Q70 / 170 · Manipal (MET) 2015 · Mathematics

The functions $u=e^x \sin x$ and $v=e^x \cos x$ satisfy the equation

  • A
    $v \frac{d u}{d x}-u \frac{d v}{d x}=u^2+v^2$
  • B
    $\frac{d^2 u}{d x^2}=2 v$
  • C
    $\frac{d^2 v}{d x^2}=-2 u$
  • Correct answer
    D
    All of these

Solution

We have, $u=e^x \sin x$ and $v=e^x \cos x$ On differentiating both sides w.r.t. $x$, we get $\frac{d u}{d x}=e^x \sin x+e^x \cos x=u+v$ and $\frac{d v}{d x}=e^x \cos x-e^x \sin x=v-u$ $\therefore v \frac{d u}{d x}-u \frac{d v}{d x}=v(u+v)-u(v-u)=u^2+v^2$ Now, $\quad \frac{d^2 u}{d x^2}=\frac{d u}{d x}+\frac{d v}{d x}=u+v+v-u=2 v$ and $\quad \frac{d^2 v}{d x^2}=\frac{d v}{d x}-\frac{d u}{d x}$ $=(v-u)-(v+u)=-2 u$

Hence, the correct answer is option: 'D'.

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