Discuss the methods of undetermined coefficients and variation of parameters can be combined to solve the given DE:

 \f$y''-2y'+y=4x^2-3+\frac{1}{x} e^x\f$

The auxiliary equation  \f$m^2-2m+1=0\f$  yields  \f$ m=1,1\f$  . Therefore  \f$y_c=c_1 e^x+c_2 xe^x\f$ .

Next the presence of   \f$4x^2-3\f$  in g(x) suggests that the particular solution includes a linear polynomial and solved by undetermined coefficients.

\f$\frac{1}{x}e^x\f$  is solved by Variation of Parameters

 

Method of Undetermined Coefficients:

\f$y_{p1} = Ax^2+Bx+C\f$

\f$y'_{p1} = 2Ax+B\f$

\f$y''_{p1} = 2A\f$

\f$y''_{p1}-2y'_{p1}+y_{p1}=2A-2(2Ax)+Ax^2+Bx+C= Ax^2+(B-4A)x+2A-2B+C\f$

\f$Ax^2+(B-4A)x+2A-2B+C=4x^2-3\f$

\f$A=4, B=16, C=27\f$

Therefore  \f$y_{p1} = 4x^2+16x+27\f$

 

Variation of Parameters:

To find  \f$y_{p2}\f$

We know   \f$y_c=c_1 e^x+c_2 xe^x\f$ \f$y_1=e^x, y_2=xe^x\f$

(a) Show that the square wave function E(t) given in Figure 7.4.4 can be written \f[E(t)=\sum_{k=0}^{\infty}{(-1)^kU(t-k)}\f]

Figure 7.4.4 can be found on page 314 of our textbooks. If anyone knows how to transfer pictures/graphs from the textbook, let me know! I’ll post it in here.

The square function can be written as

\f[E(t) = \left\{\begin{matrix} 1 & {2n\leq t<2(n+1)}\\  0 & {2n+1\leq t<2n} \end{matrix}\f]

I don’t know why it’s separating like that 🙁

Sine E(t) function is for integer values of n.

If we rewrite the this using the step function, we have

\f[E(t)=1-U(t-1)+U(t-2)-U(t-3)+U(t-4)+...=U(t-0)-U(t-1)+U(t-2)-U(t-3)+U(t-4)+...\f]

Writing the above as a summation instead, we get

\f[E(t)=\sum_{k=0}^{\infty}{(-1)^kU(t-k)}\f]

\f$y_p=y_{p1} +y_{p2}\f$

Therefore the general solution for  \f$y''-2y'+y=4x^2-3+\frac{1}{x} e^x\f$ using Method of Undetermined Coefficients and Variation of Parameters is  \f$y(x)= c_1 e^x +c_2 x e^x +4x^2+16x+27-xe^x +ln x(xe^x)\f$

Functions will only be Linearly independent when they meet these requirements:

1. When there are more than one function, for it to be linearly independent, the functions must not be capable of equating to the other.

In the case of our functions mentioned in the beginning, they are linearly dependent because they are a multiple of each other, and they equate to a specific solution e.g.:

\f$e^{x+2}=e^xe^2=e^{x-3}e^5\f$