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Solve x+ydydx=sec(x2+y2)

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Question

Solve `x + y (dy)/(dx) = sec(x^2 + y^2)`

Sum
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Solution

Given, `x + y (dy)/(dx) = sec(x^2 + y^2)`  .....(i)

Put x2 + y2 = t

Differentiating w.r.t. x, we get

`2x + 2y (dy)/(dx) = (dt)/(dx)`

⇒ `x + y (dy)/(dx) = 1/2 (dt)/(dx)`

Put this value in (i), we get

`1/2 (dt)/(dx)` = sec t

⇒ `(dt)/sec t` = 2 dx

Integrating on both sides, we get

`int cost  dt = 2 int  dx`

⇒ sin t = 2x + c

⇒ sin(x2 + y2) = 2x + C   ......(∵ t = x2 + y2

is the required solution.

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2021-2022 (March) Set 1

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