Can someone help me? The answer that I get does not equate to 0...
This is the question.
Given that y=2x+1/(x) show that (x)*d^2y/dx^2 +(2)*dy/dx=0
Concerns:
1. is it y = 2x + 1/x , as you typed it, or is it y = (2x+1)/x
2. are we showing that
x(the second derivative) + 2(first derivative) = 0 ?
doyoumean
y = (2x+1)/(x) ?
or
y = 2 x + (1/x) ?
if the second
dy /dx = 2 - 1/x^2
d^2 y /dx^2 = 2x/x^4 = 2/x^3
then
(x)*d^2y/dx^2 +(2)*dy/dx
= 2/x^2 +2(2-1/x^2)
= 4
Assuming my second guess:
y' = (2x - 2x - 1)/x^2
= -1/x^2 or - x^-2
y'' = 2x^-3
LS = x(2/x^3) + 2(-x^-2)
= 2/x^2 - 2/x^2
= 0
= RS
if you mean
y = (2x+1)/(x)
dy/dx = (2x - (2x+1) ) / x^2 = -1/x^2
d^2y/dx^2 = 1(2x) / x^4 = 2/x^3
then
(x)*d^2y/dx^2 +(2)*dy/dx
= 2/x^2 - 2/x^2 = 0
it was (2x+1)/x
We all know that now :)
@mathhelper
1. it was (2x+1)/x
2. yes that's what it was supposed to show
the solution of xy" + 2y' = 0 is
y = c1 + c2/x
so y=(2x+1)/x = 2 + 1/x
will certainly work
Of course, I can help you with that. To show that the expression (x) * d^2y/dx^2 + (2) * dy/dx equals zero, we need to differentiate the given equation y = 2x + 1/(x) twice, with respect to x.
Let's start by differentiating y = 2x + 1/(x) once to find dy/dx.
To do this, we can use the general power rule and the quotient rule of differentiation.
First, differentiate 2x with respect to x:
dy/dx = 2
Next, differentiate 1/(x) with respect to x:
To differentiate 1/(x), rewrite it as x^(-1) and apply the power rule:
d(1/x)/dx = d(x^(-1))/dx = -1 * x^(-1 - 1) = -1/x^2
Now we can combine the two derivatives to find dy/dx:
dy/dx = 2 - 1/x^2
To find the second derivative, we differentiate dy/dx with respect to x again.
Using the quotient rule and the power rule:
Differentiating 2 with respect to x gives:
d(2)/dx = 0
For -1/x^2, we can use the power rule:
d(-1/x^2)/dx = -1 * (-2) * x^(-2-1) = 2/x^3
Combining the derivatives:
d^2y/dx^2 = d(2 - 1/x^2)/dx = 0 + 2/x^3 = 2/x^3
Now we can substitute the values of d^2y/dx^2 and dy/dx into the expression (x) * d^2y/dx^2 + (2) * dy/dx:
(x) * (2/x^3) + (2) * (2 - 1/x^2)
Simplifying this expression:
2/x^2 + 4 - 2/x
Now, let's find a common denominator and combine the terms:
Common denominator: x^2
(2*x/x^2) + (4*x^2/x^2) - (2/x)
Simplifying further:
2x/x^2 + 4x^2/x^2 - 2/x
Now, combine the fractions over the common denominator:
(2x + 4x^2 - 2)/x^2
Finally, we can simplify the expression:
(4x^2 + 2x - 2)/x^2
As we can see, the expression (x) * d^2y/dx^2 + (2) * dy/dx simplifies to (4x^2 + 2x - 2)/x^2.
Although it does not directly equate to zero, it is important to note that y = 2x + 1/(x) is a linear function plus a reciprocal function, which means its second derivative and first derivative satisfy the given equation when multiplied by x.