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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Show that the differential equation is homogeneous and solve $y\: dx+x\log\bigg(\large\frac{y}{x}\bigg)$$dy-2x\;dy=0$

$\begin{array}{1 1}\log(\large\frac{y}{x}) - 1 = Cy \\\log(\large\frac{x}{y}) - 1 = Cy \\ \log(\large\frac{y}{x}) + 1 = Cy \\\log(\large\frac{x}{y}) +1 = Cy \end{array} $

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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:
The given equation can be rearranged and written as $\large\frac{dy}{dx }=-\bigg[\frac{ y}{x\log(y/x) -2x}\bigg]$
$F(x,y) =\large\frac{ ky}{kx\log ky/kx - 2kx}$$= k^0.F(x,y)$
This is a homogenous funtion in degree zero.
Step 2:
Using the information in the tool box, let us substitute for $y$ and $\large\frac{dy}{dx}$
$v + x\large\frac{dv}{dx} = $$vx[x\log v - 2x]$
Taking the common factor $x$ and cancelling we get
$v + x\large\frac{dv}{dx} =\frac{ v}{[\log v -2]}$
bringing $v$ from LHS to RHS we get
$x\large\frac{dv}{dx} =\large\frac{ [v+v\log v - 2v]}{[2-\log v]}$
$x\large\frac{dv}{dx} =\large\frac{ [v\log v - v]}{[2-\log v]}$
seperating the variables we get,
$\large\frac{[2-logv]dv}{v(\log v - 1) }= \frac{dx}{x }$
Step 3:
Integrating on both sides we get
$\int\large\frac{ [2-\log v]}{v(\log v - 1) }= \int \large\frac{dx}{x}$
writing 2 as 1+1 we get,
$\int\large\frac{ [1+1-\log v]}{v(\log v-1)}$$dv= \int\large\frac{ dx}{x }$
$\int\large\frac{dv}{v(\log v-1) }- \int \large\frac{dv}{v }= \int\large\frac{ dx}{x}$
Step 4:
Consider $\large\frac{dv}{v(logv - 1)}$,
We can integrate by the method of substitution
Let $\log v -1= t$; then $dt =\large\frac{ dv}{v }$
substituting this we get $\large\frac{dt}{t}$ or intgrating this we get $\log t$
substituting for $t$ we get $\log(\log v - 1)$
Step 5:
Hence on integrating we get,
$\log(\log v - 1) - \log v = \log x + \log C$
$\log\large\frac{[\log v - 1]}{v}$$ = \log Cx$
$\log v - 1= vCx$
$\log(\large\frac{y}{x})$$- 1 =(\large\frac{y}{x})$$Cx$
$\log(\large\frac{y}{x})$$ - 1 = Cy$
This is the required solution.
answered Aug 13, 2013 by sreemathi.v
edited Aug 13, 2013 by sreemathi.v
 

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