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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Show that the differential equation is homogeneous and solve $x^2\large\frac{dy}{dx}$$=x^2-2y^2+xy$

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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:
We can rewrite the equation as $\large\frac{dy}{dx} =\frac{ (x^2-2y^2+xy)}{x^2}$
$F(x,y) = \large\frac{(x^2 - 2y^2 + xy)}{x^2}$
$F(kx,ky) = \large\frac{(k^2X^2 - 2k^2y^2 + kxky)}{k^2x^2}$$ = k^0.F(x,y)$
Hence the given equation is a homogenous equation of degree zero.
Step 2:
Now using the information in the tool box , let us substitute $y = vx$. Hence
$v+x\large\frac{dv}{dx} =\frac{ (x^2 - 2v^2x^2 + xvx)}{x^2}$
Taking $x^2$ as a common factor and cancelling we get
$v+x\large\frac{dv}{dx} =$$ 1 - 2v^2 +v$
Bringing $v$ from LHS to RHS and cancelling v we get
$x\large\frac{dv}{dx }= $$1-2v^2$
Seperating the variables we get
$\large\frac{dv}{(1-2v^2) }= \frac{dx}{x}$
Step 3:
On integrating both sides we get
$\int\large\frac{ dv}{(1-2v^2)} = \int\large\frac{dx}{x }$
$\int\large\frac{ dv}{2(1/2 - v^2) }= \int\large\frac{ dx}{x}$
$\int\large\frac{dv}{(1/\sqrt{ 2^2 - v^2 })}$ is of the form $\int\large\frac{ dx}{a^2 - x^2} = \large\frac{1}{2a}$$\log\large\frac{ |a+x|}{|a-x|}$
Here in the place of $a$ we have $\sqrt{ \large\frac{1}{2 }}$and $x$ as $v$
$\Rightarrow (\large\frac{1}{2}). \frac{2.1}{\sqrt 2}$$\log\Large\frac{\frac{1}{\sqrt{ 2 +v}}}{\large\frac{1}{\sqrt{2 - v}}}$
$\large\frac{1}{2\sqrt 2}$$ \log\Large\frac{\large\frac{1}{\sqrt 2 +v}}{\large\frac{1}{\sqrt 2 - v} }$$= \log x + C$
Step 4:
On substituting $v =\large\frac{ y}{x }$ we get
$(\large\frac{1}{2\sqrt 2}) $$\log\Large\frac{(\large\frac{1}{\sqrt 2}) + (\large\frac{y}{x})}{(\large\frac{1}{\sqrt 2}) - (\large\frac{y}{x})} $$= \log x + C$
$(\large\frac{1}{2\sqrt 2})$$ \log\large\frac{x+\sqrt y}{x - \sqrt 2.y }$$= \log x + C$
This is the required solution of the equation.
answered Aug 13, 2013 by sreemathi.v
 

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