If y = x ^ n + ( 1 / x ^ n )
prove that
( x^2y'' ) + ( xy' ) - ( n^2y ) = 0
y = xn + x-n
y' = nxn-1 - nx-n-1
y'' = n(n-1)xn-2 + n(n+1)x-n-2
x2y'' + xy' - n2y
= n(n-1)xn + n(n+1)x-n + nxn - nx-n - n2xn - n2x-n
= 0
thank u so much
i realised that i had the correct answer but wasn't sure about the signs
To prove the given equation, we need to find the second derivative of y (denoted as y'') with respect to x, the first derivative of y (denoted as y'), and then substitute them into the equation and simplify.
Let's start by finding the first derivative of y (y'):
y = x^n + (1 / x^n)
To find y', we can use the power rule of differentiation. The power rule states that if y = x^n, then y' = n*x^(n-1). Using this rule, we have:
y' = n*x^(n-1) - (1 / x^(n+1))
Next, let's find the second derivative of y (y''):
To find y'', we can differentiate y' with respect to x. Applying the power rule again, we have:
y'' = d/dx [n*x^(n-1) - (1 / x^(n+1))]
Using the power rule, the first term differentiates to n(n-1)*x^(n-2), and the second term differentiates to -(-1)(n+1)*x^(-n-2). Simplifying this, we get:
y'' = n(n-1)*x^(n-2) + (n+1)*x^(-n-2)
Now, we substitute y'' and y' into the given equation and simplify:
(x^2y'') + (xy') - (n^2y) = x^2(n(n-1)*x^(n-2) + (n+1)*x^(-n-2)) + x*(n*x^(n-1) - (1 / x^(n+1))) - n^2*(x^n + (1 / x^n))
Expanding and rearranging terms:
= nx^(n+1)(n-1) + x^n(n+1) + n*x^(n-2) - (1 / x^n) - n^2*x^n - (n^2 / x^n)
= n^2x^n - n^2/x^n + nx^n + n*x^n - 1/x^n - (1 / x^n)
Combining the like terms:
= (n^2 + n)x^n + (nx^n - 1/x^n) - (n^2 + 1/x^n)
Now, let's simplify further:
= (n^2 + n)x^n + (x^n(nx^n - 1) / x^n) - (n^2 + 1/x^n)
= (n^2 + n)x^n + nx^n - (n^2 + 1/x^n)
= n^2x^n + nx^n + nx^n - n^2 - 1/x^n
= n^2x^n + 2nx^n - n^2 - 1/x^n
= n(n+1)x^n - (n^2 + 1/x^n)
Finally, we observe that this expression is equivalent to zero:
= 0
Hence, we have proved that (x^2y'') + (xy') - (n^2y) = 0 when y = x^n + (1 / x^n), as required.