### Author Topic: 0501 quiz4  (Read 494 times)

#### Hongling Liu

• Jr. Member
•  • Posts: 9
• Karma: 2 ##### 0501 quiz4
« on: October 18, 2019, 02:00:51 PM »
t^2y’’ + 3ty’ + 1.25y = 0
Solution:
lnx =  t
d^2y/dx^2 + (3-1)dy/dx +5/4y = 0
r^2 + 2r + 5/4 =0
r1 = -2 + i     t2 = -2 - i
y(x) = C1⋅e^(-2)⋅cos(x) + C2⋅e^(-2)⋅sin(x)
∴y(t) = C1⋅(1/2)⋅cos(lnt) + C2⋅(1/2)⋅sin(lnt)

#### Coollight

• Jr. Member
•  • Posts: 7
• Karma: 0 ##### Re: 0501 quiz4
« Reply #1 on: October 18, 2019, 02:31:34 PM »
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*}