Differential Equations Question 247
Question: The solution of $ (y+x+5)dy=(y-x+1)dx $ is
Options:
A) $ \log ({{(y+3)}^{2}}+{{(x+2)}^{2}})+{{\tan }^{-1}}\frac{y+3}{y+2}+C $
B) $ \log ({{(y+3)}^{2}}+{{(x+2)}^{2}})+{{\tan }^{-1}}\frac{y-3}{y-2}=C $
C) $ \log ({{(y+3)}^{2}}+{{(x+2)}^{2}})+2{{\tan }^{-1}}\frac{y+3}{y+2}=C $
D) $ \log ({{(y+3)}^{2}}+{{(x+2)}^{2}})-2{{\tan }^{-1}}\frac{y+3}{y+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}. $ This is a homogeneous equation. Putting $ y=vX, $ we get $ X\frac{dv}{dX}=-\frac{v^{2}+1}{v+1} $
$ \Rightarrow ( -\frac{v}{v^{2}+1}-\frac{1}{v^{2}+1} )dv=\frac{dX}{X} $
$ \Rightarrow -\frac{1}{2}\log ( v^{2}+1 )-{{\tan }^{-1}}v=\log | X |+cons\tan t $
$ \Rightarrow \log (Y^{2}+X^{2})+2ta{n^{-1}}\frac{Y}{X}= $ constant
$ \Rightarrow \log {{(y+3)}^{2}}+{{(x+2)}^{2}})+2{{\tan }^{-1}}\frac{y+3}{x+2}=C $
BETA
