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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
The function $y=f(x)$ is the solution of the differential equation $\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}}$ in $(-1,1)$ satisfying $f(0)=0$. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$ is
  • A
    $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
  • ✓
    $\frac{\pi}{3}-\frac{\sqrt{3}}{4}$
  • C
    $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$
  • D
    $\frac{\pi}{6}-\frac{\sqrt{3}}{2}$
✓

Answer

Correct option: B.
$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$
b
$\text { I.F. }=e^{\int \frac{x}{x^2-1} d x}=e^{\frac{1}{2} \int \frac{2 x}{x^2-1} d x}=e^{\frac{1}{2} \ln \left(x^2-1\right.}=e^{\frac{1}{2} \ln \left(1-x^2\right)}=\sqrt{1-x^2} $

$\therefore \quad y \sqrt{1-x^2}=\int \frac{x^4+2 x}{\sqrt{1-x^2}} \times \sqrt{1^2-x^2} d x+c$

$y \sqrt{1^2-x^2}=\frac{x^5}{5}+x^2+c $

$x=0, y=0 \Rightarrow c=0 $

$y=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}} $

$\therefore \quad I=\int_0^{\frac{\sqrt{3}}{2}}\left(\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}+\frac{\frac{-x^5}{5}+x^2}{\sqrt{1-x^2}}\right) d x=2 \int_0^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} d x $

$x=\sin \theta $

$d x=\cos \theta d \theta=2 \int_0^{\frac{\pi}{3}} \frac{\sin ^2 \theta \cos \theta}{\cos \theta} d \theta $

$=\int_0^{\frac{\pi}{3}}(1-\cos 2 \theta) d \theta=\left.\left(\theta-\frac{1}{2} \sin 2 \theta\right)\right|_0 ^{\frac{\pi}{3}}=\frac{\pi}{3}-\frac{1}{2} \times \sin \frac{2 \pi}{3}=\frac{\pi}{3}-\frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\pi}{3}-\frac{\sqrt{3}}{4} $

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