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Question Number 17322 by tawa tawa last updated on 04/Jul/17

Solve: (dy/dx) + (1/2)y = (3/2) with y(0) = 4

$$\mathrm{Solve}:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{with}\:\:\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{4} \\ $$

Answered by mrW1 last updated on 04/Jul/17

2(dy/dx)=3−y −(dy/(y−3))=(dx/2) ∫(dy/(y−3))=−∫(dx/2) ln ∣y−3∣=−(x/2)+C y(0)=4 ln (1)=0+C ⇒C=0 ln ∣y−3∣=−(x/2) y−3=±e^(−(x/2)) y=3±e^(−(x/2)) since y(0)=4 ⇒ y=3+e^(−(x/2))

$$\mathrm{2}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{3}−\mathrm{y} \\ $$$$−\frac{\mathrm{dy}}{\mathrm{y}−\mathrm{3}}=\frac{\mathrm{dx}}{\mathrm{2}} \\ $$$$\int\frac{\mathrm{dy}}{\mathrm{y}−\mathrm{3}}=−\int\frac{\mathrm{dx}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:\mid\mathrm{y}−\mathrm{3}\mid=−\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{C} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)=\mathrm{4} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}\right)=\mathrm{0}+\mathrm{C} \\ $$$$\Rightarrow\mathrm{C}=\mathrm{0} \\ $$$$\mathrm{ln}\:\mid\mathrm{y}−\mathrm{3}\mid=−\frac{\mathrm{x}}{\mathrm{2}} \\ $$$$\mathrm{y}−\mathrm{3}=\pm\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$\mathrm{y}=\mathrm{3}\pm\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$ \\ $$$$\mathrm{since}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{y}=\mathrm{3}+\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}} \\ $$

Commented by tawa tawa last updated on 04/Jul/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by tawa tawa last updated on 04/Jul/17

How is C = 0 sir ???

$$\mathrm{How}\:\mathrm{is}\:\mathrm{C}\:=\:\mathrm{0}\:\mathrm{sir}\:??? \\ $$

Commented by mrW1 last updated on 04/Jul/17

i changed my answer. there are 2 possible solutions.

$$\mathrm{i}\:\mathrm{changed}\:\mathrm{my}\:\mathrm{answer}.\:\mathrm{there}\:\mathrm{are}\:\mathrm{2} \\ $$$$\mathrm{possible}\:\mathrm{solutions}. \\ $$

Commented by tawa tawa last updated on 04/Jul/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$ \\ $$

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