Find the general solution of the differential equation: $e^{\frac{dy}{dx}} = x^2$.

$y = 2x(\log x - 1) + C$

The correct answer is Option (1) → $y = 2x(\log x - 1) + C$ ##

Given differential equation is

$e^{\frac{dy}{dx}} = x^2$

Taking log both sides, we get

$\frac{dy}{dx} \log_e e = 2 \log x$

$\Rightarrow \frac{dy}{dx} = 2 \log x \quad [∵\log_e e = 1]$

$\Rightarrow dy = 2 \log x \, dx$

On integrating both sides, we get

$\int dy = 2 \int \log x \, dx$

$\Rightarrow y = 2 \int 1 \cdot \log x \, dx$

$\Rightarrow y = 2 \left[ \log x \int 1 dx - \int \left( \frac{d}{dx}(\log x) \int 1 dx \right) dx \right]$

[Using integration by parts]

$\Rightarrow y = 2 \left[ \log x(x) - \int \frac{1}{x}(x) dx \right]$

$\Rightarrow y = 2[x \log x - x] + C$

$\Rightarrow y = 2x(\log x - 1) + C$