If #y= sinx/ x^2#, find #dy/dx# and #(d^2y)/(dx^2)#. Then prove that #x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y = 0# ?

Question

If #y= sinx/ x^2#, find #dy/dx# and #(d^2y)/(dx^2)#.  Then prove that #x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y = 0# ?

Ezra Adams · Accepted Answer

We seek to show that: # x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y = 0  # where #y= sinx/x^2# Using the quotient rule then differentiating #y= sinx/x^2# wrt #x# we have: # dy/dx = { (x^2)(cosx)-(2x)(sinx) } / (x^2)^2 #
# \ \ \ \ \ \ = (xcosx-2sinx)/x^3 # And differentiating a second time, we have: # (d^2y)/(dx^2) = { (x^3)(-xsinx+cosx-2cosx) - (3x^2)(xcosx-2sinx) } / (x^3)^2  #  # \ \ \ \ \ \ \  = { -x^2sinx+xcosx-2xcosx - 3xcosx+6sinx) } / (x^4) #  # \ \ \ \ \ \ \  = ( -x^2sinx-4xcosx +6sinx ) / (x^4) #  And so, considering the LHS of the given expression: # LHS = x^2 (d^2y)/(dx^2) + 4x dy/dx + (x^2 + 2) y # # \ \ \ \ \ \ \ \ = x^2 {( -x^2sinx-4xcosx +6sinx ) / (x^4)} #
# \ \ \ \ \ \ \ \ \ \ \ \ + 4x {(xcosx-2sinx)/x^3} #
# \ \ \ \ \ \ \ \ \ \ \ \ + (x^2 + 2) {sinx/x^2} # # \ \ \ \ \ \ \ \ = {( -x^2sinx-4xcosx +6sinx ) / (x^2)} #
# \ \ \ \ \ \ \ \ \ \ \ \ + 4 {(xcosx-2sinx)/x^2} #
# \ \ \ \ \ \ \ \ \ \ \ \ + (x^2 + 2) {sinx/x^2} # # \ \ \ \ \ \ \ \ = 1/x^2 {-x^2sinx-4xcosx +6sinx + 4xcosx-8sinx + (x^2 + 2)sinx} # # \ \ \ \ \ \ \ \ = 0 \ \ \  # QED

Emily Ammerman · Answer

undefined To find $ \frac{dy}{dx} $, differentiate $ y = \frac{\sin(x)}{x^2} $ with respect to $ x $ using the quotient rule:
$$ \frac{dy}{dx} = \frac{(x^2 \cdot \cos(x) - 2x \cdot \sin(x))}{x^4} $$

To find $ \frac{d^2y}{dx^2} $, differentiate $ \frac{dy}{dx} $ with respect to $ x $:
$$ \frac{d^2y}{dx^2} = \frac{(x^4 \cdot (-\sin(x)) - 4x^3 \cdot \cos(x) - (2x \cdot \cos(x) - 2 \sin(x)))}{x^8} $$

Then, to prove $ x^2 \frac{d^2y}{dx^2} + 4x \frac{dy}{dx} + (x^2 + 2)y = 0 $, substitute $ \frac{dy}{dx} $ and $ \frac{d^2y}{dx^2} $ into the expression and simplify. You should obtain $ 0 $ when the expression is simplified.