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Show that the given differential equation is homogeneous
and solve each of them.

Q1.    (x^2 + xy)dy = (x^2 + y^2)dx

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The given differential equation can be written as
\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}
Let  F(x,y)=\frac{x^{2}+y^{2}}{x^{2}+xy}
Now, F(\lambda x,\lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}
                                =\frac{x^{2}+y^{2}}{x^{2}+xy} = \lambda ^{0}F(x,y)   

Hence, it is a homogeneous equation.

To solve it, put y = vx 
Diff
erentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}
                                 
Substitute this value in equation (i)

\\v +x\frac{dv}{dx} = \frac{x^{2}+(vx)^{2}}{x^{2}+x(vx)}\\ v +x\frac{dv}{dx} = \frac{1+v^{2}}{1+v}

x\frac{dv}{dx} = \frac{(1+v^{2})-v(1+v)}{1+v} = \frac{1-v}{1+v}

( \frac{1+v}{1-v})dv = \frac{dx}{x}

( \frac{2}{1-v}-1)dv = \frac{dx}{x}
Integrating on both sides, we get:
\\-2\log(1-v)-v=\log x -\log k\\ v= -2\log (1-v)-\log x+\log k\\ v= \log\frac{k}{x(1-v)^{2}}\\
Again substitute the value y = \frac{v}{x} ,we get:

\\\frac{y}{x}= \log\frac{kx}{(x-y)^{2}}\\ \frac{kx}{(x-y)^{2}}=e^{y/x}\\ (x-y)^{2}=kxe^{-y/x}
This is the required solution to the given differential equation. 

 

 

Posted by

manish

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