Every curve represented by the general solution of $\frac{d y}{d x}=\frac{x \log x}{y^3 e^{y^2-5}}=0$ cuts…
Every curve represented by the general solution of $\frac{d y}{d x}=\frac{x \log x}{y^3 e^{y^2-5}}=0$ cuts every curve represented by the general solution of $\frac{d y}{d x}+\frac{y^3 e^{y^2-5}}{x \log x}=0$ at angle $\theta$. Then, $4 \theta-\frac{\pi}{2}=$
$\frac{\pi}{2}$
$2 \pi$
$\frac{3 \pi}{2}$
$\pi$
Solution
Given, $\frac{d y}{d x}=\frac{x \log x}{y^3 e^{y^2-5}}$
$\therefore$ Slope of this curve is $m_1=\frac{x \log x}{y^3 e^{y^2-5}}$ and $\frac{d y}{d x}+\frac{y^3 e^{y^2-5}}{x \log x}=0$
$
\Rightarrow \quad \frac{d y}{d x}=-\frac{y^3 e^{y^2-5}}{x \log x}
$
$\therefore$ Slope of this curve is
$
m^2=\frac{-y^3 e^{y^2-5}}{x \log x}
$
Now, $m_1 \times m_2=\frac{x \log x}{y^3 e^{y^2-5}} \times\left(\frac{-y^3 e^{y^2-5}}{x \log x}\right)=-1$
$
\Rightarrow \quad \theta=\frac{\pi}{2}
$
Now, $4 \theta-\frac{\pi}{2}=(4)\left(\frac{\pi}{2}\right)-\frac{\pi}{2}=2 \pi-\frac{\pi}{2}=\frac{3 \pi}{2}$