If u = log (x^2 + y^2)/(x + y) Then prove that x ∂^2u/∂x + y ∂u/∂y = 1.
![If u = log (x^2 + y^2)/(xy) Then prove that x ∂^2u/∂x + y ∂u/∂y = 0.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZEUM-QsZvhV7z55Eng7g2eL_fIXOkSWobwqE5LOB8KRkABEXnvx2syeWZZTpbqAtWUd6s6i2t4DS2R62ywYWUq6XeLMdvixik-6LX3k_GZymHRVoWbbl-qinL0CvHv2sfHzsNz8hTOyNm5Ey-HF6zIwbQNS9GGksXy5FCqXdeZYn2IXHgGRd_qpQDyt8/s1600-rw/If%20u%20=%20log%20%28x%5E2%20+%20y%5E2%29%28xy%29%20Then%20prove%20that%20x%20%E2%88%82%5E2u%E2%88%82x%20+%20y%20%E2%88%82u%E2%88%82y%20=%200..png)
Solution:
\(u =\log\left(\frac{x^2 + y^2}{xy}\right)\)
\(e^u = e^{\log}\left(\frac{x^2 + y^2}{xy}\right)\)
\(f\)⇒\(e^u = \left(\frac{x^2 + y^2}{xy}\right)\)
\(f(x,y) = \frac{x^2 +y^2}{x y}\)
\(f(xt,yt) = \frac{(xt)^2 +(yt)^2}{x tyt}\)
\(= \frac{t^2(x^2 + y^2)}{t^2 (xy)}\)
\(f(xt, yt) = t^{2-2}.f(x, y)\)
\(f(xt, yt) = t^0.f(x, y)\)
This is the form of f(xt, yt) = tn f(x, y) n=0.
f is a homogeneous function of degree n = 0.
Euler's theorem:
\(x\frac {\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n f \)
\(x\frac {\partial }{\partial x} (e^u)+ y\frac{\partial }{\partial y} (e^u)= 0\times f\)
\(x \,e^u \frac{\partial u}{\partial x} + y\,e^u \frac{\partial u}{\partial y}
= 0\)
\(e^u \left(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial
y}\right) =0\)
\(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} =0\)