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If y = y(x) and it follows the relation ...

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

A

`y'(0)=-1`

B

`y''(0)=2`

C

`y'(0)=3//2`

D

`y''(0)=-2`

Text Solution

Verified by Experts

The correct Answer is:
A, B
`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`
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