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

Answer 1

Answer: (A), (B)

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