Differential Equations Question 186
Question: The solution of (y + x + 5) dy = (y - x + 1) dx is
Options:
A) $ \log ({{(y+3)}^{2}}+{{(x+2)}^{2}})+ta{n^{-1}}\frac{y+3}{y+2}+C $
B) $ \log ({{(y+3)}^{2}}+{{(x-2)}^{2}})+ta{n^{-1}}\frac{y-3}{x-2}=C $
C) $ \log ({{(y+3)}^{2}}+{{(x+2)}^{2}})+2ta{n^{-1}}\frac{y+3}{x+2}=C $
D) $ \log ({{(y+3)}^{2}}+{{(x+2)}^{2}})-2ta{n^{-1}}\frac{y+3}{x+2}=C $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The intersection of $ y-x+1=0 $ and $ y+x+5=0 $ is $ ( -2,-3 ) $ .
Put $ x=X-2,y=Y-3. $ The given equation reduces to $ \frac{dy}{dx}=\frac{Y-X}{Y+X} $ .
Putting $ Y=vX, $ we get $ X\frac{dv}{dX}=-\frac{v^{2}+1}{v+1} $ Or $ ( -\frac{v}{v^{2}+1}-\frac{1}{v^{2}+1} )dv=\frac{dx}{X} $
Or $ -\frac{1}{2}\log ( v^{2}+1 )-{{\tan }^{-1}}v=\log | X |+ $ constant Or $ \log (Y^{2}+X^{2})+2ta{n^{-1}}\frac{Y}{X}=constant $
Or $ \log ({{(y+3)}^{3}}+{{(x+2)}^{2}})+2ta{n^{-1}}\frac{y+3}{x+2}=C $
BETA
