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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Show that the differential equation is homogeneous and 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} $

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1 Answer

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Toolbox:
  • 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 Aug 19, 2013 by sreemathi.v
 

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