Solution of Linear Differential Equation – JEE Main 24 January 2025 Shift 1

Solution:

Rearrange the differential equation:
$$ (1+x^2)dy = -(xy – 5x^2\sqrt{1+x^2})dx $$
$$ \frac{dy}{dx} = -\frac{xy}{1+x^2} + \frac{5x^2\sqrt{1+x^2}}{1+x^2} $$
$$ \frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}} $$

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$.
Integrating Factor (I.F.):
$$ I.F. = e^{\int \frac{x}{1+x^2} dx} = e^{\frac{1}{2} \ln(1+x^2)} = \sqrt{1+x^2} $$

Solution is given by:
$$ y(I.F.) = \int Q(I.F.) dx + C $$
$$ y\sqrt{1+x^2} = \int \frac{5x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} dx + C $$
$$ y\sqrt{1+x^2} = \int 5x^2 dx + C $$
$$ y\sqrt{1+x^2} = \frac{5x^3}{3} + C $$

Using initial condition $y(0)=0$:
$$ 0 \cdot \sqrt{1} = 0 + C \Rightarrow C = 0 $$

The particular solution is:
$$ y = \frac{5x^3}{3\sqrt{1+x^2}} $$

Find $y(\sqrt{3})$:
$$ y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+(\sqrt{3})^2}} = \frac{5(3\sqrt{3})}{3\sqrt{4}} $$
$$ = \frac{15\sqrt{3}}{3(2)} = \frac{5\sqrt{3}}{2} $$

Ans. (1)