(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).

Q: (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).

∻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) $.⚹◊What is the Abel's formula that can be used to find the coefficient $p(x)$ in the differential equation provided?⍭ Generate me a similar question◊

Question

(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).

Accepted Answer

∻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) $.⚹◊What is the Abel's formula that can be used to find the coefficient $p(x)$ in the differential equation provided?⍭ Generate me a similar question◊

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