Author Topic: TUT0801  (Read 502 times)

ZYR

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TUT0801
« on: October 19, 2019, 12:49:32 AM »
$t^2y" + ty' + y = 0$

Let $x = lnt$, $t >0$, then $\frac{\partial x}{\partial t}  = \frac{\partial }{\partial t}(lnt) = \frac{1}{t}$
Then we have $\frac{\partial y}{\partial t}  = \frac{\partial y}{\partial x} \frac{1}{t}$
$\frac{\partial^2 y}{\partial t^2} = \frac{\partial }{\partial dt}(\frac{\partial y}{\partial x} \frac{1}{t}) = \frac{1}{t^2}(\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x}) $

 When we substitute these to the original equation, we have :
 $t^2 (\frac{1}{t^2}(\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x})) + t\frac{\partial y}{\partial x} \frac{1}{t} + y = 0$
 
 $\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x} + \frac{\partial y}{\partial x} + y = 0$
 
$\frac{\partial^2 y}{\partial x} + y = 0$
Then we have a homogeneous equation, $y''+ y = 0$
$r^2 = -1$, $r = \pm i$, since $\lambda = 0$, $\mu = 1$
 Then the general solution of this differential equation $y(x) = c_1 cos(x) + c_2 sin(x)$
 And then substitute $x = lnt$, we get $y(t) = c_1 cos(lnt) + c_2 sin(lnt)$

ZYR

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Re: TUT0801
« Reply #1 on: October 19, 2019, 01:03:59 AM »
The question says that there is no need to change the variables, dose anyone can refer to one problem from section 3.1?

Yichen Ji

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Re: TUT0801
« Reply #2 on: October 19, 2019, 03:38:57 PM »
Which section does this question belong to?