一題微分方程

Question

dv/dt=Acoswt-Bv

其中A B w為正實數常數

求解謝謝

Answer 1

最佳解答

dv/dt=Acoswt-Bv,其中A B w為正實數常數,求解謝謝v'+B*v=A*cos(wt)積分因子: ln(F)=∫B*dt=B*tF=e^(Bt)[v*e^(Bt)]'=A*cos(wt)*e^(Bt)v*e^(Bt)-c=A∫cos(wt)*e^(Bt)*dt=aa積分演算如下:a=A/B*∫cos(wt)*d[e^(Bt)]=A/B*{cos(wt)*e^(Bt)+w∫e^(Bt)sin(wt)dt}=A/B*{cos(wt)*e^(Bt)+w/B*[sin(wt)*e^(Bt)-w∫e^(Bt)*cos(wt)dt]}=A/B*{cos(wt)*e^(Bt)+w/B*[sin(wt)*e^(Bt)-w*a/A]}=A/B*{cos(wt)*e^(Bt)+w/B*sin(wt)*e^(Bt)-w^2*a/AB}=Ae^(Bt)/B*{cos(wt)+w/B*sin(wt)}-w^2*a/B^2a(1+w^2/B^2)=A*e^(Bt)/B*{cos(wt)+w*sin(wt)/B}a(w^2+B^2)/B^2=A*e^(Bt)/B^2*{B*cos(wt)+w*sin(wt)}a=e^(Bt)*{B*cos(wt)+w*sin(wt)}A/(w^2+B^2)代入上式: v*e^(Bt)=e^(Bt)*{B*cos(wt)+w*sin(wt)}A/(w^2+B^2)+cv={B*cos(wt)+w*sin(wt)}A/(w^2+B^2)+c/e^(Bt)......ans

Answer 2

很基本的一階常係數線性微分方程.

大一微積分教本中一階微分方程的解法,

或微分方程教本中常係數線性微分方程多種解法可用.