Question Number 34051 by adil last updated on 29/Apr/18
$$\mathrm{2}{dy}/{dx}+{y}=\mathrm{0}\:\:{y}\left(\mathrm{0}\right)=−\mathrm{3} \\ $$
Answered by candre last updated on 30/Apr/18
$$\mathrm{2}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{{y}}{\mathrm{2}} \\ $$$$\frac{{dy}}{{y}}=−\frac{{dx}}{\mathrm{2}} \\ $$$$\int\frac{{dy}}{{y}}=−\int\frac{{dx}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:{y}=−\frac{{x}}{\mathrm{2}}+{K} \\ $$$${y}={ce}^{−{x}/\mathrm{2}} \\ $$$${x}=\mathrm{0}\Rightarrow{c}=−\mathrm{3}\Rightarrow{y}=−\mathrm{3}{e}^{−{x}/\mathrm{2}} \\ $$
Commented by candre last updated on 30/Apr/18
$$\underset{−\mathrm{3}} {\overset{{y}} {\int}}\frac{{dy}}{{y}}=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{{x}} {\int}}{dx} \\ $$$$\mathrm{ln}\:{y}−\mathrm{ln}\:\left(−\mathrm{3}\right)=\mathrm{ln}\:\frac{{y}}{−\mathrm{3}}=−\frac{{x}}{\mathrm{2}} \\ $$$${y}=−\mathrm{3}{e}^{−\frac{{x}}{\mathrm{2}}} \\ $$