JEE Main 2024 — Mathematics Calculus

Given,

$$({x}^{2}-4)\mathrm{dy}-({y}^{2}-3y)\mathrm{dx}=0$$(x2−4)dy−(y2−3y)dx=0

$$\Rightarrow \int \frac{\mathrm{dy}}{{y}^{2}-3y}=\int \frac{\mathrm{dx}}{{x}^{2}-4}$$⇒∫y2−3ydy​=∫x2−4dx​

$$\Rightarrow \frac{1}{3}\int \frac{y-(y-3)}{y(y-3)}\mathrm{dy}=\int \frac{\mathrm{dx}}{{x}^{2}-4}$$⇒31​∫y(y−3)y−(y−3)​dy=∫x2−4dx​

$$\Rightarrow \frac{1}{3}\int (\frac{1}{y-3}-\frac{1}{y})\mathrm{dy}=\int \frac{\mathrm{dx}}{{x}^{2}-{2}^{2}}$$⇒31​∫(y−31​−y1​)dy=∫x2−22dx​

$$\Rightarrow \frac{1}{3}(\mathrm{log}|y-3|-\mathrm{log}|y|)=\frac{1}{4}\mathrm{log}|\frac{x-2}{x+2}|+C$$⇒31​(log∣y−3∣−log∣y∣)=41​log∣x+2x−2​∣+C

$$\Rightarrow \frac{1}{3}\mathrm{log}|\frac{y-3}{y}|=\frac{1}{4}\mathrm{log}|\frac{x-2}{x+2}|+C$$⇒31​log∣yy−3​∣=41​log∣x+2x−2​∣+C

At $x=4,y=\frac{3}{2}$x=4,y=23​

$$\Rightarrow \frac{1}{3}\mathrm{log}|\frac{\frac{3}{2}-3}{\frac{3}{2}}|=\frac{1}{4}\mathrm{log}|\frac{4-2}{4+2}|+C$$⇒31​log∣23​23​−3​∣=41​log∣4+24−2​∣+C

$$\Rightarrow 0=\frac{1}{4}\mathrm{log}|\frac{1}{3}|+C$$⇒0=41​log∣31​∣+C

$$\Rightarrow 0=-\mathrm{log}{|3|}^{\frac{1}{4}}+C$$⇒0=−log∣3∣41​+C

$$\Rightarrow C=\frac{1}{4}\mathrm{log}(3)$$⇒C=41​log(3)

$$\Rightarrow \frac{1}{3}\mathrm{log}|\frac{y-3}{y}|=\frac{1}{4}\mathrm{log}|\frac{x-2}{x+2}|+\frac{1}{4}\mathrm{log}(3)$$⇒31​log∣yy−3​∣=41​log∣x+2x−2​∣+41​log(3)

Putting, $x=10$x=10

$$\Rightarrow \frac{1}{3}\mathrm{log}|\frac{y-3}{y}|=\frac{1}{4}\mathrm{log}|\frac{2}{3}|+\frac{1}{4}\mathrm{log}(3)$$⇒31​log∣yy−3​∣=41​log∣32​∣+41​log(3)

$$\Rightarrow \frac{1}{3}\mathrm{log}|\frac{y-3}{y}|=\frac{1}{4}\mathrm{log}|2|$$⇒31​log∣yy−3​∣=41​log∣2∣

$$\Rightarrow \mathrm{log}|\frac{y-3}{y}|=\frac{3}{4}\mathrm{log}2$$⇒log∣yy−3​∣=43​log2

$$\Rightarrow \mathrm{log}|\frac{y-3}{y}|=\mathrm{log}({2}^{\frac{3}{4}})$$⇒log∣yy−3​∣=log(243​)

$$\Rightarrow -y+3={8}^{\frac{1}{4}}\cdot y$$⇒−y+3=841​⋅y

$$\Rightarrow 3=({8}^{\frac{1}{4}}+1)y$$⇒3=(841​+1)y

$$\Rightarrow \frac{3}{1+{(8)}^{\frac{1}{4}}}=y$$⇒1+(8)41​3​=y