JEE Main 2023 — Mathematics Calculus

Given differential equation can be rewritten as,

$$\frac{dy}{dx}-\frac{y}{x}={y}^{3}(1+{\mathrm{log}}_{e}x)$$dxdy​−xy​=y3(1+loge​x)

$$\frac{1}{{y}^{3}}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{{\mathrm{xy}}^{2}}=1+{\mathrm{log}}_{e}x$$y31​dxdy​−xy21​=1+loge​x

Let $(-\frac{1}{{y}^{2}})=t\Rightarrow \frac{2}{{y}^{3}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$(−y21​)=t⇒y32​dxdy​=dxdt​

$$\therefore \frac{\mathrm{dt}}{2\mathrm{dx}}+\frac{t}{x}=(1+{\mathrm{log}}_{e}x)$$∴2dxdt​+xt​=(1+loge​x)

$$\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}+\frac{2t}{x}=2(1+{\mathrm{log}}_{e}x)..........(1)$$⇒dxdt​+x2t​=2(1+loge​x)..........(1)

We know the solution of the differential equation $\frac{dy}{dx}+Py=Q$dxdy​+Py=Q is given by,

$${\mathrm{ye}}^{\int \mathrm{Pdx}}=\int {\mathrm{Qe}}^{\int \mathrm{Pdx}}+C(\mathrm{Where}I.F.={e}^{\int \mathrm{Pdx}})$$ye∫Pdx=∫Qe∫Pdx+C(WhereI.F.=e∫Pdx)

Therefore, on solving equation(1) we get,

I.F.$\displaystyle ={e}^{\int \frac{2}{x}dx}={x}^{2}$=e∫x2​dx=x2

$$\frac{-{x}^{2}}{{y}^{2}}=\frac{2}{3}((1+{\mathrm{log}}_{e}x){x}^{3}-\frac{{x}^{3}}{3})+C......(2)$$y2−x2​=32​((1+loge​x)x3−3x3​)+C......(2)

Given,$y(1)=3$y(1)=3

$$\Rightarrow \frac{1}{9}=-2(\frac{1}{3}-\frac{1}{9})+C$$⇒91​=−2(31​−91​)+C

$$\therefore C=\frac{5}{9}$$∴C=95​

Now putting value of C in equation(2) we get,

$$\frac{{y}^{2}}{9}=\frac{{x}^{2}}{5-2{x}^{3}(2+{\mathrm{log}}_{e}{x}^{3})}$$9y2​=5−2x3(2+loge​x3)x2​