logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

General solution of differential equation : $(1+y^2)dx =(\tan^{-1}y -x)dy$

$(a)\;x= \tan ^{-1} x+1+ce^{\tan ^{-1} x} \\ (b)\;y= \tan ^{-1} x+1+ce^{\tan ^{-1} x} \\ (c)\;x= \tan ^{-1} -1+ce^{\tan ^{-1} x} \\ (d)\;y= \tan ^{-1} x-1+ce^{\tan ^{-1} x} $
Can you answer this question?
 
 

1 Answer

0 votes
$(1+y^2)\large\frac{dx}{dy}$$=\tan ^{-1} y-x$
$(1+y^2) \large\frac{dx}{dy} $$+x=\tan ^{-1}y$
$\large\frac{dx}{dy} +\frac{x}{1+y^2}=\large\frac{\tan ^{-1}y}{1+y^2}$
$I.P= e^{\int pdy} $
$\qquad= e^{\int \Large\frac{1}{1+y}.dy}$
$\qquad= e^{\tan ^{-1}y}$
Thus the solution is $x(I.F) =\int \Large\frac{\tan ^{-1} y}{1+y^2} $ $I.F\;dy+c$
$xe^{\tan ^{-1}y} =\int \large\frac{\tan ^{-1}y}{1+y^2} e^{\tan ^{-1}y} $$dy+c$
$\tan ^{-1}y=t$
$\large\frac{1}{1=t^2}$$dt=dt$
$xe^{\tan ^{-1}y} =\int t.e^{t}dt+c$
$xe^{\tan ^{-1}y} = t.e^{t} -\int e^t.dt+c$
$xe^{\tan ^{-1}y} = t.e^{t}-e^t+c$
$xe^{\tan ^{-1}y} =\tan ^{-1}y e^{\tan ^{-1}y}-e^{\tan ^{-y}}+c$
$x= \tan ^{-1} -1+ce^{\tan ^{-1} x}$
Hence c is the correct answer.
answered Feb 4, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...