X= tan logy then prove that (1+x^2)yn + 1 + (2nx - 1)yn + n(n -1 ) yn -1 = 0 solve using lebnitz theorem
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Answer:
X= tan logy then prove that (1+x^2)yn + 1 + (2nx - 1)yn + n(n -1 ) yn -1 = 0 solve using lebnitz theorem
Answer:
Given the equation X = tan(log y), and using the chain rule, we can find the derivative of X with respect to y.
Using the Leibniz theorem, we can find the derivative of the product of two functions:
(1+x^2)yn + 1 + (2nx - 1)yn + n(n -1 ) yn -1 = 0
The Leibniz theorem states that the derivative of a product of two functions is equal to the product of the first function multiplied by the derivative of the second function plus the product of the second function multiplied by the derivative of the first function.
Applying this to the given equation, we get:
(1+x^2)yn * yn' + (1+ (2nx - 1)yn + n(n -1 ) yn -1 ) * yn' = 0
Rearranging the terms and simplifying, we get:
(1+x^2 + 2nxyn + n(n-1) yn^2) yn' = 0
We can see that the equation is true when yn' = 0 or when (1+x^2 + 2nxyn + n(n-1) yn^2) = 0
The first case is true when yn' = 0, meaning that the derivative of yn is zero, and the second case is true when (1+x^2 + 2nxyn + n(n-1) yn^2) = 0, meaning that the equation is true when the product of the two functions is equal to zero.
In order to find the value of yn, we will have to solve this equation for yn. This requires more information about the specific functions and values involved, as well as knowledge of advanced mathematical techniques.
To learn more about Leibniz theorem from the link below
https://brainly.in/question/35342136
To learn more about derivative from the link below
https://brainly.com/question/23819325
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