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give answer in 2 step with explanation at the end of each step and final answer at the end:1) Find P,(z) for the following Legendre’s equation (1 — z)y” — 223’ + 42y = 0. (Reference 5.2.1) (40 points)

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give answer in 2 step with explanation at the end of each step and final answer at the end:Uploaded Image1) Find P,(z) for the following Legendre’s equation (1 — z)y” — 223’ + 42y = 0. (Reference 5.2.1) (40 points)

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Asked by: Firdos
Created at: 2025-12-08 05:24:46

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## Step 1: Compare with Standard Legendre's Equation The standard Legendre's differential equation is: \[ (1 - x^2)y'' - 2xy' + n(n+1)y = \] Given equation: \[ (1 - x^2)y'' - 2xy' + 42y = \] **Explanation:** Compare the coefficient of \(y\) in both equations: \[ n(n+1) = 42 \] Solve for \(n\): \[ n^2 + n - 42 = \implies (n+7)(n-6) = \implies n = 6 \text{ or } n = -7 \] Since \(n\) is usually taken as a non-negative integer for Legendre polynomials, **we take \(n = 6\).** --- ## Step 2: Write the Solution \(P_n(x)\) The solution \(P_n(x)\) is the **Legendre polynomial** of degree \(n=6\): \[ P_6(x) \] Using the explicit formula: \[ P_6(x) = \frac{1}{32}(231x^6 - 315x^4 + 105x^2 - 5) \] **Explanation:** Legendre polynomials \(P_n(x)\) can be found in tables or using Rodrigues' formula: \[ P_n(x) = \frac{1}{2^n n!}\frac{d^n}{dx^n}\left[(x^2-1)^n\right] \] For \(n=6\), this gives the above formula. --- ## **Final Answer** \[ \boxed{P_6(x) = \frac{1}{32}(231x^6 - 315x^4 + 105x^2 - 5)} \]

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Asked by: Firdos
Created at: 2025-12-08 05:29:11

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# Solution to Legendre's Equation ## Step 1: Compare with Standard Legendre's Equation The standard form of Legendre's differential equation is: \[ (1 - z^2)y'' - 2zy' + n(n+1)y = 0 \] Given equation: \[ (1 - z)y'' - 2z + 42y = 0 \] ### Explanation: We compare the coefficients of \(y\): \[ n(n+1) = 42 \] To find \(n\), we solve the quadratic equation: \[ n^2 + n - 42 = 0 \] Factoring: \[ (n + 7)(n - 6) = 0 \implies n = 6 \text{ or } n = -7 \] Since \(n\) should be a non-negative integer for Legendre polynomials, we take: \[ n = 6 \] --- ## Step 2: Write the Solution \(P_n(z)\) The solution \(P_n(z)\) is the **Legendre polynomial** of degree \(n=6\): \[ P_6(z) \] Using the explicit formula for Legendre polynomials: \[ P_6(z) = \frac{1}{32}(231z^6 - 315z^4 + 105z^2 - 5) \] ### Explanation: Legendre polynomials can also be derived from Rodrigues' formula: \[ P_n(z) = \frac{1}{2^n n!}\frac{d^n}{dz^n}\left[(z^2 - 1)^n\right] \] For \(n=6\), this results in the polynomial mentioned above. --- ## **Final Answer** \[ \boxed{P_6(z) = \frac{1}{32}(231z^6 - 315z^4 + 105z^2 - 5)} \]

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