```
(Reconstruction of an ODE) A group of scientists at USYD have done countless experiments,
which all point to a single homogeneous linear 2nd order ODE on I = R,
𝑦
′′ + 𝑝(𝑥)𝑦
′ + 𝑞(𝑥)𝑦 = 0.
However, they have not been able to determine the coefficients 𝑝(𝑥) and 𝑞(𝑥) of the differential
equation. Their experiments indicate that there exist solutions 𝑦1(𝑥) and 𝑦2 (𝑥) such that
• the Wronskian of 𝑦1 and 𝑦2 is identically equal to 1,
• 𝑦1 is also a solution to the 1st order differential equation 𝑦
′ + sin(𝑥)𝑦 = 0.
Can you help the scientists find the coefficients 𝑝(𝑥) and 𝑞(𝑥)?
Hint: first use Abel’s formula to find 𝑝(𝑥).
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

Given the Wronskian of $y_1$ and $y_2$ is identically equal to 1, we use Abel's formula to find $p(x)$. Abel's formula states that the Wronskian $W(y_1, y_2)$ of two solutions of the second-order linear differential equation $y'' + p(x)y' + q(x)y = 0$ is given by $W(y_1, y_2) = c e^{-\int p(x) \, dx}$

step 2

Since the Wronskian is 1, we have $1 = c e^{-\int p(x) \, dx}$. Therefore, $c = 1$ and $e^{-\int p(x) \, dx} = 1$. This implies $\int p(x) \, dx = 0$, so $p(x) = 0$

step 3

Next, we use the fact that $y_1$ is a solution to the first-order differential equation $y' + \sin(x)y = 0$. Solving this, we get $y_1(x) = c_1 e^{-\cos(x)}$

step 4

To find $q(x)$, we substitute $y_1(x) = e^{-\cos(x)}$ into the original second-order differential equation $y'' + p(x)y' + q(x)y = 0$. Since $p(x) = 0$, the equation simplifies to $y'' + q(x)y = 0$

step 5

Compute $y_1''(x)$:
$y_1'(x) = e^{-\cos(x)} \sin(x)$
$y_1''(x) = e^{-\cos(x)} (\sin^2(x) - \cos(x))$
Substitute $y_1(x)$ and $y_1''(x)$ into the equation:
$e^{-\cos(x)} (\sin^2(x) - \cos(x)) + q(x) e^{-\cos(x)} = 0$
$\sin^2(x) - \cos(x) + q(x) = 0$
Thus, $q(x) = \cos(x) - \sin^2(x)$

Answer

The coefficients are $p(x) = 0$ and $q(x) = \cos(x) - \sin^2(x)$.

Key Concept

Abel's formula and solving differential equations

Explanation

Abel's formula helps determine the coefficient $p(x)$ by using the Wronskian, and solving the given first-order differential equation provides the form of $y_1$. Substituting $y_1$ into the original second-order differential equation allows us to find $q(x)$.

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