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