Let y=y(x) is a solution of dxdy=f(x) where f(xy)=f(x)f(y)∀x,y∈R and f(0)=0 and y(0)=21, then y(41)+y(43) is equal to

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Let $y = y (x)$ is a solution of $\frac{d y}{d x} = f (x)$ where $f (x y) = f (x) f (y) \forall x, y \in R$ and $f (0) \neq = 0$ and $y (0) = \frac{1}{2}$ , then $y (\frac{1}{4}) + y (\frac{3}{4})$ is equal to

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Lucas Discussed

Let

y = y (x)

is a solution of

\frac{d y}{d x} = f (x)

where

f (x y) = f (x) f (y) \forall x, y \in R

and

f (0) \neq = 0

and

y (0) = \frac{1}{2}

, then

y (\frac{1}{4}) + y (\frac{3}{4})

is equal to

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Question Text	Let $y = y (x)$ is a solution of $\frac{d y}{d x} = f (x)$ where $f (x y) = f (x) f (y) \forall x, y \in R$ and $f (0) \neq = 0$ and $y (0) = \frac{1}{2}$ , then $y (\frac{1}{4}) + y (\frac{3}{4})$ is equal to
Answer Type	Video solution: 1
Upvotes	150