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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve the differential equation $ y.e^{\large\frac{x}{y}} dx=(x.e^{\large\frac{x}{y}}+y^2)dy ; y \neq 0$

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1 Answer

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Toolbox:
  • A differential equation of the form $\large\frac{dy}{dx }$$= F(x,y)$ is said to be homogenous if $F(x,y)$ is a homogenous function of degree zero.
  • To solve this type of equations substitute $y = vx$ and $\large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
$ye^{\Large\frac{x}{y}}dx=(xe^{\Large\frac{x}{y}}+y^2)dy$
This can be written as
$\large\frac{dx}{dy}=\frac{xe^{\Large x/y}+y^2}{ye^{\Large x/y}}$
Step 2:
Clearly this is a homogeneous differential equation .
Hence put $x=vy$
$\large\frac{dx}{dy}=$$v+y\large\frac{dv}{dy}$
$\therefore v+y\large\frac{dv}{dy}=\frac{vye^v+y^2}{ye^v}$
$\Rightarrow \large\frac{ ve^v+y}{e^v}$
$\Rightarrow y\large\frac{dv}{dy}=\frac{ve^v+y}{e^v}$$-v$
$y\large\frac{dv}{dy}=\frac{y}{e^v}$
Step 3:
Now separating the variables we get,
$e^vdv=dy$
On integrating we get
$\int e^vdv=\int dy$
(i.e) $e^v=y+c$
Substituting for $v=\large\frac{x}{y}$ we get,
$e^{\Large\frac{x}{y}}=$$y+c$
This is the required solution.
answered Nov 7, 2013 by sreemathi.v
 

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