Question

The general solution of the differential equations \(\frac{dy}{dx} = (9x+y+5)^2\) is...

• A. \(tan^{-1}(\frac{9x+y+5}{2}) = -2x+c\)
• B. \(tan^{-1}(\frac{9x+y+5}{2}) = 2x+c\)
• C. \(tan^{-1}(\frac{9x+y+5}{3}) = -3x+c\)
• D. \(tan^{-1}(\frac{9x+y+5}{3}) = 3x+c\)

Hint:

A vector coplanar with a and b is a combination of them; impose the perpendicularity.

Answer 1

The Correct Option is D

Step 1: Setup:
Let \(\vec a = \hat i + \hat j + 2\hat k\), \(\vec b = \hat i + 2\hat j + \hat k\), \(\vec c = \hat i + \hat j + \hat k\). Any vector coplanar with a and b is \(\vec v = \alpha\vec a + \beta\vec b\).

Step 2: Perpendicular to c:
\(\vec a\cdot\vec c = 1 + 1 + 2 = 4\) and \(\vec b\cdot\vec c = 1 + 2 + 1 = 4\). So \(\vec v\cdot\vec c = 4\alpha + 4\beta = 0\), giving \(\beta = -\alpha\).

Step 3: Unit vector:
\(\vec v \propto \vec a - \vec b = -\hat j + \hat k\). Its magnitude is \(\sqrt2\). So the unit vector is \(\pm\frac{1}{\sqrt2}(\hat j - \hat k)\).
Check: \((\hat j - \hat k)\cdot(\hat i + \hat j + \hat k) = 1 - 1 = 0\).

Final Answer:
The unit vector is \(\pm\frac{1}{\sqrt2}(\hat j - \hat k)\), option (C). \[ \boxed{\pm\frac{1}{\sqrt{2}}(\hat{j} - \hat{k})} \]