Toronto Math Forum

Correct me, if I am wrong, but I think the solution should be following:

Solution:

 $\text{According to Problem 34, if we have an Euler equation in the form of } t^2 y''+aty' + by = 0 \\
 \text{Then by changing variables } x = \ln(t) \text{, we can transform the equation into } \frac{d^2y}{dx^2} + (a-1)\frac{dy}{dx} + by = 0$
\begin{align*}
 \text{In our case, } t^2 y''+3ty' + 1.25y = 0 \text{ is a Euler eqaution where a=3, b=1.25 } \\
 \text{Then by changing variables } x = \ln(t), \\
 \text{we can transform the equation into } y(x)'' + (3-1)y(x)' + 1.25y = 0\\
 Then \ y(x)'' +2y(x)'+ 1.25y = 0 \\
 \text{To solve }y(x)'' +2y(x)'+ 1.25y = 0, \text{assume } y =e^{rx} \text{ is a solution,}\\
 \text{and we will have } y'=re^{rx}, y''&=r^2e^{rx}\\
 Then \ y(x)'' +2y(x)'+ 1.25y = 0 &\Rightarrow r^2e^{rx}+2re^{rx} +1.25e^{rx}=0\\
 e^{rx}(r^2+2r+1.25) &= 0\\
 \text{Since }e^{rx} \neq 0, \text{then } r^2+2r+1.25 = 0\\
 r &= \frac{-2 \pm \sqrt{4-(4 \times 1.25)}}{2} = -1 \pm \frac{1}{2}i  \text{ where }\lambda =-1 \ and \ \mu =\frac{1}{2} \\
 y(x) &= c_{1} e^{\lambda x} cos(\mu x) + c_{2} e^{\lambda x} sin(\mu x) \\
 &= c_{1} e^{-x} cos(\frac{x}{2}) + c_{2} e^{-x} sin(\frac{x}{2})\\
 &=  e^{-x}(c_{1}cos(\frac{x}{2}) + c_{2} sin(\frac{x}{2}) \\
 \text{Since we let }x = \ln(t) \text{ at the beginning, }\\
 \text{Then we will have, } \\
 y(t) &= e^{- \ln(t)}(c_{1}cos(\frac{ \ln(t)}{2}) + c_{2} sin(\frac{ \ln(t)}{2}) \\
 &= t^{-1}(c_{1}cos(\frac{ \ln(t)}{2}) + c_{2} sin(\frac{ \ln(t)}{2})\\
 &\text{as the general solution.} \\
\end{align*}

if I am wrong, please correct me.

\begin{align*}
\ r^{2} + 3r = 0 \ should \ be \\
\ r(r+3) &= 0 \\
\ r_{1} = 0 ,& \ r_{2} = -3 \\
\ so \ the \ general \ solution \ should \ be \\
\ y &= c_{1} + c_{2} e^{-3t} \\
\ since \ y(0) = -2 \\
\ c_{1}+ c_{2} e^{0} &= -2 \\
\ c_{1}+ c_{2}  &= -2 \\
\ since \ y'(0) = 3 \\
\ y' &= -3c_{2} e^{-3t} \\
\ y' &= -3c_{2} e^{0} \\
\ -3c_{2} &= 3 \\
\ so \ c_{1} =-1 ,& \ c_{2} = -1 \\
\ so \ the \  solution \ will \ be \\
\ y &= -1 - e^{-3t}
\end{align*}

Find the solution of the initial value problem
\begin{align*}
\ 2y”- 3y’+y=0, \ y(0)=2, \ y’(0)=\frac{1}{2} \\
\end{align*}
Solution:
\begin{align*}
\ Assume \ y=e^{rt} \ a \ solution \ of \ the \ equation. \\
\ Then \ we \ will \ have \\
\ y&=e^{rt} \\
\ y’&=re^{rt} \\
\ y”&=r^{2}e^{rt} \\
\ Then \ 2y”-3y’+y=0 \ will \ be \\
\ 2(r^{2}e^{rt})-3(re^{rt})+e^{rt}&=0 \\
\ e^{re}(2r^{2}-3r+1)&=0 \ where \ e^{rt} \neq 0 \\
\ 2r^2-3r+1 &=0 \\
\ (2r-1)(r-1)&=0\\
\ r_{1}=\frac{1}{2}&,\ r_{2}=1 \\
\ Then \ we \ will \ have \ the \ general \ solution \\
\ y=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}&=c_{1}e^{\frac{1}{2}t}+c_{2}e^{t} \\
\ Since \ y(0)=2 \\
\ c_{1}e^0 + c_{2}e^0 &= 2 \\
\ c_{1} + c_{2} &=2 \\
\ Since \ y’(0)=\frac{1}{2} \\
\ y’&=\frac{1}{2}c_{1}e^{\frac{1}{2}t}+c_{2}e^t \\
\ \frac{1}{2}c_{1}e^0 + c_{2}e^0 &= \frac{1}{2}\\
\ \frac{1}{2}c_{1}+c_{2}&= \frac{1}{2}\\
\ So \ solve \ for \ c_{1}, \ c_{2}\ to \ get: \\
\ c_{1}= 3, \ c_{2}=-1\\
\ So \ the \ solution \ will \ be\ y&=3e^{\frac{1}{2}t}-e^{t} \\
\end{align*}