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प्रश्न
The solution of the differential equation y2 dx + (x2 − xy + y2)dy = 0 is ______.
पर्याय
`tan^-1 (x/y) + ln y + C = 0`
`2 tan^-1 (x/y) + ln x + C`
`ln(v + sqrt(x^2 + y^2)) + ln y + C = 0`
`ln (x + sqrt(x^2 + y^2)) + C = 0`
MCQ
रिकाम्या जागा भरा
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उत्तर
The solution of the differential equation y2 dx + (x2 − xy + y2)dy = 0 is `bbunderline(tan^-1 (x/y) + ln y + C = 0)`.
Explanation:
`(dx)/(dy) + (x^2)/(y^2) - x/y + 1 = 0`
Put x = vy
⇒ `(dx)/(dy) = v = y (dv)/(dy)`
∴ `v + y (dv)/(dy) + v^2 - v + 1 = 0`
⇒ `-y (dv)/(dy) = (1 + v^2)`
⇒ `(dv)/(v^2 + 1) = -(dy)/y`
Integrating, we get
tan−1 v + C = −In y
⇒ `tan^-1 (x/y) + ln y + C = 0`
where C is arbitrary constant
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