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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
Question
Let $y=y(x)$ be the solution of the differential equation $\left(1-x^2\right) d y=\left[x y+\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}\right] d x$ $-1 < x < 1, y(0)=0$. If $y\left(\frac{1}{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}, \mathrm{m}$ and $\mathrm{n}$ are coprime numbers, then $\mathrm{m}+\mathrm{n}$ is equal to . . . . . . . . . .
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Answer

c
$ \frac{d y}{d x}-\frac{x y}{1-x^2}=\frac{\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}}{1-x^2} $

$ I F=e^{-\int \frac{x}{1-x^2} d x}=e^{+\frac{1}{2} \ln \left(1-x^2\right)}=\sqrt{1-x^2} $

$ y \sqrt{1-x^2}=\sqrt{3} \int\left(x^3+2\right) d x $

$ y \sqrt{1-x^2}=\sqrt{3}\left(\frac{x^4}{4}+2 x\right)+c $

$ \Rightarrow y(0)=0 \quad \therefore c=0 $

$ y\left(\frac{1}{2}\right)=\frac{65}{32}=\frac{m}{n} $

$ m+n=97$

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