Step-by-step explanation:
Given
dx
dy
=1+x
2
+y
2
+x
2
y
2
⇒
dx
dy
=(1+x
2
)+y
2
(1+x
2
)
⇒
dx
dy
=(1+x
2
)(1+y
2
)
⇒(1+x
2
)dx=
(1+y
2
)
dy
Integrating both sides, we get
∫(1+x
2
)dx=∫
(1+y
2
)
dy
⇒∫dx+∫x
2
dx=∫
(1+y
2
)
dy
⇒x+
3
x
3
+C=tan
−1
y
Putting y=1 and x=0, we get
tan
−1
(1)=0+0+C
⇒C=tan
−1
(1)=
4
π
Therefore, required particular solution is
tan
−1
y=x+
3
x
3
+
4
π
Author:
aryannaa9by
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