宣佑 asked in 科學數學 · 1 decade ago

跪求!!工程數學解答 希望有詳解 20點

Chapter 2 Second-Order Linear ODEs

2.6 Existence and Uniqueness of Solutions. Wronskian

Bases of solutions. Corresponding ODEs. Wronskians

Find an ODE (1) for which the given functions are solutions. Show linear independence (a) by considering quotients, (b) by Theorem 2.

第一題 x0.25 , x0.25 lnx 因為打不出來 解釋一下 0.25都是X的次方

第二題 cos( 2 lnx) , sin( 2 inx)

Chapter 2 Second-Order Linear ODEs

2.7 Nonhomogeneous ODEs

General solutions of nonhomogeneous ODEs

Find a (real) general solution. Which rule are you using? (Show each step of your calculation.)

第一題 y'' + 3y' + 2y = 30e(2x) 因為打不出來 解釋一下(2x)是30的次方

第二題 y'' + 6y' + 73y = 80e(x) cos4x 因為打不出來 解釋一下(x)是80e的次方

1 Answer

Rating
  • Chaos
    Lv 7
    1 decade ago
    Favorite Answer

    因為看不懂你的式子 解釋如何解如下

    令二個解分別為y1, y2,Wronskians方法是看下面矩陣行列式是否為零 來決定其是否線性相依或獨立

    | y1 y2 |

    W= | | = y1*y2'-y2*y1'

    | y1' y2' |

    若你式子是y1=x^(0.25) 那y1'=0.25*x^(-0.75)

    y2=(x^0.25)* lnx 那y2'=0.25*x^(-0.75)*lnx+=(x^0.25)/x

    帶入上式求即可

    y'' + 3y' + 2y = 30e(2x)

    利用特徵方程式先解齊次解yh=c1*e(-1x)+c2*e(-2x)

    利用未定係數法求特別解yp=a*e(2x)

    yp'=2a*e(2x); yp"=4a*e(2x) 代回原式

    4a*e(2x)+3*[2a*e(2x)]+2* a*e(2x)=30*e(2x)

    ==> 12*a*e(2x)=30*e(2x) ==> a=2.5

    因此general solution is y=yh+yp= c1*e(-1x)+c2*e(-2x)+2.5*e(2x)

    同理解第二題

Still have questions? Get your answers by asking now.