Solve. (1 - x^2)y'' - 2xy' + 2y = 0
logistic_guy Full Member Joined Apr 17, 2024 Messages 901 Jun 23, 2025 #1 Solve. (1−x2)y′′−2xy′+2y=0\displaystyle (1 - x^2)y'' - 2xy' + 2y = 0(1−x2)y′′−2xy′+2y=0
K khansaheb Senior Member Joined Apr 6, 2023 Messages 1,077 Jun 23, 2025 #2 logistic_guy said: Solve. (1−x2)y′′−2xy′+2y=0\displaystyle (1 - x^2)y'' - 2xy' + 2y = 0(1−x2)y′′−2xy′+2y=0 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem
logistic_guy said: Solve. (1−x2)y′′−2xy′+2y=0\displaystyle (1 - x^2)y'' - 2xy' + 2y = 0(1−x2)y′′−2xy′+2y=0 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem
logistic_guy Full Member Joined Apr 17, 2024 Messages 901 Jun 23, 2025 #3 khansaheb said: Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem Click to expand... Thank you Sir khan.
khansaheb said: Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem Click to expand... Thank you Sir khan.