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प्रश्न
Solve the differential equation: `x+ydy/dx=sec(x^2+y^2)` Also find the particular solution if x = y = 0.
उत्तर
`x+ydy/dx=sec(x^2+y^2)...........(i)`
`put x^2+y^2=t`
Differentiating w.r.t. x, we get
`2x+2ydy/dx=dt/dx`
`x+ydy/dx=1/2dt/dx`
`1/2 dt/dx=sect`
`dt/sect=2dx`
Integrating on both sides, we get
`intcostdt=2intdx`
sin t = 2x + c
sin (x2 + y2) = 2x + c [1]
When x = y = 0
sin (0 + 0) = 2 (0) + c
c = 0
Particular solution is sin (x2 + y2) = 2x
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