# 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.

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.)

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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