```python
import loader
from sympy import *
init_printing()
from root.solver import *
```

#### Given that $y_1 = \frac{4}{t}$ is a solution of the differential equation $t^2 y'' + 3t y' + y = 0$, find the other solution using reduction of order.


```python
t = Symbol('t')
y, p = red_order(4 / t, *to_std(t**2, 3*t, 1), 0, t)
p.display_ipython()
```


$\displaystyle \text{Solution }y_2\text{ takes the form }y_2 = v(t)y_1$



$\displaystyle y_{2} = \frac{4 v{\left (t \right )}}{t}$



$\displaystyle \text{Calculate the derivatives}$



$\displaystyle \begin{align*}&\frac{d}{d t} \frac{4 v{\left (t \right )}}{t} = \frac{4}{t} \frac{d}{d t} v{\left (t \right )} - \frac{4 v{\left (t \right )}}{t^{2}}\\&\frac{d^{2}}{d t^{2}} \frac{4 v{\left (t \right )}}{t} = \frac{4}{t} \frac{d^{2}}{d t^{2}} v{\left (t \right )} - \frac{8}{t^{2}} \frac{d}{d t} v{\left (t \right )} + \frac{8 v{\left (t \right )}}{t^{3}}\\\end{align*}$



$\displaystyle \text{Plug in the derivatives and simplify LHS of }y'' + p(t)y' + q(t)y = g(t)$



$\displaystyle \begin{align*}&\frac{3}{t} \left(\frac{4}{t} \frac{d}{d t} v{\left (t \right )} - \frac{4 v{\left (t \right )}}{t^{2}}\right) + \frac{4}{t} \frac{d^{2}}{d t^{2}} v{\left (t \right )} - \frac{8}{t^{2}} \frac{d}{d t} v{\left (t \right )} + \frac{12 v{\left (t \right )}}{t^{3}} = 0\\&\frac{4}{t} \frac{d^{2}}{d t^{2}} v{\left (t \right )} + \frac{4}{t^{2}} \frac{d}{d t} v{\left (t \right )} = 0\\\end{align*}$



$\displaystyle \text{Given that }y_1(t)\text{ satisfies the homogeneous equation } y_1''(t) + p(t)y_1' + q(t)y_1 = 0$



$\displaystyle \frac{4}{t} \frac{d^{2}}{d t^{2}} v{\left (t \right )} + \frac{4}{t^{2}} \frac{d}{d t} v{\left (t \right )} = 0$



$\displaystyle \text{Let }w(t) = v'(t)\text{ and convert to standard form}$



$\displaystyle \begin{align*}&\frac{4}{t} \frac{d}{d t} w{\left (t \right )} + \frac{4 w{\left (t \right )}}{t^{2}} = 0\\&\frac{d}{d t} w{\left (t \right )} + \frac{w{\left (t \right )}}{t} = 0\\\end{align*}$



$\displaystyle \text{Solve the first order linear differential equation in }w(t)$



$\displaystyle w{\left (t \right )} = \frac{C_{1}}{t}$



$\displaystyle v(t) = \int \frac{C_{1}}{t}\, dt = C_{1} \ln{\left (t \right )} + C_{2}$



$\displaystyle y_{2} = \frac{4 C_{1}}{t} \ln{\left (t \right )} + \frac{4 C_{2}}{t} = \frac{1}{t} \left(C_{1} \ln{\left (t \right )} + C_{2}\right)$


#### Given that $y_1 = e^{-\frac{b}{2a}}$ is a solution of the differential equation $a y'' + b y' + c y = 0$, find the other solution using reduction of order.


```python
a, b, c, t = symbols('a b c t', real=True)
y1 = exp(-b / (2*a) * t)
y, p = red_order(y1, *to_std(a, b, c), 0, t)
p.display_ipython()
```


$\displaystyle \text{Solution }y_2\text{ takes the form }y_2 = v(t)y_1$



$\displaystyle y_{2} = v{\left (t \right )} e^{- \frac{b t}{2 a}}$



$\displaystyle \text{Calculate the derivatives}$



$\displaystyle \begin{align*}&\frac{\partial}{\partial t} v{\left (t \right )} e^{- \frac{b t}{2 a}} = e^{- \frac{b t}{2 a}} \frac{d}{d t} v{\left (t \right )} - \frac{b v{\left (t \right )} e^{- \frac{b t}{2 a}}}{2 a}\\&\frac{\partial^{2}}{\partial t^{2}} v{\left (t \right )} e^{- \frac{b t}{2 a}} = e^{- \frac{b t}{2 a}} \frac{d^{2}}{d t^{2}} v{\left (t \right )} - \frac{b e^{- \frac{b t}{2 a}}}{a} \frac{d}{d t} v{\left (t \right )} + \frac{b^{2} v{\left (t \right )} e^{- \frac{b t}{2 a}}}{4 a^{2}}\\\end{align*}$



$\displaystyle \text{Plug in the derivatives and simplify LHS of }y'' + p(t)y' + q(t)y = g(t)$



$\displaystyle \begin{align*}&e^{- \frac{b t}{2 a}} \frac{d^{2}}{d t^{2}} v{\left (t \right )} + \frac{b}{a} \left(e^{- \frac{b t}{2 a}} \frac{d}{d t} v{\left (t \right )} - \frac{b v{\left (t \right )} e^{- \frac{b t}{2 a}}}{2 a}\right) - \frac{b e^{- \frac{b t}{2 a}}}{a} \frac{d}{d t} v{\left (t \right )} + \frac{c e^{- \frac{b t}{2 a}}}{a} v{\left (t \right )} + \frac{b^{2} v{\left (t \right )} e^{- \frac{b t}{2 a}}}{4 a^{2}} = 0\\&\left(\frac{c e^{- \frac{b t}{2 a}}}{a} - \frac{b^{2} e^{- \frac{b t}{2 a}}}{4 a^{2}}\right) v{\left (t \right )} + e^{- \frac{b t}{2 a}} \frac{d^{2}}{d t^{2}} v{\left (t \right )} = 0\\\end{align*}$



$\displaystyle \text{Given that }y_1(t)\text{ satisfies the homogeneous equation } y_1''(t) + p(t)y_1' + q(t)y_1 = 0$



$\displaystyle e^{- \frac{b t}{2 a}} \frac{d^{2}}{d t^{2}} v{\left (t \right )} = 0$



$\displaystyle \text{Let }w(t) = v'(t)\text{ and convert to standard form}$



$\displaystyle \begin{align*}&e^{- \frac{b t}{2 a}} \frac{d}{d t} w{\left (t \right )} = 0\\&\frac{d}{d t} w{\left (t \right )} = 0\\\end{align*}$



$\displaystyle \text{Solve the first order linear differential equation in }w(t)$



$\displaystyle w{\left (t \right )} = C_{1}$



$\displaystyle v(t) = \int C_{1}\, dt = C_{1} t + C_{2}$



$\displaystyle y_{2} = C_{1} t e^{- \frac{b t}{2 a}} + C_{2} e^{- \frac{b t}{2 a}} = \left(C_{1} t + C_{2}\right) e^{- \frac{b t}{2 a}}$


6ecd3ab2d29bc19dafb28e919330256520cbeb4a
ipynb
Jupyter Notebook
notebooks/reduction-of-order.ipynb
kaiyingshan/ode-solver
30c6798efe9c35a088b2c6043493470701641042
MIT
2019-02-17T23:15:20.000Z
2019-02-17T23:15:27.000Z
notebooks/reduction-of-order.ipynb
kaiyingshan/ode-solver
30c6798efe9c35a088b2c6043493470701641042
MIT
notebooks/reduction-of-order.ipynb
kaiyingshan/ode-solver
30c6798efe9c35a088b2c6043493470701641042
MIT
Qwen/Qwen-72B
1. YES
2. YES