx² − 2·dy/dx + 2y = 0, exactly the given equation.

In shortVerified: the function satisfies the equation.

(iii) y = x sin 3x

1. Differentiate y = x sin 3x: dy/dx = sin 3x + 3x cos 3x.
2. Differentiate again: d²y/dx² = 3 cos 3x + 3 cos 3x − 9x sin 3x = 6 cos 3x − 9y.
3. So d²y/dx² + 9y − 6 cos 3x = 0, exactly the given equation.

In shortVerified: the function satisfies the equation.

(iv) x² = 2y² log y

1. Differentiate x² = 2y² log y with respect to x: 2x = (4y log y + 2y)·dy/dx = 2y(2 log y + 1)·dy/dx, so dy/dx = x / [y(2 log y + 1)].
2. From the given relation, x² + y² = 2y² log y + y² = y²(2 log y + 1).
3. So (x² + y²)·dy/dx = y²(2 log y + 1) · x / [y(2 log y + 1)] = xy.
4. Hence (x² + y²)·dy/dx − xy = 0, exactly the given equation.

In shortVerified: the function satisfies the equation.

Watch this explained “Testing a candidate”, 7:39 into What makes an equation differential, and why the answer is a curve not a number

Open NCERT p. 334One way to think about it

1. Given: x² − y² = c(x² + y²)² and the differential equation (x³ − 3xy²) dx = (y³ − 3x²y) dy.
2. Differentiate x² − y² = c(x² + y²)² with respect to x, treating c as a constant: 2x − 2y·dy/dx = 2c(x² + y²)(2x + 2y·dy/dx), i.e. x − y·dy/dx = 2c(x² + y²)(x + y·dy/dx).
3. Eliminate c. From the given relation, c = (x² − y²)/(x² + y²)². Substituting: x − y·dy/dx = 2(x² − y²)(x + y·dy/dx)/(x² + y²).
4. Multiply by (x² + y²): (x − y·dy/dx)(x² + y²) = 2(x² − y²)(x + y·dy/dx).
5. Expand and collect the dy/dx terms on one side: x³ + xy² − 2x³ + 2xy² = y·dy/dx·(x² + y² + 2x² − 2y²), i.e. 3xy² − x³ = (3x²y − y³)·dy/dx.
6. Multiply both sides by −1: x³ − 3xy² = (y³ − 3x²y)·dy/dx, which is (x³ − 3xy²) dx = (y³ − 3x²y) dy — the given equation.
7. So the relation is a solution. It carries one arbitrary constant, c, and the equation is of first order, so it is the general solution.

In shortProved: x² − y² = c(x² + y²)² satisfies the equation and carries one arbitrary constant, so it is the general solution.

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Open NCERT p. 334Matches NCERT’s answer

1. Given: dy/dx + √((1 − y²)/(1 − x²)) = 0.
2. Write the equation as dy/√(1−y²) = −dx/√(1−x²); the variables are already separated.
3. Integrate both sides: sin⁻¹y = −sin⁻¹x + C.
4. Rearranged, this is sin⁻¹x + sin⁻¹y = C.

Answersin⁻¹x + sin⁻¹y = C.

Watch this explained “The shortest complete one”, 7:36 into Pushing all the y's one side and all the x's the other, then integrating

Open NCERT p. 334One way to think about it

1. Separate the variables: dy/(y² + y + 1) = −dx/(x² + x + 1).
2. Complete the square: y² + y + 1 = (y + ½)² + (√3/2)², and the same for x. Using ∫du/(u² + k²) = (1/k) tan⁻¹(u/k): (2/√3) tan⁻¹((2y + 1)/√3) = −(2/√3) tan⁻¹((2x + 1)/√3) + C₁.
3. Multiply by √3/2 and rename the constant: tan⁻¹((2x + 1)/√3) + tan⁻¹((2y + 1)/√3) = C.
4. Take tan of both sides, using tan(P + Q) = (tan P + tan Q)/(1 − tan P·tan Q) with tan P = (2x + 1)/√3 and tan Q = (2y + 1)/√3: [(2x + 1)/√3 + (2y + 1)/√3] / [1 − (2x + 1)(2y + 1)/3] = tan C.
5. Simplify: the top is 2(x + y + 1)/√3 and the bottom is (3 − 4xy − 2x − 2y − 1)/3 = 2(1 − x − y − 2xy)/3, so the left side is √3(x + y + 1)/(1 − x − y − 2xy).
6. So √3(x + y + 1) = tan C·(1 − x − y − 2xy), i.e. x + y + 1 = A(1 − x − y − 2xy), where A = (tan C)/√3 is a parameter.

In shortShown: the general solution is (x + y + 1) = A(1 − x − y − 2xy).

Watch this explained “The shortest complete one”, 7:36 into Pushing all the y's one side and all the x's the other, then integrating

Open NCERT p. 334Matches NCERT’s answer

1. Separate the variables by dividing by cos x cos y: tan x dx + tan y dy = 0.
2. Integrate: −log|cos x| − log|cos y| = C₁, i.e. cos x·cos y = C (a constant).
3. Use the point (0, π/4): cos 0·cos(π/4) = 1×(√2/2) = √2/2, so C = √2/2.

AnswerThe curve is cos x·cos y = √2/2 (equivalently √2·cos x·cos y = 1).

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Open NCERT p. 334Matches NCERT’s answer

1. Given: (1 + e²ˣ) dy + (1 + y²) eˣ dx = 0.
2. Separate the variables: dy/(1+y²) = −eˣdx/(1+e²ˣ).
3. The left side integrates to tan⁻¹y. For the right side, substitute u = eˣ (du = eˣdx): −∫du/(1+u²) = −tan⁻¹(eˣ).
4. So tan⁻¹y + tan⁻¹(eˣ) = C.
5. Use y = 1 when x = 0: tan⁻¹1 + tan⁻¹1 = π/4 + π/4 = π/2, so C = π/2.

Answertan⁻¹y + tan⁻¹(eˣ) = π/2.

Watch this explained “The shortest complete one”, 7:36 into Pushing all the y's one side and all the x's the other, then integrating

Open NCERT p. 334Matches NCERT’s answer

1. Given: y e^(x/y) dx = (x e^(x/y) + y²) dy, (y ≠ 0).
2. Write the equation as dx/dy = (x e^(x/y) + y²)/(y e^(x/y)); the right side is homogeneous of degree 0, so substitute x = vy (v = x/y).
3. Then dx/dy = v + y·dv/dy, and the equation becomes v + y·dv/dy = v + y·e⁻ᵛ, i.e. dv/dy = e⁻ᵛ.
4. Separate and integrate: eᵛ dv = dy gives eᵛ = y + C.
5. Put back v = x/y: e^(x/y) = y + C.

Answere^(x/y) − y = C.

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Open NCERT p. 335Matches NCERT’s answer

1. Following the hint, let t = x − y so dt = dx − dy; also let s = x + y so ds = dx + dy.
2. The equation (x−y)(dx+dy) = dx−dy becomes t·ds = dt.
3. This separates: ds = dt/t, so s = log|t| + C, i.e. x + y = log|x − y| + C.
4. At x = 0, y = −1, we have x − y = 1 > 0, so the bars are not needed here. 0 + (−1) = log(1) + C = C, so C = −1.

Answerx + y + 1 = log(x − y).

Open NCERT p. 335Matches NCERT’s answer

1. Given: [e^(−2√x)/√x − y/√x] dx/dy = 1, (x ≠ 0).
2. Rearrange: multiplying out gives dy/dx + y/√x = e^(−2√x)/√x — a linear equation in y with P = 1/√x, Q = e^(−2√x)/√x (taking x > 0, since √x appears and x ≠ 0).
3. Integrating factor = e^{∫(1/√x) dx} = e^2√x.
4. Multiply through and integrate: y·e^2√x = ∫e^2√x·e^−2√x/√x dx = ∫dx/√x = 2√x + C.

Answery = (2√x + C)·e^−2√x.

Watch this explained “Three steps”, 8:37 into Multiplying through by an integrating factor to make the left side a single derivative

Open NCERT p. 335Matches NCERT’s answer

1. Given: dy/dx + y cot x = 4x cosec x, (x ≠ 0), with y = 0 when x = π/2.
2. This is linear: dy/dx + y cot x = 4x cosec x, with P = cot x, Q = 4x cosec x.
3. Integrating factor = e^{∫cot x dx} = e^{log(sin x)} = sin x.
4. Multiply through and integrate: y·sin x = ∫4x cosec x·sin x dx = ∫4x dx = 2x² + C.
5. Use y = 0 at x = π/2: 0 = 2(π/2)² + C = π²/2 + C, so C = −π²/2.

Answery·sin x = 2x² − π²/2.

Watch this explained “Two integrals that annihilate”, 13:05 into Multiplying through by an integrating factor to make the left side a single derivative

Open NCERT p. 335Matches NCERT’s answer

1. Given: (x + 1) dy/dx = 2e^(−y) − 1.
2. Separate the variables: dy/(2e⁻ʸ−1) = dx/(x+1).
3. Multiply the top and bottom of the left side by eʸ: eʸdy/(2−eʸ) = dx/(x+1). Substitute u = eʸ (du = eʸdy): du/(2−u) = dx/(x+1).
4. Integrate: −log|2−eʸ| = log|x+1| + C₁, i.e. (x+1)(2−eʸ) = C.
5. Use y = 0 at x = 0: (0+1)(2−1) = 1 = C.

Answer(x + 1)(2 − eʸ) = 1.

Watch this explained “A point picks a member”, 8:23 into Pushing all the y's one side and all the x's the other, then integrating

Open NCERT p. 335Matches NCERT’s answer

1. Multiply through by y: y dx − x dy = 0, so y dx = x dy, i.e. dy/dx = y/x — the variables separate directly.
2. Separate and integrate: dy/y = dx/x gives log y = log x + log C, i.e. y = Cx.

Answery = Cx — option (C).

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Open NCERT p. 335Matches NCERT’s answer

1. Given: dx/dy + P₁x = Q₁.
2. Here x is being treated as the unknown and y as the independent variable, so the equation is linear in x of the form dx/dy + P₁x = Q₁.
3. Its integrating factor is e^{∫P₁ dy} (integrated with respect to y, since we are solving for x), and multiplying through makes the left side d/dy(x·e^{∫P₁ dy}).
4. Integrating both sides gives x·e^{∫P₁ dy} = ∫(Q₁·e^{∫P₁ dy}) dy + C.

AnswerOption (C).

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Open NCERT p. 335Matches NCERT’s answer

1. Given: eˣ dy + (y eˣ + 2x) dx = 0.
2. Divide the whole equation by eˣ: dy/dx + y = −2x·e⁻ˣ — a linear equation with P = 1, Q = −2x·e⁻ˣ.
3. Integrating factor = e^{∫1 dx} = eˣ.
4. Multiply through and integrate: y·eˣ = ∫−2x·e⁻ˣ·eˣ dx = ∫−2x dx = −x² + C, i.e. y·eˣ + x² = C.

Answery·eˣ + x² = C — option (C).

Watch this explained “Three steps”, 8:37 into Multiplying through by an integrating factor to make the left side a single derivative