- #1

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## Homework Statement

[tex]\frac{{dy}}{{dx}} + 2xy = 0[/tex]

## Homework Equations

Back of Book: [tex]y\left( x \right) = C\exp \left( { - x^2 } \right)[/tex]

## The Attempt at a Solution

[tex]\begin{array}{l}

\frac{{dy}}{{dx}} = - 2xy\,\,\,\, \Rightarrow \,\,\,\,\frac{{dy}}{y} = - 2x\,dx\,\,\,\, \Rightarrow \,\,\,\,\frac{1}{y}dy = - 2x\,dx \\

\\

\int_{}^{} {\frac{1}{y}dy} = \int_{}^{} { - 2x\,dx} \\

\\

\ln \left( y \right) = \frac{{ - 2x^2 }}{2} + C\,\,\,\, \Rightarrow \,\,\,\,\ln \left( y \right) = - x^2 + C\,\,\,\, \Rightarrow \,\,\,\, \\

\\

\exp \left( {\ln \left( {y + C} \right)} \right) = \exp \left( { - x^2 + C} \right)\,\,\,\, \Rightarrow \,\,\,\, \\

\\

y + C = \exp \left( { - x^2 + C} \right)\,\,\,\, \Rightarrow \,\,\,\,y = \exp \left( { - x^2 + C} \right) - C \\

\end{array}[/tex]

The C from the y part is not necessarily the same C as from the x part, is it? How did they end up with only 1 C, and how did they get their C out of the parenthesis to end up multiplying by the right side?

Thanks in advance!