d x d y = sin y − x + y x y − x 3 − x 2 − ln y
Given that the slope of a curve passing through ( 0 , 2 π ) is as shown above.
The equation of curve can be represented as
b x a + d x c − e x y − f cos ( g y ) + h x ln ( i y ) = 0 ,
where a , b , c , d , e , f , g , h , and i are positive integers .
Find a + b + c + d + e + f + g + h + i .
Original.
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Good technique sir
Its a simple variable separation method.
Cross mutliply
to get
d(x) denotes derivative
d(xy)-(x^3)dx-(x^2)dx = sinydy +d(xlny)
Integrate directly to get required equation
xy - x^4 / 4 - x^3 / 3 = -cosy +xlny
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d x d ( b x a + d x c − e x y − f cos ( g y ) + h x ln ( i y ) ) b a x a − 1 + d c x c − 1 − e y − e x d x d y + g f sin ( g y ) d x d y + h ln ( i y ) + i y i h x d x d y = 0 = 0
⟹ d x d y = g f sin ( g y ) − e x − y h x e y − b a x a − 1 − d c x c − 1 − h ln ( i y ) = sin y − x + y x y − x 3 − x 2 − ln y
Comparing the coefficients, we have:
⟹ a = b = 4 , c = d = 3 , and e = f = g = h = i = 1 , therefore, a + b + c + d + e + f + g + h + i = 4 + 4 + 3 + 3 + 1 + 1 + 1 + 1 + 1 = 1 9 .