If $\ln u = \frac{x^3 + y^3}{3x + 4y}$, then the value of $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}$ is given by:
यदि $\ln u = \frac{x^3 + y^3}{3x + 4y}$ हो, तो $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}$ का मान है:
Let $f(x, y) = \ln u = \frac{x^3 + y^3}{3x + 4y}$.
- Testing homogeneity of $f(x, y)$:
$$f(tx, ty) = \frac{(tx)^3 + (ty)^3}{3(tx) + 4(ty)} = \frac{t^3 (x^3 + y^3)}{t(3x + 4y)} = t^2 f(x, y)$$
Thus, $f(x, y) = \ln u$ is a homogeneous function of degree $n = 2$.
- By Euler's Theorem for homogeneous functions:
$$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n f = 2\ln u$$ - Since $f = \ln u$, $\frac{\partial f}{\partial x} = \frac{1}{u}\frac{\partial u}{\partial x}$ and $\frac{\partial f}{\partial y} = \frac{1}{u}\frac{\partial u}{\partial y}$:
$$\frac{1}{u}\left(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}\right) = 2\ln u$$
$$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 2u \ln u$$
Hence, Option 3 is correct.
मान लीजिए $f = \ln u = \frac{x^3 + y^3}{3x + 4y}$:
यह घात $n = 3 - 1 = 2$ का एक समघात फलन (homogeneous function) है।
ऑयलर के प्रमेय से:
अतः विकल्प 3 सही उत्तर है।
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