JEE Main 2022 — Differential Equations Question with Solution

$$\underset{3}{\overset{x}{\int }}f(x)dx={\left(\frac{f(x)}{x}\right)}^3$$

$$x^3.\underset{3}{\overset{x}{\int }}f(x)dx=f^3(x)$$

Differentiate w.r.t. x

$$x^{3}f(x)+3x^2.\frac{f^3(x)}{x^3}=3f^2(x)f^{\prime }(x)$$

$$\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y+\frac{3y^3}{x}$$

$$3xy\frac{dy}{dx}=x^4+3y^2$$

Let $$y^2=t$$

$$\frac{3}{2}\frac{dt}{dx}=x^3+\frac{3t}{x}$$

$$\frac{dt}{dx}-\frac{2t}{x}=\frac{2x^3}{3}$$

$$I.F.=.e^{\int -\frac{2}{x}dx}=\frac{1}{x^2}$$

Solution of differential equation

$$t.\frac{1}{x^2}=\int \frac{2}{3}xdx$$

$$\frac{y^2}{x^2}=\frac{x^2}{3}+C$$

$$y^2=\frac{x^4}{3}+C{x}^2$$

Curve passes through $$(3,3)\Rightarrow C=-2$$

$$y^2=\frac{x^4}{3}-2x^2$$

Which passes through $$\left(\alpha ,6\sqrt{10}\right)$$

$$\frac{{\alpha }^4-6{\alpha }^2}{3}=360$$

$${\alpha }^4-6{\alpha }^2-1080=0$$

$$\alpha =6$$