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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
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
✓

Answer

Correct option: D.
All of these
d
$u=e^{x} \sin x, v=e^{x} \cos x$

$\frac{d u}{d x}=\frac{d}{d x}\left(e^{x} \sin x\right), \quad \frac{d v}{d x}=\frac{d}{d x}\left(e^{x} \cos x\right)$

$\frac{d u}{d x}=e^{x}(\cos x+\sin x), \quad \frac{d v}{d x}=e^{x}(\cos x-\sin x)$

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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