Question

If u = sin^-1 (x^1/2 + y ^1/2 /x^1/3+ y ^1/3) , show that x(du/dx) + y (du/dy) = 1/12 tan u

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{u=sin^{-1}\left(\dfrac{x^\frac{1}{2}+y^\frac{1}{2}}{x^\frac{1}{3}+y^\frac{1}{3}}\right)}$

$\underline{\textbf{To prove:}}$

$\mathsf{x\,\dfrac{\partial\,u}{\partial\,x}+y\,\dfrac{\partial\,u}{\partial\,y}=\dfrac{1}{6}tanu}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Euler's theorem:}}$

$\textsf{If f is a homogeneous function of degree n in x and y}$

$\mathsf{then\;x\,\dfrac{\partial\,u}{\partial\,x}+y\,\dfrac{\partial\,u}{\partial\,y}=n\,f}$

$\mathsf{Consider,}$

$\mathsf{u=sin^{-1}\left(\dfrac{x^\frac{1}{2}+y^\frac{1}{2}}{x^\frac{1}{3}+y^\frac{1}{3}}\right)}$

$\mathsf{sin\,u=\dfrac{x^\frac{1}{2}+y^\frac{1}{2}}{x^\frac{1}{3}+y^\frac{1}{3}}=f(x,y)\;(say)}$

$\mathsf{f(tx,ty)=\dfrac{(tx)^\frac{1}{2}+(ty)^\frac{1}{2}}{(tx)^\frac{1}{3}+(ty)^\frac{1}{3}}}$

$\mathsf{f(tx,ty)=\dfrac{t^\frac{1}{2}x^\frac{1}{2}+t^\frac{1}{2}y^\frac{1}{2}}{t^\frac{1}{3}x^\frac{1}{3}+t^\frac{1}{3}y^\frac{1}{3}}}$

$\mathsf{f(tx,ty)=\dfrac{t^\frac{1}{2}(x^\frac{1}{2}+y^\frac{1}{2})}{t^\frac{1}{3}(x^\frac{1}{3}+y^\frac{1}{3})}}$

$\mathsf{f(tx,ty)=t^{\frac{1}{2}-\frac{1}{3}}\left(\dfrac{x^\frac{1}{2}+y^\frac{1}{2}}{x^\frac{1}{3}+y^\frac{1}{3}}\right)}$

$\mathsf{f(tx,ty)=t^\frac{3-2}{6}\left(\dfrac{x^\frac{1}{2}+y^\frac{1}{2}}{x^\frac{1}{3}+y^\frac{1}{3}}\right)}$

$\mathsf{f(tx,ty)=t^\frac{1}{6}\,f(x,y)}$

$\therefore\mathsf{f\;is\;a\;homogeneous\;function\;of\;degree\;\dfrac{1}{6}}$

$\textsf{By Euler's theorem,}$

$\mathsf{x\,\dfrac{\partial\,f}{\partial\,x}+y\,\dfrac{\partial\,f}{\partial\,y}=\dfrac{1}{6}f}$

$\mathsf{x\,\dfrac{\partial(sin\,u)}{\partial\,x}+y\,\dfrac{\partial(sin\,u)}{\partial\,y}=\dfrac{1}{6}sin\,u}$

$\mathsf{x\,cos\,u\dfrac{\partial\,u}{\partial\,x}+y\,cos\,u\dfrac{\partial\,u}{\partial\,y}=\dfrac{1}{6}sin\,u}$

$\mathsf{x\,\dfrac{\partial\,u}{\partial\,x}+y\,\dfrac{\partial\,u}{\partial\,y}=\dfrac{1}{6}\dfrac{sin\,u}{cos\,u}}$

$\boxed{\mathsf{x\,\dfrac{\partial\,u}{\partial\,x}+y\,\dfrac{\partial\,u}{\partial\,y}=\dfrac{1}{6}\,tanu}}$

$\underline{\textbf{Find more:}}$

If u = cos ^(-1) x+y/√x+√y, prove that , x du/dx + y dy/dx + 1/2 cot u =0​

https://brainly.in/question/36360810  

Answer 2

Step-by-step explanation:

AC = BD ( Diagonals bisect each other )

→ OA = y + 4 and OC = 2y - 3 ( given )

→ ( y + 4 ) = ( 2y - 3 )

→ 4 + 3 = 2y - y

→ 7 = y

Therefore , the value of y = 7