Let f:[$$0$$,$$1$$]→R be such that $$f(xy)=f(x)f(y)$$∀x,y∈[$$0$$,$$1$$] and $$f(0)$$≠$$0$$ . If $$y=y(x)$$ satisfies the differential equation $$dydx=f(x)$$ with $$y_(0)=1$$ then $$y(1/4)+y(3/4)$$ is

$$f(x)=1$$⇒$$dydx=f(x)=1$$ given ∫$$dy=$$∫$$dxy=x+1$$⇒$$y(0)=1$$→$$c=1y=x+1$$⇒$$y(1/4)+y(3/4)=14+34+2=3$$

Introduction to Differential equations

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$Let f : [0, 1] \to R be such that f (xy) = f (x) f (y) \forall x, y \in [0, 1] and f (0) \neq 0 . If y = y (x)$ $satisfies the differential equation \frac{dy}{dx} = f (x) with \underline{y} (0) = 1 then y (1 / 4) + y (3 / 4) is$

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Solution

$\begin{array}{l} f (x) = 1 \Rightarrow \frac{dy}{dx} = f (x) = 1 given \\ \int d y = \int d x \\ y = x + 1 \Rightarrow y (0) = 1 \to c = 1 \\ y = x + 1 \Rightarrow y (1 / 4) + y (3 / 4) = \frac{1}{4} + \frac{3}{4} + 2 = 3 \end{array}$

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