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Question
For the differential equation find a particular solution satisfying the given condition:
`2xy + y^2 - 2x^2 dy/dx = 0; y = 2` when x = 1
Solution
The given equation
`2xy + y^2 - 2x^2 dy/dx = 0`
or `dy/dx = (2xy + y^2)/(2x^2)`
`= y/x + 1/2(y/x)^2` ....(i)
Clearly, this equation is a differential equation.
∴ putting y = vx
`dy/dx = v + x (dv)/dx` ....(in equation (i))
`v + x (dv)/dx = v + 1/2 v^2`
`=> x (dv)/dx = 1/2 v^2`
`=> 2 xx 1/v^2 (dv) = 1/x dv`
On integrating,
`2 int 1/v^2 dv = int 1/x dv - 2/v = log |x| + C`
So, on substituting `(y/x)` in place of v,
`- (2x)/y = log |x| + C` ....(ii)
Given y = 2 if x = 1 ...[from equation (ii)]
`(- 2)/2 = log1 + C`
- 1 = 0 + C
⇒ C = - 1
On putting C = – 1 in equation (ii)
`(- 2x)/y = log |x| - 1`
y = `(2x)/(1 - log |x|)`
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Based on the above, answer the following questions:
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