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Question
Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.
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Solution
Given equation is `"dy"/"dx"` = 1 + x + y2 + xy2
⇒ `"dy"/"dx"` = 1(1 + x) + y2(1 + x)
⇒ `"dy"/"dx"` = (1 + x)(1 + y2)
⇒ `"dy"/(1 + y^2)` = (1 + x)dx
Integrating both sides, we get
`int "dy"/(1 + y^2) = int(1 + x)"d"x`
⇒ `tan^-1y = x + x^2/2 + "c"`
Put x = 0 and y = 0
We get tan–1(0) = 0 + 0 + c
⇒ c = 0
∴ tan–1y = `x + x^2/2`
⇒ y = `tan(x + x^2/2)`
Hence, the required solution is y = `tan(x + x^2/2)`.
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