Find the particular solution of the differential equation (dy)/(dx) = 2y ^ 2, given y = 1 when x = 1​

Answers 1

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

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