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प्रश्न
The solution of `dy/dx = 1 + y + y^2 + x + xy + xy^2` is ______
पर्याय
`2/sqrt3 tan^-1((2y + 1)/sqrt3) = x + x^2 + c`
`4tan^-1((2y + 1)/sqrt3) = sqrt3/2(2x + x^2) + c`
`sqrt3tan^-1((3y + 1)/sqrt3) = 4(1 + x + x^2) + c`
`4tan^-1((2y + 1)/sqrt3) = sqrt3(2x + x^2) + c`
MCQ
रिकाम्या जागा भरा
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उत्तर
The solution of `dy/dx = 1 + y + y^2 + x + xy + xy^2` is `underline(4tan^-1((2y + 1)/sqrt3) = sqrt3(2x + x^2) + c)`.
Explanation:
`dy/dx = 1 + y + y^2 + x + xy + xy^2`
⇒ `dy/dx = (1 + y + y^2) (x + 1)`
Integrating on both sides, we get
`intdy/(1 + y + y^2) = int(x + 1)dx + c_1`
⇒ `intdy/((y + 1/2)^2 + 3/4) = x^2/2 + x + c_1`
⇒ `1/(sqrt3"/"2) . tan^-1((y + 1/2)/(sqrt3"/"2)) = x^2/2 + x + c_1`
⇒ `4tan^-1((2y + 1)/sqrt3) = sqrt3(x^2 + 2x) + c`,
where c = `2sqrt3 c_1`
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