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Differential Equations · Question of the Day

Differential Equations Question of the Day

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Friday, October 23, 2026

The differential equation (x2+1)y′+2xy=cos⁡(x)(x^2 + 1)y' + 2xy = \cos(x) is a first-order linear equation. What is its general solution for y(x)y(x)?

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The differential equation (x2+1)y′+2xy=cos⁡(x)(x^2 + 1)y' + 2xy = \cos(x) is a first-order linear equation. What is its general solution for y(x)y(x)?

  1. y=−sin⁡(x)+Cx2+1y = \frac{-\sin(x) + C}{x^2 + 1}
  2. y=sin⁡(x)x2+1+Cy = \frac{\sin(x)}{x^2 + 1} + C
  3. y=sin⁡(x)+Cx2+1y = \sin(x) + \frac{C}{x^2+1}
  4. y=sin⁡(x)+Cx2+1y = \frac{\sin(x) + C}{x^2 + 1} (correct answer)

Explanation: First, put the equation in standard form y′+P(x)y=Q(x)y' + P(x)y = Q(x) by dividing by (x2+1)(x^2 + 1): y′+2xx2+1y=cos⁡(x)x2+1y' + \frac{2x}{x^2+1}y = \frac{\cos(x)}{x^2+1}. The integrating factor is μ(x)=e∫2xx2+1dx=eln⁡(x2+1)=x2+1\mu(x) = e^{\int \frac{2x}{x^2+1} dx} = e^{\ln(x^2+1)} = x^2+1. Multiplying the standard form equation by μ(x)\mu(x) gives d/dx((x2+1)y)=cos⁡(x)d/dx((x^2+1)y) = \cos(x). Integrating both sides yields (x2+1)y=∫cos⁡(x)dx=sin⁡(x)+C(x^2+1)y = \int \cos(x) dx = \sin(x) + C. Finally, solving for yy gives y=sin⁡(x)+Cx2+1y = \frac{\sin(x) + C}{x^2 + 1}.