Question

Every curve represented by the general solution of \[ \frac{dy}{dx}-\frac{x\log x}{y^3e^{y^2-5}}=0 \] cuts every curve represented by the general solution of \[ \frac{dy}{dx}+\frac{y^3e^{y^2-5}}{x\log x}=0 \] at angle \(\theta\). Then \[ 4\theta-\frac{\pi}{2}= \]

• A. \(\frac{\pi}{2}\)
• B. \(2\pi\)
• C. \(\frac{3\pi}{2}\)
• D. \(\pi\)

Hint:

If the product of slopes of two curves at their point of intersection is \(-1\), then the curves cut each other orthogonally, so the angle between them is \(\frac{\pi}{2}\).

Answer 1

The Correct Option is C

Step 1: Find the slope of the first family of curves.
From \[ \frac{dy}{dx}-\frac{x\log x}{y^3e^{y^2-5}}=0 \] we get \[ \frac{dy}{dx}=\frac{x\log x}{y^3e^{y^2-5}} \] So, the slope of the first family is \[ m_1=\frac{x\log x}{y^3e^{y^2-5}} \]

Step 2: Find the slope of the second family of curves.
From \[ \frac{dy}{dx}+\frac{y^3e^{y^2-5}}{x\log x}=0 \] we get \[ \frac{dy}{dx}=-\frac{y^3e^{y^2-5}}{x\log x} \] So, the slope of the second family is \[ m_2=-\frac{y^3e^{y^2-5}}{x\log x} \]

Step 3: Multiply the slopes.
Now, \[ m_1m_2= \frac{x\log x}{y^3e^{y^2-5}} \left(-\frac{y^3e^{y^2-5}}{x\log x}\right) \] \[ m_1m_2=-1 \] Since the product of slopes is \(-1\), the two curves cut each other orthogonally.
Therefore, \[ \theta=\frac{\pi}{2} \]

Step 4: Calculate the required value.
We need \[ 4\theta-\frac{\pi}{2} \] Substitute \[ \theta=\frac{\pi}{2} \] \[ 4\theta-\frac{\pi}{2} = 4\left(\frac{\pi}{2}\right)-\frac{\pi}{2} \] \[ = 2\pi-\frac{\pi}{2} \] \[ = \frac{3\pi}{2} \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{3\pi}{2}} \]