MCQ
The functions $u = e^x sin x$ ; $v = e^x cos x$ satisfy the equation :
- A$v\frac{{d^2u}}{{dx}} - u\frac{{dv}}{{dx}} = u^2 + v^2$
- B$\frac{{{d^2}u}}{{d{x^2}}}= 2 v$
- C$\frac{{{d^2}v}}{{d{x^2}}}= - 2 u$
- ✓All of these
$\frac{d u}{d x}=\frac{d}{d x}\left(e^{x} \sin x\right), \quad \frac{d v}{d x}=\frac{d}{d x}\left(e^{x} \cos x\right)$
$\frac{d u}{d x}=e^{x}(\cos x+\sin x), \quad \frac{d v}{d x}=e^{x}(\cos x-\sin x)$
For $v \frac{d u}{d x}-u \frac{d v}{d x}=u^{2}+v^{2}$
$e^{x} \cos x\left(e^{x}(\cos x+\sin x)\right)-e^{x} \sin x\left(e^{x}(\cos x-\sin x)\right)=\left(e^{x} \sin x\right)^{2}+\left(e^{x} \cos x\right)^{2}$
$\left(e^{x}\right)^{2} \cos ^{2} x+\left(e^{x}\right)^{2} \cos x \sin x-\left(e^{x}\right)^{2} \cos x \sin x+\left(e^{x}\right)^{2} \sin ^{2} x=\left(e^{x}\right)^{2}\left(\sin ^{2} x+\cos ^{2} x\right)$
$\left(e^{x}\right)^{2}=\left(e^{x}\right)^{2}$
For $\frac{d^{2} u}{d x^{2}}=2 v$
$\frac{d^{2} u}{d x^{2}}=\frac{d}{d x}\left(e^{x}(\cos x+\sin x)\right)$
$=e^{x}(-\sin x+\cos x)+e^{x}(\cos x+\sin x)$
$=2 e^{x} \cos x=2 v$
For $\frac{d^{2} v}{d x^{2}}=-2 u$
$\frac{d^{2} v}{d x^{2}}=\frac{d}{d x}\left(e^{x}(\cos x-\sin x)\right)$
$=e^{x}(-\sin x-\cos x)+e^{x}(\cos x-\sin x)$
$=-2 e^{x} \sin x=-2 u$
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$\left[\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]$ is invertible.
$(A)$ $(4,2 \sqrt{2})$ $(B)$ $(9,3 \sqrt{2})$ $(C)$ $\left(\frac{1}{4}, \frac{1}{\sqrt{2}}\right)$ $(D)$ $(1, \sqrt{2})$