Differential Equations Question 179
Question: The solution of the differential equation $ \frac{dy}{dx}=\frac{1-3y-3x}{1+x+y} $ is
Options:
A) $ x+y-\ell n| x+y |=c $
B) $ 3x+y+2\ell n| 1-x-y |=c $
C) $ x+3y-2\ell n| 1-x-y |=c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] This is the form in which $ \frac{a_1}{a_2}=\frac{b_1}{b_2} $ The given equation can be rewritten as $ \frac{dy}{dx}=\frac{1-3(x+y)}{1+(x+y)}=f(x+y) $ Substitute $ x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}. $ The equation then become. $ \frac{dz}{dx}-1=\frac{1-3z}{1+z}\Rightarrow \frac{dz}{dx}=\frac{1-3z+1+z}{1+z}=\frac{2-2z}{1+z} $
$ \Rightarrow \frac{1+z}{2(1-z)}dz=dx. $
On integrating we get $ \frac{1}{2}\int{\frac{1+z}{1-z}dz=\int{dx+a\Rightarrow \frac{1}{2}\int{[ \frac{2}{1-z}-1 ]}dz=x+a}} $
$ \Rightarrow -\ell n| 1-z |-\frac{1}{2}z=x+a $
$ \Rightarrow -\ell n| 1-x-y |-\frac{1}{2}(x+y)=x+a $
$ \Rightarrow -2\ell n| 1-x-y |-3x-y=2a $
$ \Rightarrow 3x+y+2\ell n| 1-x-y |=c $ Where c = - 2a
BETA
