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The equation of curve passing through (0...

The equation of curve passing through (0,1) which is a solution of differential equation `(1 + y^(2)) dx + (1 + x^(2)) dy = 0` is given by

A

`Tan^(-1)(x) + Tan^(-1)(y) = 0`

B

`Tan^(-1)(x) + Tan^(-1)(y) = pi//4`

C

`Sin h^(-1)(x) + Sin h^(-1)(y) = 0`

D

`Sin h^(-1)(x) + Sin h^(-1)(y) log (1+sqrt(2)) = 0`

Text Solution

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The correct Answer is:
B
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AAKASH SERIES-DIFFERENTIAL EQUATIONS-Practice Exercise
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  8. Solve : x dy= (y+x " cos"^(2)(y)/(x))dx

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  9. Solve : x " sin"(y)/(x).(dy)/(dx)=y " sin"(y)/(x)-x

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  11. The solution of (x^(2) - y^(2)) dx + 2xy dy = 0 is

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  14. The solution of (dy)/(dx) = (x-2y + 3)/(2x-y + 5) is

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  19. The solution of 2x(dy)/(dx) + y = 2x^(3) is

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