If e^y+x y=e the ordered pair ( d y d x , d^2 y d x^2 ) at x=0 is equal to

Step-by-step solution

Given, c^ +x y=c Differentiating w.r.t. x , we get aligned & e^y d y d x +x d y d x +1 y=0 & d y d x = -y (e^y+x ) aligned Again, differentiating Eq. (ii) w.r.t. x , we get aligned & e^y d^2 y d x^2 +e^v ( d y d x )^2+x d^2 y d x^2 + d y d x + d y d x =0 & (e^y+x ) d^2 y d x^2 +e^y ( d y d x )^2+2 d y d x =0 aligned Now, on putting x=0 in Eq. (i), we get e^y+0 y=e e^y=e^1 y=1 On putting x=0, y=1 in Eq. (iii), we get d y d x = -1 e+0 =- 1 e Now, putting x=0, y=1 and d y d x = -1 e in Eq. (iv), we get aligned & (e^1+