Differential Equations — Practice

Question 1

Solve the differential equation: dy/dx = (1 + y²) / (1 + x²), given y = 0 when x = 0.

Accepted Answer

Separate variables: dy/(1+y²) = dx/(1+x²). Integrate both sides: tan⁻¹(y) = tan⁻¹(x) + C. Apply initial condition y = 0, x = 0: tan⁻¹(0) = tan⁻¹(0) + C → 0 = 0 + C → C = 0. PARTICULAR SOLUTION: tan⁻¹(y) = tan⁻¹(x) → y = x.

Question 2

Solve: x·dy/dx − y = √(x² + y²).

Accepted Answer

Rewrite: dy/dx = [y + √(x²+y²)] / x. Check homogeneity: f(tx,ty) = [ty + √(t²x²+t²y²)]/(tx) = t[y + √(x²+y²)]/(tx) = [y + √(x²+y²)]/x. Homogeneous of degree 0. ✓ SUBSTITUTION: Let y = vx. Then dy/dx = v + x(dv/dx). Substituting: v + x(dv/dx) = [vx + √(x²+v²x²)]/x = v + √(1+v²). x(dv/dx) = √(1+v²). Separate: dv/√(1+v²) = dx/x. Integrate: ln|v + √(1+v²)| = ln|x| + ln|C| = ln|Cx|. v + √(1+v²) = Cx. Back-substitute v = y/x: y/x + √(1+y²/x²) = Cx → y/x + √(x²+y²)/x = Cx → y + √(x²+y²) = Cx². GENERAL SOLUTION: y + √(x²+y²) = Cx².

Question 3

Solve: (1 + x²) dy/dx + 2xy = cos x. Also find the particular solution when y = 0 at x = 0.

Accepted Answer

Divide by (1+x²): dy/dx + [2x/(1+x²)]·y = cos x/(1+x²). This is linear: P = 2x/(1+x²), Q = cos x/(1+x²). INTEGRATING FACTOR: IF = e^∫P dx = e^∫[2x/(1+x²)] dx = e^(ln(1+x²)) = (1+x²). Multiply both sides by IF: d/dx[(1+x²)y] = (1+x²)·cos x/(1+x²) = cos x. Integrate: (1+x²)y = ∫cos x dx = sin x + C. GENERAL SOLUTION: y = (sin x + C)/(1+x²). PARTICULAR SOLUTION: y=0 when x=0: 0 = (sin 0 + C)/(1+0) = C. So C = 0. PARTICULAR SOLUTION: y = sin x/(1+x²).

Practice — Differential Equations

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