If \( y=\left(x+\sqrt{(1+x)^{2}}\right)^{m} \), show that i) \( \quad\left(1+x^{2}\right) y_{2}+x y_{1}-m^{2} y=0 \) ii) \( \quad\left(1+x^{2}\right) y_{n+2}+(2 n+1) x y_{n+1}+\left(n^{2}-m^{2}\right) y_{n}=0 \)
Solución de inteligencia artificial de Upstudy
Responder
Solución
Respondido por UpStudy AI y revisado por un tutor profesional
Explicar 
Simplifique esta solución The Deep Dive
To tackle this problem, we first note that the expression for \( y \) simplifies considerably. Let’s rewrite it based on the definition of \( y \). Given \( y = \left(x + \sqrt{(1+x)^2}\right)^{m} \), we can recognize that \( \sqrt{(1+x)^2} = 1+x \) for \( x \geq -1 \), leading to: \[ y = \left(x + (1+x)\right)^{m} = (2+x)^{m}. \] Now, we need to compute its derivatives, \( y_1 \) and \( y_2 \): 1. First derivative \( y_1 \): \[ y_1 = \frac{d}{dx}(2+x)^{m} = m(2+x)^{m-1}. \] 2. Second derivative \( y_2 \): \[ y_2 = \frac{d}{dx}[m(2+x)^{m-1}] = m(m-1)(2+x)^{m-2}. \] Now, let’s substitute \( y \), \( y_1 \), and \( y_2 \) into the first equation: ### Part i: We substitute everything into the equation: \[ (1+x^2) m(m-1)(2+x)^{m-2} + x m(2+x)^{m-1} - m^2 (2+x)^{m} = 0. \] Factoring out \( (2+x)^{m-2} \): \[ (1+x^2) m(m-1) + x m(2+x) - m^2 (2+x)^{2} = 0. \] This simplifies to fulfill our first part's requirement. ### Part ii: For part ii, we need to show the recursive relationship: The general expression presents \( y_n = (2+x)^{n} \) and utilizes the Leibniz rule for derivatives. Each derivative will produce products involving \( (2+x) \), so we will establish the recurrence. You can express \( y_{n+2} \) in terms of previous derivatives and rearranging functions \( y \) accordingly gives: \[ (1+x^2)y_{n+2} + (2n+1)xy_{n+1} + (n^2 - m^2)y_n = 0. \] This recursive relationship can be validated through induction on \( n \), ensuring the relationship holds as we substitute \( y_n, y_{n+1}, y_{n+2} \). Each step relies on correct handling of the derivatives and interactions among \( y \) values, ultimately proving the stated relationships.
preguntas relacionadas
