If `y = y(x)` and it follows the relation `e^(xy) + y cos x = 2`, then find (i) `y'(0)` and (ii) `y"(0)`.

Options

A. `y'(0)=-1`
B. `y''(0)=2`
C. `y'(0)=3//2`
D. `y''(0)=-2`

Answer

A, B

Explanation

`e(xy)+ycosx=2" (1)"`
Putting x = 0, we get
`e^(0)+y cos x=2`
`therefore" "y=1`
Differentiating (1) w.r.t. x, we get
`e^(xy)(x(dy)/(dx)+y)-ysinx+cosx(dy)/(dx)=0`
`therefore" "(xe^(xy)+cosx)(dy)/(dx)=y(sinx-e^(xy))" (2)"`
`therefore" "(dy)/(dx)=(ysinx-ye^(xy))/(xe^(xy)+cosx)`
`therefore" "(dy)/(dx):|_("(0,1)")=(1+sin0-1e^(0))/(0+cos0)`
`=-1`
Differentiating (2), w.r.t. x, we get
`(xe^(xy)+cosx)(d^(2)y)/(dx^(2))+[e^(xy)+xe^(xy)(x(dy)/(dx)+y)-sinx](dy)/(dx)`
`=(dy)/(dx)(sinx-e^(xy))+y[cosx-e^(xy)(y+x(dy)/(dx))]`
Putting x = 0 and y = 1
`therefore" "(0+1)(d^(2)y)/(dx^(2))+(e^(0))(-1)`
`=(-1)(0-1)+1.(1-1(1))`
`therefore" "(d^(2)y)/(dx^(2))=2`