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Question
State the type of the differential equation for the equation. xdy – ydx = `sqrt(x^2 + y^2) "d"x` and solve it
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Solution
Given equation can be written as xdy = `(sqrt(x^2 + y^2) + y) "d"x`
i.e., `"dy"/"dx" = (sqrt(x^2 + y^2) + y)/x` ......(1)
Clearly R.H.S of (1) is a homogeneous function of degree zero.
Therefore, the given equation is a homogeneous differential equation.
Substituting y = vx, we get from (1)
`"v" + x "dv"/"dx" = (sqrt(x^2 + "v"^2 + x^2) + vx)/x`
i.e. `"v" + x "dv"/"dx" = sqrt(1 + "v"^2) + "v"`
`x "dv"/"dx" = sqrt(1 + "v"^2)`
⇒ `"dv"/sqrt(1 + "v"^2) = "dx"/x` ......(2)
Integrating both sides of (2), we get
`log("v" + sqrt(1 + "v"^2))` = logx + logc
⇒ `"v" + sqrt(1 + "v"^2)` = cx
⇒ `y/x + sqrt(1 + y^2/x^2)` = cx
⇒ `y + sqrt(x^2 + y^2)` = cx2
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