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Solve the differential equation dydxdydx = 1 + x + y2 + xy2, when y = 0, x = 0. - Mathematics

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Question

Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.

Sum
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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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Chapter 9: Differential Equations - Exercise [Page 193]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 9 Differential Equations
Exercise | Q 9 | Page 193

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