`y*(x+1) + y' = 0 , y(-2) = 1` Find the particular solution that satisfies the initial condition

Tuesday, October 25, 2016

This differential equation may be expressed as

$\displaystyle\frac{{{y}'}}{{y}}=-{x}-{1}$

and then integrated:

$\displaystyle{\ln}{\left|{y}\right|}=-\frac{{{x}}^{{2}}}{{2}}-{x}+{C},{\quad\text{or}\quad}{y}={C}{{e}}^{{-\frac{{{x}}^{{2}}}{{2}}-{x}}},$

where $\displaystyle{C}$ is any constant.

To determine the specific constant, we use the given condition $\displaystyle{y}{\left(-{2}\right)}={1},$ which gives

$\displaystyle{1}={C}{{e}}^{{-{2}+{2}}}={C},$

so the final answer is $\displaystyle{y}{\left({x}\right)}={{e}}^{{-\frac{{{x}}^{{2}}}{{2}}-{x}}}.$

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