Author Topic: TUT0801  (Read 433 times)

suyichen

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TUT0801
« on: October 18, 2019, 02:00:01 PM »
Find the solution for $t^{2} y^{\prime \prime}+t y^{\prime}+y=0$

$\displaystyle t^{2} \frac{d y}{d t^{2}}+\alpha t \frac{d y}{d t}+\beta y=0$ is the general form of
Euler equation.

So , we assume $x=\ln t$

$$
\begin{array}{l}{\displaystyle\frac{d y}{d t}=\frac{d y}{d x} \cdot \frac{d x}{d t}=\frac{d y}{d x} \cdot \frac{1}{t}} \\ {\displaystyle\frac{d^{2} y}{d t^{2}}=\frac{1}{t^{2}}\left(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}\right)}\end{array}
$$

$$
\begin{array}{l}{\therefore t^{2}\left[\frac{1}{t^{2}}\left(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}\right)\right]+t\left(\frac{1}{t} \frac{d y}{d x}\right)+y=0} \\ {\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0}\end{array}
$$

$$
\therefore \frac{d y}{d x^{2}}+y=0
$$

$$
\begin{array}{l}{\Rightarrow r^{2}+1=0} \\ {r=\sqrt{-1}} \\ {r=\pm 1 i} \\ {x=0 \quad \mu=1}\end{array}
$$

$$
\therefore y(x)=c_{1} e^{0} \cos x+c_{2} e^{0} \sin x
$$

Then, substitute $x=\ln t$.

The solution for the equation is

$$
y(t)=c_{1} \cos (\ln t)+c_{2} \sin (\ln t).
$$