If dydx=y+3>0 and y(0)=2 then y(log2)=

dydx−y=3I.F=e∫−1 dx =e−xSolution isy(I.F)=∫3(I.F) dx=3∫e−xdxy e−x=−3 e−x+cput x=0y(0)e− 0=−3 e− 0+c⇒ 2= −3+c⇒ c=5⇒ y e−x=−3 e−x+5⇒ y= −3+5ex⇒ y+3=5exPut x= log 2⇒ y(log 2)+3=5e log 2=10⇒ y(log 2)=7

If dydx=y+3>0 and y(0)=2 then y(log2)=

dydx−y=3I.F=e∫−1 dx =e−xSolution isy(I.F)=∫3(I.F) dx=3∫e−xdxy e−x=−3 e−x+cput x=0y(0)e− 0=−3 e− 0+c⇒ 2= −3+c⇒ c=5⇒ y e−x=−3 e−x+5⇒ y= −3+5ex⇒ y+3=5exPut x= log 2⇒ y(log 2)+3=5e log 2=10⇒ y(log 2)=7

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If $\frac{dy}{dx} = y + 3 > 0$ and $y (0) = 2$ then $y (\log 2) =$

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} - y = 3 \\ I . F = e^{\int - 1 d x} = e^{- x} \\ S o l u t i o n i s \\ y (I . F) = \int 3 (I . F) d x \\ = 3 \int e^{- x} d x \\ y e^{- x} = - 3 e^{- x} + c \\ p u t x = 0 \end{array}$

$\begin{array}{l} y (0) e^{- 0} = - 3 e^{- 0} + c \\ \Rightarrow 2 = - 3 + c \Rightarrow c = 5 \\ \Rightarrow y e^{- x} = - 3 e^{- x} + 5 \\ \Rightarrow y = - 3 + 5 e^{x} \\ \Rightarrow y + 3 = 5 e^{x} \\ P u t x = \log 2 \\ \Rightarrow y (\log 2) + 3 = 5 e^{\log 2} = 10 \\ \Rightarrow y (\log 2) = 7 \end{array}$