Question

The functions $u=e^{x} \sin x ; v=e^{x} \cos x$ satisfy the equation

• A. $v \frac{ du }{ dx }- u \frac{ dv }{ dx }= u ^{2}+ v ^{2}$
• B. $\frac{ d ^{2} u }{ dx ^{2}}=2 v$
• C. $\frac{ d ^{2} v }{ dx ^{2}}=-2 u$
• D. $\frac{ du }{ dx }+\frac{ dv }{ dx }=2 v$

Answer 1

Answer: (A), (B)

$u=e^{x} \sin x, v=e^{x} \cos x $
$v \frac{d u}{d x}-u \frac{d v}{d x}=v\left(e^{x} \cos x+e^{x} \sin x\right)-u\left(e^{x} \cos x-e^{x} \sin x\right) $
$=e^{x} \sin x(v+u)+e^{x} \cos x(v-u) $
$=u(v+u)+v(v-u) $
$=v^{2}+u^{2} $
$ \frac{d u}{d x}=e^{x} \sin x+e^{x} \cos x$
again $\frac{d^{2} u}{d x^{2}}=e^{x} \sin x+e^{x} \cos x+e^{x} \cos x-e^{x} \sin x$
$\frac{d^{2} u}{d x^{2}} =2 v$
similarly other options can be checked.