Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Solve $ \bigg( 1 + e^{\Large\frac{x}{y}} \bigg) dx + e^{\Large\frac{x}{y}} \bigg( 1 -\large \frac{x}{y} \bigg) $$dy= 0 $

$\begin{array}{1 1} (A)\;x+ye^{\Large\frac{x}{y}}=0 \\ (B)\;x-e^{\Large\frac{x}{y}}=0 \\ (C)\;x+e^{\Large\frac{2x}{y}}=0\\ (D)\;x+e^{\Large\frac{x}{2y}}=0 \end{array} $

Can you answer this question?

1 Answer

0 votes
  • If the homogenous differential equation is in the form $\large\frac{dx}{dy} =$$ F(x,y)$, where F(x,y) is homogenous function of degree zero, then we make substitution $x = vy$, hence $\large\frac{dx}{dy}$$ = v + y\large\frac{dv}{dy}.$
Step 1:
To check whether the given problem is a homogenous function of degree zero
we rewrite the given equation as $\large\frac{dx}{dy} = -\large\frac{ e^{\Large\frac{x}{y}}(1 - \large\frac{x}{y})}{1 + e^{\Large\frac{x}{y}}}$
Let $F(x,y) = -\large\frac{ e^{\Large\frac{x}{y}}(1 - \large\frac{x}{y})}{ 1+ e^{\Large\frac{x}{y}}}$
$F(kx,ky) =\large\frac{ e^{\Large\frac{-kx}{ky}}(1-\large\frac{kx}{ky})}{1 + e^{\Large kx/ky} }= -\large\frac{ e^x/y(1 - x/y)}{1 + e^x/y}$$= k^0.F(x,y) $
Hence this is a homogenous differential equation.
Step 2:
To solve this put $x = vy$ and $\large\frac{dx}{dy}$$ = v + y\large\frac{dv}{dy}$
So $v + y\large\frac{dv}{dy }= -e^{\Large\frac{vy}{y}}$$(1- vy/y)/1+e^{\Large\frac{ vy}{y}}$
Cancelling the common factor y we get
$v + v \large\frac{dv}{dy} = - e^v\large\frac{(1 - v)}{1+e^v}$
$v\large\frac{dv}{dy}= -e^v\large\frac{(1 - v)}{1+e^v}$$ -v$
$v\large\frac{dv}{dy} = v+\large\frac{ e^v}{1+e^v}$
Step 3:
Bringing v and dv to the LHS and x and dx to the RHS we get,
$[\large\frac{1+e^v}{v+e^v} ] = \frac{- dy}{y}$
Integrating on both sides we get,
$\log(v+e^y) = - \log y + \log C$
Step 4:
Substituting back for v we get
$\large\frac{x}{y} + e^{\Large\frac{x}{y }}= \frac{C}{y}$
On cancelling y throughout,we get the solution as
$x + ye^{\Large\frac{x}{y}} $$= C$
answered Mar 13, 2014 by balaji.thirumalai

Related questions

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