MAT244-2014F > MT

MT problem 2

(1/1)

Victor Ivrii:
Find the general solution of
\begin{equation*}
z''+ 2z' + z = e ^{- x} \ln x \ , \qquad  x > 0 \ .
\end{equation*}
Also, find a solution satisfying $\ z(1)=-2\$.

Roro Sihui Yap:
First find the solution to the homogeneous equation.

$r^2 + 2r + 1 = 0 \\$
$(r + 1)^2 = 0 \\$
$r = -1, -1 \\$

The solution to the homogeneous equation is
$z = c_1e^{-x} + c_2xe^{-x}$

Use the method of variation of parameters
let $z_p = u_1e^{-x} + u_2xe^{-x}$

\begin{gather}
u_1'e^{-x} + u_2'xe^{-x} = 0,\label{eq-1} \\
-u_1'e^{-x} + u_2'(e^{-x}-xe^{-x}) = e^{-x}\ln x  \label{eq-2} \\
\end{gather}

Add equation (\ref{eq-1}) and (\ref{eq-2})  together,
$u_2'e^{-x} = e^{-x}\ln x \\$
$u_2' = \ln x$
$u_2 = x\ln x - x$

From equation 1,
$u_1' = -xu_2'$
$u_1' = -x\ln x$
$u_1 = x^2/4 - x^2\ln x/2$

$z_p = x^2e^{-x}/4 - x^2e^{-x}\ln x/2 + x^2e^{-x}\ln x - x^2e^{-x} \\$
$z_p = -3x^2e^{-x}/4 + x^2e^{-x}\ln x/2$

The solution is $z = c_1e^{-x} + c_2xe^{-x} -3x^2e^{-x}/4 + x^2e^{-x}\ln x/2$
When x = 1, z = -2
$-2 = c_1e^{-1} + c_2e^{-1} - (3/4)e^{-1} \\$
$-2e + 3/4 = c_1 + c_2$
let $c_1 = -2e$ and $c_2 = 3/4$

The solution satisfying z(1) = -2 is
$z = -2e^{1-x} + 3xe^{-x}/4 -3x^2e^{-x}/4 + \frac{1}{2}x^2e^{-x}\ln x$

Victor Ivrii:
Very nice. Small remarks: you should write

--- Code: ---\ln x
--- End code ---
learn \label - \ref mechanism and avoid excessive vertical spacing

Shengnan Li:
can we use method on handout 8 to solve this question?
if we can, what value m should we choose?

Thanks.

Victor Ivrii:
You cannot use the method of undetermined coefficients here and thus cannot use the method of handout 8