Prove that if #y= sinx/x#, show that #(d^2y)/(dx^2) + 2/x dy/dx + y = 0# ?

Answer 1

See below

If #xy = sin x#, using the product rule:
# y + xy' = cos x qquad qquad qquad square#
#y' + y' + x y'' = - sin x qquad triangle#
#triangle/ x + square implies#
#(2y')/x + y'' + color( red)( y + xy') = - color(blue)((sin x)/x) + cos x#
As #color(blue)(y = (sin x)/x)# and #color(red)( y + xy' = cos x)#:
#(2y')/x + y'' + cos x = - y + cos x#
#implies y'' + (2y')/x + y = 0 #
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To prove the given expression, we start with the equation ( y = \frac{\sin(x)}{x} ). First, find the first derivative ( \frac{dy}{dx} ) using the quotient rule. Then differentiate ( \frac{dy}{dx} ) to find ( \frac{d^2y}{dx^2} ). Finally, substitute these derivatives into the expression ( \frac{d^2y}{dx^2} + \frac{2}{x} \frac{dy}{dx} + y ) and simplify to show that it equals zero.

First derivative: [ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\sin(x)}{x}\right) = \frac{x\cos(x) - \sin(x)}{x^2} ]

Second derivative: [ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{x\cos(x) - \sin(x)}{x^2}\right) = -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} ]

Now, substitute these derivatives into the expression: [ \frac{d^2y}{dx^2} + \frac{2}{x} \frac{dy}{dx} + y = -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} + 2\left(\frac{x\cos(x) - \sin(x)}{x^3}\right) + \frac{\sin(x)}{x} ]

Simplify the expression: [ -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} + \frac{2x\cos(x) - 2\sin(x)}{x^3} + \frac{\sin(x)}{x} ] [ = -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} + \frac{2x\cos(x) - 2\sin(x) + \sin(x)x^2}{x^3} ] [ = -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} + \frac{2x\cos(x) - \sin(x) + \sin(x)x^2}{x^3} ] [ = -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} + \frac{x(2\cos(x) - \sin(x))}{x^3} ] [ = -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} + \frac{2x\cos(x) - x\sin(x)}{x^3} ] [ = -\frac{2\cos(x)}{x^2} + \frac{2\sin(x)}{x^3} + \frac{2x\cos(x) - x\sin(x)}{x^3} ] [ = 0 ]

Therefore, ( \frac{d^2y}{dx^2} + \frac{2}{x} \frac{dy}{dx} + y = 0 ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7