JEE Main 2024 — Mathematics Calculus

Given,

$$(1-{x}^{2})dy=[xy+({x}^{3}+2)\sqrt{3(1-{x}^{2})}]dx$$(1−x2)dy=[xy+(x3+2)3(1−x2)​]dx

$$\Rightarrow \frac{dy}{dx}-\frac{xy}{1-{x}^{2}}=\frac{({x}^{3}+2)\sqrt{3(1-{x}^{2})}}{1-{x}^{2}}$$⇒dxdy​−1−x2xy​=1−x2(x3+2)3(1−x2)​​

The above equation is a linear differential equation,

So, $\displaystyle \mathrm{IF}={e}^{-\int \frac{x}{1-{x}^{2}}dx}={e}^{\frac{1}{2}\mathrm{ln}(1-{x}^{2})}=\sqrt{1-{x}^{2}}$IF=e−∫1−x2x​dx=e21​ln(1−x2)=1−x2​

Now, the solution of the differential equation is given by,

$$y\sqrt{1-{x}^{2}}=\int \frac{({x}^{3}+2)\sqrt{3(1-{x}^{2})}}{1-{x}^{2}}\times \sqrt{(1-{x}^{2})}dx$$y1−x2​=∫1−x2(x3+2)3(1−x2)​​×(1−x2)​dx

$$\Rightarrow y\sqrt{1-{x}^{2}}=\sqrt{3}\int ({x}^{3}+2)dx$$⇒y1−x2​=3​∫(x3+2)dx

$$\Rightarrow y\sqrt{1-{x}^{2}}=\sqrt{3}(\frac{{x}^{4}}{4}+2x)+c$$⇒y1−x2​=3​(4x4​+2x)+c

Now, using $y(0)=0$y(0)=0 we get, $c=0$c=0

Now, finding $y(\frac{1}{2})$y(21​)

$$y\sqrt{1-{(\frac{1}{2})}^{2}}=\sqrt{3}(\frac{1}{{2}^{4}\cdot 4}+2\cdot \frac{1}{2})$$y1−(21​)2​=3​(24⋅41​+2⋅21​)

$$\Rightarrow y=(\frac{1}{32}+2)$$⇒y=(321​+2)

$$\Rightarrow y(\frac{1}{2})=\frac{65}{32}=\frac{m}{n}$$⇒y(21​)=3265​=nm​

Hence, $m+n=97$m+n=97