題目 Problem (N) Solve the differential equation tdydt+2y=ett\frac{\mathrm{d}y}{\mathrm{d}t}+2y=e^ttdtdy+2y=et with the initial condition y(1)=2y(1)=2y(1)=2. 解答 解法一 在 t>0t>0t>0 上先除以 ttt： dydt+2ty=exp(t)t.\frac{\mathrm{d}y}{\mathrm{d}t}+\frac2t y=\frac{\exp(t)}t.dtdy+t2y=texp(t). 積分因子為 t2t^2t2，所以 ddt(t2y)=texp(t).\frac{\mathrm{d}}{\mathrm{d}t}(t^2y)=t\exp(t).dtd(t2y)=texp(t). 積分並代入 y(1)=2y(1)=2y(1)=2： t2y=exp(t)(t−1)+C,C=2.t^2y=\exp(t)(t-1)+C,\qquad C=2.t2y=exp(t)(t−1)+C,C=2. 故解為 y=exp(t)(t−1)+2t2.\boxed{y=\frac{\exp(t)(t-1)+2}{t^2}}.y=t2exp(t)(t−1)+2.
