JEE Main 2020 — Mathematics Calculus

$$\frac{dy}{dx}=\frac{y}{x}+\frac{{y}^{2}}{2{x}^{2}}$$dxdy​=xy​+2x2y2​

$$\Rightarrow {y}^{-2}\frac{dy}{dx}-\frac{1}{y}\cdot \frac{1}{x}=\frac{1}{2{x}^{2}}$$⇒y−2dxdy​−y1​⋅x1​=2x21​

Put $-\frac{1}{y}=t\Rightarrow \frac{1}{{y}^{2}}\frac{dy}{dx}=\frac{dt}{dx}$−y1​=t⇒y21​dxdy​=dxdt​

$$\Rightarrow \frac{dt}{dx}+(\frac{1}{x})t=\frac{1}{2{x}^{2}}$$⇒dxdt​+(x1​)t=2x21​

This is a linear differential equation,

with Integrating Factor : $\displaystyle {e}^{\int \frac{1}{x}dx}={e}^{\mathrm{ln}(x)}=x$e∫x1​dx=eln(x)=x

So, solution of the linear differential equation is $\displaystyle tx=\int \frac{1}{2{x}^{2}}\cdot xdx+C$tx=∫2x21​⋅xdx+C

$$\Rightarrow -\frac{x}{y}=\frac{1}{2}\mathrm{ln}(x)+C$$⇒−yx​=21​ln(x)+C

The curve passes through $(1,2)$(1,2)

$$\Rightarrow -\frac{1}{2}=\frac{1}{2}\mathrm{ln}(1)+C\Rightarrow C=-\frac{1}{2}$$⇒−21​=21​ln(1)+C⇒C=−21​

Hence, the particular solution to the differential equation is $-\frac{x}{y}=\frac{1}{2}\mathrm{ln}(x)-\frac{1}{2}$−yx​=21​ln(x)−21​

$$\Rightarrow \frac{x}{y}=\frac{1-\mathrm{ln}(x)}{2}\Rightarrow y=\frac{2x}{1-\mathrm{ln}(x)}$$⇒yx​=21−ln(x)​⇒y=1−ln(x)2x​

$$\Rightarrow f(\frac{1}{2})=\frac{2\times \frac{1}{2}}{1-\mathrm{ln}(\frac{1}{2})}=\frac{1}{1+\mathrm{ln}(2)}=\frac{1}{1+{\mathrm{log}}_{e}(2)}$$⇒f(21​)=1−ln(21​)2×21​​=1+ln(2)1​=1+loge​(2)1​