Q74 / 125 · BITSAT 2014 · Mathematics
Solution of differential equation
$$
x^{2}=1+\left(\frac{x}{y}\right)^{-1} \frac{d y}{d x}+\frac{\left(\frac{x}{y}\right)^{-2}\left(\frac{d y}{d x}\right)^{2}}{2 !}+\frac{\left(\frac{x}{y}\right)^{-3}\left(\frac{d y}{d x}\right)^{3}}{3 !}+\ldots \text { is }
$$
- Correct answerA$y^{2}=x^{2}\left(\ln x^{2}-1\right)+C$
- B$y=x^{2}(\ln x-1)+C$
- C$y^{2}=x(\ln x-1)+\mathrm{C}$
- D$y=x^{2} e^{x^{2}}+\mathrm{C}$
Solution
$x^{2}=e^{\left(\frac{x}{y}\right)^{-1}\left(\frac{d y}{d x}\right)} \Rightarrow x^{2}=e^{\left(\frac{y}{x}\right)\left(\frac{d y}{d x}\right)}$
$\Rightarrow \ln x^{2}=\frac{y}{x} \frac{d y}{d x}$ or $\int x \ln x^{2} d x=\int y d y$
Put $x^{2}=t \Rightarrow 2 x d x=d t \therefore \frac{1}{2} \int \ln t d t=\frac{y^{2}}{2}$
$C+t \ln t-t=y^{2}$ or $y^{2}=x^{2}\left(\ln x^{2}-1\right)+C$
Hence, the correct answer is option: 'A'.