工程數學問題 (如何解 通解與特解)

1.y" + 4y = sec(x)Ans:特徵方程式為:D^2+4=0 => D=+-2j得到齊次方程式解為:yh(x)=c1*cos(2x)+c2*sin(2x)使用參變數法最簡單,設特殊解為: yp(x)=A*y1+B*y2=A*cos(2x)+B*sin(2x)yp'=-2A*sin(2x)+2B*cos(2x)A=-∫y2*f(x)dx/(y1*y2'-y2*y1')=-∫sin(2x)*sec(x)/(cos^2(2x)+sin^2(2x))=-∫2sin(x)cos(x)dx/cos(x)=-∫2sin(x)dx=2cos(x)B=∫y1*f(x)dx/(y1*y2'-y2*y1')=∫cos(2x)sec(x)dx/1=∫[2cos^2(x)-1]sec(x)dx=∫2cos(x)dx-∫dx/cos(x)=2sin(x)-∫cos(x)dx/cos^2(x)=2sin(x)-∫d(sinx)/(1-sin^2 x)=2sin(x)-∫du/(1+u)(1-u).....u=sinx=2sin(x)-∫du/2(1+u)-∫du/2(1-u)=2sin(x)-ln[(1+u)(1-u)]/2=2sin(x)-ln(1-u^2)/2=2sin(x)-(1/2)*ln(cos^2x)=2sin(x)-ln(cos(x))得到特殊方程式解為:yp(x)=A*cos(2x)+B*sin(2x)=2cos(x)cos(2x)+[2sin(x)-ln(cos(x))]sin(2x)通解為:y(x)=yh(x)+yp(x)=[c1*cos(2x)+c2*sin(2x)]+2cos(x)cos(2x)+[2sin(x)-ln(cos(x))]sin(2x)

2.y" - (4/x)y' + (4y/x^2) = (x^2) +1Ans:x^2*y"-4xy'+4y=x^4+x^2Let y=x^m, y'=mx^(m-1), y"=m(m-1)x^(m-2)代入原式裡面:0=m(m-1)-4m+4=m^2-5m+3m=(5+-√13)/2得到齊次方程式解為:yh(x)=c1*x^(5+√13)/2+c2*x^(5-√13)/2使用係數待定法,設特殊解為: yp(x)=ax^4+bx^3+cx^2+fx+gyp'=4ax^3+3bx^2+2cx+fyp"=12ax^2+6bx+2c代入原式裡面:x^4+x^2=x^2(12ax^2+6bx+2c)-4x(4ax^3+3bx^2+2cx+f)+4(ax^4+bx^3+cx^2+fx+g)=12ax^4-2bx^3-2cx^2+4ga=1/12, b=-1/2, c=g=0得到特殊解為:yp(x)=x^4/12-x^3/2+fx通解為:y(x)=yh(x)+yp(x)=[c1*x^(5+√13)/2+c2*x^(5-√13)/2]+(x^4/12-x^3/2+fx)

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