The function `u=e^x sin x ; v=e^x cos x` satisfy the equation a.`v(d u)/(dx)-u(d v)/(dx)=u^2+v^2` b. `(d^2u)/(dx^2)=2v` c. `(d^2v)/(dx^2)=-2u` d. `(d u)/(dx)+(d v)/(dx)=2v`

Options

A. `v(du)/(dx)u(dv)/(dx)=u^(2)+v^(2)`
B. `(d^(2)u)/(dx^(2))=2v`
C. `(d^(2)v)/(dx^(2))=-2u`
D. all the above

Answer

D

Explanation

We have,
`u=e^(x)sinximplies(du)/(dx)=e^(x)sinx+e^(x)cosx=u+v`
`v=e^(x)cosximplies(dv)/(dx)=e^(x)cosx+e^(x)sinx=v-u`
`:." "v(du)/(dx)-(udv)/(dx)=v(u+v)-u(v-u)=u^(2)+v^(2)`
`(d^(2)u)/(dx^(2))=(du)/(dx)+(dv)/(dx)=u+v+v-u=2v`
and,`(d^(2)u)/(dx^(2))=(dv)/(dx)-(du)/(dx)=(v-u)-(v+u)=-2u`