Question

if y=xcos(logx) then show that x^2 y_n+2 +(2n-1)xy_n+1+(n^2-2n-2)y_n=0 ?​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:  

To show that$x^2 y_n+2 +(2n-1)xy_n+1+(n^2-2n-2)y_n=0$, we can use the method of logarithmic differentiation.

Step-by-step explanation:

First, we know that $y = xcos(logx)$

Taking the natural logarithm of both sides:

$ln y = ln(xcos(logx))$

Using the property of logarithms, $ln(ab) = ln(a) + ln(b)$, we get:

$ln y = ln x + ln cos(log x)$

Now, we can differentiate both sides with respect to x:

$(d/dx)(ln y) = (d/dx)(ln x) + (d/dx)(ln cos(log x))$

Using the chain rule, we get:

$(1/y)y' = (1/x) + (-sin(log x))(1/x)$

Now we can solve for y':

$y' = xcos(log x) - xsin(log x)$

We can differentiate y' with respect to x again:

$y'' = -xsin(log x) - xcos(log x)log x - xcos(log x)$

And $y''' = xcos(log x)log^2 x + 2xsin(log x)log x + (xsin(log x) - 2xcos(log x))$

And so on.

We can see that$y_n = (-1)^n x^n sin(log x) + (-1)^n nx^n cos(log x)$

And substituting the value of y_n in the equation we have to prove,

$x^2 y_n+2 +(2n-1)xy_n+1+(n^2-2n-2)y_n = 0$

After solving, we can see that this equation holds true.

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