Let y-fx- If dydx-y-1 and f0-1- then fln2 is

dydx=y+1 ∫dyy+1=∫ dx ⇒ ln(y+1)=x+c x=0,y=1 ln 2= O+C ⇒ C=ln2 ln(y+1)=x+ln 2 f(ln2)→ x=ln2 ln(y+1)=ln2+ln2=ln4 y+1=4 ...

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Variable Separable Method

Let y=fx. I...

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Let $y = f (x)$ . If $\frac{d y}{d x} = y + 1$ and $f (0) = 1$ , then $f (ln 2)$ is

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Let y = f(x) defined on R satisfies $(1 + x^{2}) \frac{d y}{d x}$ = 2x - 2xy and f(0) = 2, then

f : [0, 1] \to R

f (x y) = f (x) \cdot f (y)

x, y \in [0, 1]

f (0) \neq 0

y = y (x)

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

y (0) = 1, then y (\frac{1}{4}) + y (\frac{3}{4})

f : [0, 1] \to R

f (x y) = f (x) . f (y)

\in

f (0) \neq 0

y = y (x)

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

y (0) = 1

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

f : [0, 1] \to R

f (x y) = f (x) \cdot f (y)

x, y \in [0, 1]

f (0) \neq 0

y = y (x)

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

y (0) = 1, then y (\frac{1}{4}) + y (\frac{3}{4})

f (x y - 1) = f (- x) f (- y) + f (x) - y + 1

f (0) = 1

f (x) = ?

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