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Question Number 34051 by adil last updated on 29/Apr/18

2dy/dx+y=0  y(0)=−3

$$\mathrm{2}{dy}/{dx}+{y}=\mathrm{0}\:\:{y}\left(\mathrm{0}\right)=−\mathrm{3} \\ $$

Answered by candre last updated on 30/Apr/18

2(dy/dx)+y=0  (dy/dx)=−(y/2)  (dy/y)=−(dx/2)  ∫(dy/y)=−∫(dx/2)  ln y=−(x/2)+K  y=ce^(−x/2)   x=0⇒c=−3⇒y=−3e^(−x/2)

$$\mathrm{2}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{{y}}{\mathrm{2}} \\ $$$$\frac{{dy}}{{y}}=−\frac{{dx}}{\mathrm{2}} \\ $$$$\int\frac{{dy}}{{y}}=−\int\frac{{dx}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:{y}=−\frac{{x}}{\mathrm{2}}+{K} \\ $$$${y}={ce}^{−{x}/\mathrm{2}} \\ $$$${x}=\mathrm{0}\Rightarrow{c}=−\mathrm{3}\Rightarrow{y}=−\mathrm{3}{e}^{−{x}/\mathrm{2}} \\ $$

Commented by candre last updated on 30/Apr/18

∫_(−3) ^y (dy/y)=−(1/2)∫_0 ^x dx  ln y−ln (−3)=ln (y/(−3))=−(x/2)  y=−3e^(−(x/2))

$$\underset{−\mathrm{3}} {\overset{{y}} {\int}}\frac{{dy}}{{y}}=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{{x}} {\int}}{dx} \\ $$$$\mathrm{ln}\:{y}−\mathrm{ln}\:\left(−\mathrm{3}\right)=\mathrm{ln}\:\frac{{y}}{−\mathrm{3}}=−\frac{{x}}{\mathrm{2}} \\ $$$${y}=−\mathrm{3}{e}^{−\frac{{x}}{\mathrm{2}}} \\ $$

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