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प्रश्न
Solve the following homogeneous differential equation:
`x ("d"y)/("d"x) - y = sqrt(x^2 + y^2)`
उत्तर
`x ("d"y)/("d"x) - y = sqrt(x^2 + y^2)`
`x ("d"y)/("d"x) = sqrt(x^2 + y^2) + y`
`("d"y)/("d"x) = sqrt(x^2 + y^2 + y)/x` ........(1)
It is a homogeneous differential equation, same degree in x and y
Put y = vx and `("d"y)/("d"x) = "v" + x "dv"/("d"x)`
Equation (1)
⇒ `"v" + x "dv"/("d"x) = (sqrt(x^2 + ("v"x)^2) + "v"x)/x`
`"v" + x "dv"/("d"x) = (sqrt(x^2 (1 + "v"^2)) + "v"x)/x`
`"v" + x "dv"/("d"x) = (x[sqrt((1 + "v"^2)) + "v"])/x`
`"v" + x "dv"/("d"x) = sqrt((1 + "v"^2)) + "v"`
⇒ `x "dv"/("d"x) = sqrt(1 + "v"^2)`
`"dv"/sqrt(1 + "v"^2) = 1/x "d"x`
`log("v" + sqrt(1 + "v"^2)) = log x + log "c"`
`log("v" + sqrt(1 + "v"^2)) = log x "c"`
⇒ `"v" + sqrt(1 + "v"^2)` = xc
`y/x + sqrt(1 + y^2/x^2)` = xc
`y/x + sqrt((x^2 + y^2)/x^2)` = xc
⇒ `y/x + sqrt(x^2 + y^2)/x^2` = xc
`1/x [y + sqrt(y^2 + y^2)]` = xc
⇒ `y + sqrt(x^2 + y^2)` = x2c
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