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Question
For the differential equation find a particular solution satisfying the given condition:
x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
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Solution
Given differential equation,
x2dy + (xy + y2) dx = 0
or `dy/dx = (- (xy + y^2))/x^2 = - (y/x) - (y/x)^2`
Clearly, this is a homogeneous differential equation.
∴ Putting y = vx
`dy/dx = v + x (dv)/dx` ...(in equation (i))
`v + x (dv)/dx = - v - v^2`
`=> x (dv)/dx = - 2v - v^2`
`= - (2v + v^2)`
`=> (dv)/(v(v + 2)) = 1/x dx`
On integrating
`=> 1/2 int 1/v dv - 1/2 int 1/(v + 2) dv = - int 1/x dy`
`=> 1/2 [log v - log (v + 2)] = - log x + log C`
`=> log (v/(v + 2)) = 2 log (C/x)`
`=> log (v/(v + 2)) = log (C/x)^2`
`=> v/(v + 2) = (C/x)^2`
On substituting `y/x` in place of v
`(y/x)/(y/x + 2) = C^2/x^2`
x2y = C2(y + 2x) ...(ii)
Given, Putting x = 1, y = 1 in equation (ii),
1 = C2 (1 + 1) ⇒ C2 = `1/3`
Putting this value of C2 in equation (ii),
`x^2y = 1/3(y + 2x)`
y + 2x = 3x2y
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