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  •  Let the solution curve   of the differential equation <img alt="\mathrm{\frac{d y}{d x}+\frac{

 Let the solution curve \mathrm{y=f(x)}  of the differential equation \mathrm{\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x \in(-1,1)}  pass through the origin. Then \mathrm{\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x} is equal to

Option: 1

\mathrm{\frac{\pi}{3}-\frac{1}{4}}


Option: 2

\mathrm{\frac{\pi}{3}-\frac{\sqrt{3}}{4}}


Option: 3

\mathrm{\frac{\pi}{6}-\frac{\sqrt{3}}{4}}


Option: 4

\mathrm{\frac{\pi}{6}-\frac{\sqrt{3}}{2}}


Answers (1)

best_answer

Linear differential equation

To get I.F

\mathrm{ \int \frac{x}{x^{2}-1} d x} \\

\mathrm{= \frac{1}{2} \int \frac{2 x}{x^{2}-1} d x}

\mathrm{=\frac{1}{2} \log \left|x^{2}-1\right|} \\

\mathrm{=\log \sqrt{1-x^{2}} \quad(\text { as } x \in(-1,1))} \\

\mathrm{\text { I.F. }=\sqrt{1-x^{2}}}

\therefore Solution is

\mathrm{y \cdot \sqrt{1-x^{2}}=\int\left(x^{4}+2 x\right) d x} \\

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

\therefore It passes through origin

\mathrm{0=0+0+c \Rightarrow c=0} \\

\mathrm{\therefore \quad y =\left(\frac{x^{5}}{5}+x^{2}\right) \cdot \frac{1}{\sqrt{1-x^{2}}} }\\

\text { Now } \int_{-\frac{\sqrt{3}}{2}} ^{\frac{\sqrt{3}}{2}}f(x) d x \\

\mathrm{= \int _{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{x^{5}}{5\sqrt{1-x^{2}}}+\int _{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{x^{2}}{\sqrt{1-x^{2}}}dx}

\mathrm{=0+2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x}

( Using odd and even function properties )

\mathrm{Put \;x=\sin \theta \Rightarrow d x=\cos \theta d \theta}

\mathrm{=2 \int_{0}^{\pi / 3} \frac{\sin ^{2} \theta \cdot \cos \theta d \theta}{\cos \theta}} \\

\mathrm{=\int_{0}^{\pi / 3} 2 \sin ^{2} \theta d \theta}

\mathrm{=\int_{0}^{\pi / 3}(1-\cos 2 \theta) d \theta} \\

\mathrm{=\theta-\left.\frac{\sin 2 \theta}{2}\right|_{0} ^{\pi / 3}} \\

\mathrm{=\frac{\pi}{3}-\left(\frac{\sqrt{3}}{4}\right)}

  Hence correct option is 2

 

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Anam Khan

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