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First-Order Linear Differential Equations — Full Worked Solutions

1. Solve dy/dx + 3y = 12. Fluency

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   Step 1 — Standard form: dy/dx + 3y = 12.   P(x) = 3, Q(x) = 12.

   Step 2 — Integrating factor: μ = e^{∫3 dx} = e^3x.

   Step 3 — Multiply: e^3x dy/dx + 3e^3xy = 12e^3x.

   Step 4 — Recognise: d/dx[e^3xy] = 12e^3x.

   Step 5 — Integrate: e^3xy = ∫12e^3x dx = 4e^3x + C.

   Step 6 — Divide: y = 4 + Ce^−3x

   Check: dy/dx = −3Ce^−3x, so dy/dx + 3y = −3Ce^−3x + 12 + 3Ce^−3x = 12. ✓

2. Solve dy/dx − 4y = 0, given y(0) = 5. Fluency

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   Standard form: dy/dx + (−4)y = 0.   P(x) = −4, Q(x) = 0.

   Integrating factor: μ = e^{∫(−4) dx} = e^−4x.

   Multiply and recognise: d/dx[e^−4xy] = 0.

   Integrate: e^−4xy = C, so y = Ce^4x.

   Apply initial condition: y(0) = 5:   5 = Ce⁰ = C.

   y = 5e^4x

   Note: This is separable. Alternatively: dy/y = 4 dx ⇒ ln|y| = 4x + k ⇒ y = Ae^4x.

3. Solve dy/dx + y = e^x. Fluency

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   Standard form: dy/dx + y = e^x.   P(x) = 1.

   Integrating factor: μ = e^{∫1 dx} = e^x.

   Multiply: d/dx[e^xy] = e^x × e^x = e^2x.

   Integrate: e^xy = ∫e^2x dx = e^2x/2 + C.

   Divide by e^x: y = e^x/2 + Ce^−x

   Check: dy/dx = e^x/2 − Ce^−x. Then dy/dx + y = e^x/2 − Ce^−x + e^x/2 + Ce^−x = e^x. ✓

4. Solve dy/dx + (2/x)y = x³, for x > 0. Fluency

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   Standard form: dy/dx + (2/x)y = x³.   P(x) = 2/x.

   Integrating factor: μ = e^{∫(2/x) dx} = e^{2 ln x} = x² (for x > 0).

   Multiply: d/dx[x²y] = x² × x³ = x⁵.

   Integrate: x²y = ∫x⁵ dx = x⁶/6 + C.

   Divide by x²: y = x⁴/6 + C/x²

5. Solve dy/dx + 2xy = 4x, given y(0) = 3. Understanding

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   Standard form: dy/dx + 2xy = 4x.   P(x) = 2x, Q(x) = 4x.

   Integrating factor: μ = e^{∫2x dx} = e^x².

   Multiply: d/dx[e^x²y] = 4xe^x².

   Integrate (substitution u = x², du = 2x dx):

   e^x²y = ∫4xe^x² dx = 2e^x² + C.

   Divide: y = 2 + Ce^−x².

   Apply y(0) = 3: 3 = 2 + C ⇒ C = 1.

   y = 2 + e^−x²

   As x → ±∞, e^−x² → 0, so y approaches the equilibrium solution y = 2.

6. Solve x dy/dx − y = x², for x > 0. Understanding

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   Divide by x to get standard form: dy/dx − (1/x)y = x.

   P(x) = −1/x, Q(x) = x.

   Integrating factor: μ = e^{∫(−1/x) dx} = e^{−ln x} = x⁻¹ = 1/x.

   Multiply: (1/x) dy/dx − (1/x²)y = 1, which is d/dx[y/x] = 1.

   Integrate: y/x = x + C.

   y = x² + Cx

   Check: dy/dx = 2x + C. Then x(2x+C) − (x²+Cx) = 2x²+Cx−x²−Cx = x². ✓

7. Solve dy/dx + y tan x = sec x, for −π/2 < x < π/2. Understanding

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   Standard form: dy/dx + (tan x)y = sec x.   P(x) = tan x.

   Integrating factor: μ = e^{∫tan x dx}.   Since ∫tan x dx = ln|sec x|,

   μ = e^{ln|sec x|} = sec x   (positive on (−π/2, π/2)).

   Multiply: d/dx[sec x · y] = sec x × sec x = sec² x.

   Integrate: y sec x = ∫sec² x dx = tan x + C.

   Divide by sec x: y = sin x + C cos x

   Note: C cos x is the homogeneous solution (when Q = 0); sin x is the particular solution.

8. Solve dy/dx − 2y = e^3x, given y(0) = 0. Understanding

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   Standard form: dy/dx + (−2)y = e^3x.   P(x) = −2.

   Integrating factor: μ = e^{∫(−2) dx} = e^−2x.

   Multiply: d/dx[e^−2xy] = e^−2x × e^3x = e^x.

   Integrate: e^−2xy = e^x + C.

   Divide: y = e^3x + Ce^2x.

   Apply y(0) = 0: 0 = e⁰ + C ⇒ C = −1.

   y = e^3x − e^2x

   Check: dy/dx = 3e^3x − 2e^2x. Then dy/dx − 2y = 3e^3x − 2e^2x − 2e^3x + 2e^2x = e^3x. ✓

9. An RL circuit: L dI/dt + RI = V with L = 2, R = 4, V = 12, I(0) = 0. Problem Solving

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   Rewrite in standard form: dI/dt + (R/L)I = V/L  ⇒  dI/dt + 2I = 6.

   P(t) = 2, Q(t) = 6.

   Integrating factor: μ = e^{∫2 dt} = e^2t.

   Multiply: d/dt[e^2tI] = 6e^2t.

   Integrate: e^2tI = 3e^2t + C  ⇒  I = 3 + Ce^−2t.

   Apply I(0) = 0: 0 = 3 + C ⇒ C = −3.

   I(t) = 3(1 − e^−2t) amperes

   Interpretation: At t = 0, I = 0 (no current initially). As t → ∞, I → 3 A, the steady-state current given by Ohm’s law I = V/R = 12/4 = 3 A. The time constant is τ = L/R = 2/4 = 0.5 s.

10. Solve (1 + x²) dy/dx + 2xy = cos x. Problem Solving

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    Divide by (1 + x²) to get standard form:

    dy/dx + [2x/(1+x²)]y = cos x / (1+x²).

    P(x) = 2x/(1+x²).

    Integrating factor: ∫[2x/(1+x²)] dx = ln(1+x²) (since d/dx[1+x²] = 2x).

    μ = e^ln(1+x²) = 1 + x².

    Multiply: The equation becomes

    (1+x²) dy/dx + 2xy = cos x, i.e., d/dx[(1+x²)y] = cos x.

    Integrate: (1+x²)y = ∫cos x dx = sin x + C.

    y = (sin x + C) / (1 + x²)

    Note: μ = 1 + x² > 0 for all real x, so no domain restrictions apply.