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Question:

The solution of the differential equation ydx−(x+2y²)dy=0 is x=f(y). If f(√2)=1, then f(1) is equal to: 3 1 4 2

3

1

2

4

Solution:

Given ydx−(x+2y²)dy=0
⇒ydx−xdy=2y²dy
⇒ydx−xdy/y²=2dy
⇒d(x/y)=2dy
Integrating we get,
⇒(x/y)=2y+c
⇒2y²+cy=x=f(y)
Given f(√2)=1
⇒2(√2)²−c=1
⇒c=1
⇒f(y)=2y²+y
∴f(1)=2(1)²+1=3