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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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True-or-False: The solution of the differential equation $\large\frac{dy}{dx} = \frac{x+2y}{x}$ is $x+y=kx^2$.

$\begin{array}{1 1} True \\ False \\ cannot be determined \end{array}$

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Toolbox:
  • A homogenous linear differential equation can be solved by putting $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large \frac{dv}{dx}$
  • $\int \large\frac{dx}{1+x}$$=\log |1+x|+c$
$\large\frac{dy}{dx}=\frac{x+2y}{x}$
This is a homogenous linear differential equation hence this can be solved by substituting $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large \frac{dv}{dx}$
$v+x \large \frac{dv}{dx}=\frac{x+2vx}{x}$
$v+x \large \frac{dv}{dx}=\frac{x(1+2v)}{x}$
$=>x \large \frac{dv}{dx}$$=1+2v-v$
$=>x \large \frac{dv}{dx}$$=1+v$
Now seperating the variables we get,
$\large\frac{dv}{1+v}=\frac{dx}{x}$
Now integrating on both sides we get,
$ \large \int \frac{dv}{1+v}=\int \frac{dx}{x}$
$\log |1+v|=\log x +\log k$
Substituting for $v=y/x$ we get,
$\log |1+\large\frac{y}{x}|$$=\log kx$
$=> \large \frac{x+y}{x}$$=kx$
Therefore $ x+y=kx^2$
Hence the solution of the differential equation $=\large\frac{dy}{dx}=\frac{x+2y}{x}$ is $x+y=kx^2$. is $True$
answered May 23, 2013 by meena.p
 

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