यदि ` x=a cos theta + bsin theta ,y =a sin theta -b cos theta,` दिखाइए की `y^(2)(d^(2y))/(dx^(2))-x (dy)/(dx) +y=0`

Explanation

`because " "x =acos theta +b sin theta `
`rArr (dx)/(d theta )=- a sin theta + b cos theta `
व् ` y= asin theta - bcos theta `
`rArr (dy)/(d theta ) =acos theta + b sin theta `
अब ` (dy)/(dx) =(dy//d theta )/(dx//d theta ) =(acos theta + b sin theta )/(-a sin theta +b cos theta )= (x)/(-y)`
` rArr " " y(dy)/(dx)+ x=0 " "...(1)`
x के सापेक्ष पुनः अवकलन करने पर
` " "y (d^(2)y)/(dx^(2))+(dy)/(dx) *(dy)/(dx)+ 1=0`
` rArr " "y^(2)(d^(2)y)/(dx^(2) )+ (y(dy)/(dx)) (dy)/(dx)+y=0`
` rArr " "y^(2) (d^(2y))/(dx^(2))-x (dy)/(dx)+ y=0`