Question 2 of 8: General Solution of a Linear First-Order ODE via Integrating Factor

Question 2: General Solution of a Linear First-Order ODE via Integrating Factor (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $y'+2xy=e^{-x^2}\sec(2x)$.

Find. The general solution $y(x)$.

Approach. Use the integrating factor $\mu=e^{\int2x\,dx}=e^{x^2}$, which is engineered to exactly cancel the $e^{-x^2}$ factor in the forcing, leaving an elementary secant integral.

1. Integrating factor. $\mu(x)=e^{\int2x\,dx}=e^{x^2}$. Multiplying through, $$\left(e^{x^2}y\right)'=e^{x^2}\cdot e^{-x^2}\sec(2x)=\sec(2x).$$
2. Integrate. $\displaystyle\int\sec(2x)\,dx=\tfrac12\ln\left|\sec2x+\tan2x\right|+C$, so $$e^{x^2}y=\tfrac12\ln\left|\sec2x+\tan2x\right|+C.$$
3. Solve for $y$. $$y(x)=e^{-x^2}\left[\tfrac12\ln\left|\sec2x+\tan2x\right|+C\right].$$

$$y(x)=\boxed{e^{-x^2}\left[\tfrac12\ln\left|\sec2x+\tan2x\right|+C\right]}$$