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Differential Equation
Hard
Question
The solution of the differential equation (1 + y
2
) +
= 0, is
Options
A.
xe
2tan
−1
y = e
tan
−1
y
+ k
B.
(x − 2) = Ke
tan
−1
y
C.
2xe
tan
−1
y
= e
tan
−1
y
+ K
D.
xe
tan
−1
y
= tan
−1
y + K
Solution
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Topic:
Differential Equation
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= 0, is = 0, is = 0, is