InterviewSolution
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InterviewSolution
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The function `u=e^x s in ; v=e^x cos x`satisfy the equation`v(d u)/(dx)-u(d v)/(dx)=u^2+v^2`b. `(d^2u)/(dx^2)=2v`c. `(d^2)/(dx^2)=-2u`d. `(d u)/(dx)+(d v)/(dx)=2v`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 |
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Answer» Correct Answer - D 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` |
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