Question

The complete solution of \( (mz-ny)\frac{\partial z}{\partial x} + (nx-lz)\frac{\partial z}{\partial y} = ly-mx \) is

• A. \( xyz = f(lx+my+nz) \)
• B. \( lx+my+nz = f(x+yz) \)
• C. \( x^2+y^2+z^2 = f(lx+my+nz) \)
• D. \( (lx)^2+(my)^2+(nz)^2 = f(lmnxyz) \)

Hint:

For \(Pp+Qq=R\), use Lagrange's auxiliary equations \(\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}\).
Find two independent solutions \(u(x,y,z)=c_1\) and \(v(x,y,z)=c_2\) using method of multipliers or grouping.
The general solution is then \(F(u,v)=0\) or \(u=f(v)\).

Answer 1

The Correct Option is C

Two functions \(f(x)\) and \(g(x)\) are linearly dependent if there exist constants \(c_1, c_2\), not both zero, such that \(c_1f(x) + c_2g(x) = 0\) for all \(x\). This is equivalent to one function being a constant multiple of the other (if neither is identically zero).

Let's examine option (b): \(f(x) = \sin x(4\sin^2 x - 3)\) and \(g(x) = \sin 3x\). 

We know the trigonometric identity for \(\sin 3x\): \(\sin 3x = 3\sin x - 4\sin^3 x\).

Now consider \(f(x)\): \(f(x) = \sin x(4\sin^2 x - 3) = 4\sin^3 x - 3\sin x\).

Comparing \(f(x)\) with \(\sin 3x\): \(f(x) = 4\sin^3 x - 3\sin x = -(3\sin x - 4\sin^3 x) = -\sin 3x\).

So, \(f(x) = -g(x)\), or \(f(x) + g(x) = 0\). Since we can write \(1 \cdot f(x) + 1 \cdot g(x) = 0\), with non-zero constants, the functions are linearly dependent.

Let's briefly check other options:

(a) \(e^x \sin 2x, e^x \cos 2x\): Linearly independent as \(\sin 2x\) and \(\cos 2x\) are independent.

(c) \(\cos x, x \cos x\): If \(c_1 \cos x + c_2 x \cos x = 0 \Rightarrow \cos x (c_1 + c_2 x) = 0\). For this to hold for all \(x\), \(c_1=0\) and \(c_2=0\). Linearly independent.

(d) \(e^{3x}, (x+1)e^{2x}\): Different exponential growth rates and forms. Linearly independent.

\[ \boxed{\sin x(4\sin^2 x - 3), \sin 3x} \]