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Question
Show that the given differential equation is homogeneous and solve them.
(x2 + xy) dy = (x2 + y2) dx
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Solution
Let `dy/dx = (x^2 + y^2)/(x^2 + xy) = f (xy)` ..... (i)
Now, f `(lamda x, lambda y) = (lambda^2 (x^2 + y^2))/(lambda^2 (x^2 + xy)) = lambda^0` f (x, y)
`therefore` F(x,y) is an exponential function of degree zero.
Hence the given differential equation is a homogeneous differential equation.
Now y = vx
`dy/dx = v + x (dv)/dx`
Then, from equation (i)
V + x `(dv)/dx = (x^2 + v^2 x^2)/(x^2 + vx^2)`
`=> x (dv)/dx = (1 + v^2)/(1 + v) - v`
`x (dv)/dx = (1 + v - v - v^2)/(1 + v)`
`x (dv)/dx = (dv)/dx = (1 - v)/(1 + v)`
`=> (1 + v)/(1 - v) dv = dx/x`
On integrating,
`int (1 + v)/(1 - v) dv = int 1/x dx`
`=> int (-1 + 2/(1 - v)) dv = int 1/x dx`
⇒ - v - 2 log (1 - v) = log x + log C
⇒ - v = log Cx + 2 log (1 - v)
⇒ - v = log Cx + log (1 - v)2
Cx . (1 - v)2 = e-v
On substituting `y/x` in place of v,
`C. x ((x - y)^2)/x^2 = e^(-y/x)`
`=> (x - y)^2 = Cxe^(-y/x)`
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