Play with Curves!

Calculus Level 4

d y d x = y x 3 x 2 ln y sin y x + x y \frac{dy}{dx} = \frac{ y - x^3 - x^2 - \ln y }{\sin y - x + \frac{x}{y}}

Given that the slope of a curve passing through ( 0 , π 2 ) ( 0 , \frac{\pi}{2} ) is as shown above.

The equation of curve can be represented as

x a b + x c d e x y f cos ( g y ) + h x ln ( i y ) = 0 , \frac{x^a}{b} + \frac{x^c}{d} - exy - f \cos (gy) + hx \ln (iy) = 0,

where a a , b b , c c , d d , e e , f f , g g , h h , and i i are positive integers .

Find a + b + c + d + e + f + g + h + i a + b + c + d + e + f + g + h + i .


Original.


The answer is 19.

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2 solutions

Chew-Seong Cheong
Aug 19, 2016

d d x ( x a b + x c d e x y f cos ( g y ) + h x ln ( i y ) ) = 0 a x a 1 b + c x c 1 d e y e x d y d x + g f sin ( g y ) d y d x + h ln ( i y ) + i h x i y d y d x = 0 \begin{aligned} \frac d{dx} \left( \frac{x^a}{b} + \frac{x^c}{d} - exy - f \cos (gy) + hx \ln (iy) \right) & = 0 \\ \frac{ax^{a-1}}{b} + \frac{cx^{c-1}}{d} - ey - ex \frac {dy}{dx} + gf \sin (gy) \frac {dy}{dx} + h\ln (iy) + \frac {ihx}{iy} \frac {dy}{dx} & = 0 \end{aligned}

d y d x = e y a x a 1 b c x c 1 d h ln ( i y ) g f sin ( g y ) e x h x y = y x 3 x 2 ln y sin y x + x y \begin{aligned} \implies \frac {dy}{dx} & = \frac {ey - \frac{ax^{a-1}}{b} - \frac{cx^{c-1}}{d} - h\ln (iy)}{gf \sin (gy) - ex - \frac {hx}{y}} = \frac{ y - x^3 - x^2 - \ln y }{\sin y - x + \frac{x}{y}} \end{aligned}

Comparing the coefficients, we have:

a = b = 4 \implies a=b=4 , c = d = 3 c=d=3 , and e = f = g = h = i = 1 e=f=g=h=i=1 , therefore, a + b + c + d + e + f + g + h + i = a+b+c+d+e+f+g+h+i = 4 + 4 + 3 + 3 + 1 + 1 + 1 + 1 + 1 = 19 4+4+3+3+1+1+1+1+1=\boxed{19} .

Good technique sir

Ayush Sharma - 4 years, 9 months ago
Prakhar Bindal
Aug 18, 2016

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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