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Aniya

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Asked: 2021-04-03T20:34:54+05:30 In:

if x=acostheta+bsintheta and y=asintheta-bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0

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if x=acostheta+bsintheta and y=asintheta-bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0? How can I prove this math problem?

if x acostheta bsintheta and y asintheta bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0if x acostheta bsintheta and y asintheta bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0 explanationif x acostheta bsintheta and y asintheta bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0 solutionif x equal toif x=if x=acostheta+bsintheta and y=asintheta-bcosthetaif x=acostheta+bsintheta and y=asintheta-bcostheta prove thatif x=acostheta+bsintheta and y=asintheta-bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0if x=acostheta+bsintheta and y=asintheta-bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0 how to proveif x=acostheta+bsintheta and y=asintheta-bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0 solutionmath problems
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    1 Answer

    1. Given, x=acosθ+bsinθ  & y=asinθ-bcosθ

      then dx/dθ = -asinθ+bcosθ    &    dy/dθ = acosθ+bsinθ

      dy/dx= (dy/dθ) * (dθ/dx)

      =  (acosθ+bsinθ)/(-(asinθ-bcosθ)) = x/-y

      ∴ dx/dy = -x/y

      apply derviative

      d^2y/dx^2 = 1/-y + -(-x)/y^2  * dy/dx

      by solving we can get :

      y^2 * d^2y/dx^2 -x*dy/dx +y =0

      hence proved

      You can also refer to the similar problems like:

      If x+y+z=0 teh show that x^3+y^3 + z^3 =3xyzexplanation

      https://techanswered.com/question/if-xyz0-then-show-that-x3y3z33xyz/

       

      if x=acostheta+bsintheta and y=asintheta-bcostheta then prove that y 2 d 2y/dx^2 xdy/dx + y = 0
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