równanie różniczkowe o rozdzielonych zmiennych - do spr.
: 27 cze 2011, o 11:34
Witam!
Czy ja tu coś źle robię?
\(\displaystyle{ x ^{2} \frac{dy}{dx} +y -a= 0}\)
\(\displaystyle{ x ^{2} \frac{dy}{dx}= a -y}\)
\(\displaystyle{ \int_{}^{} \frac{1}{a -y} dy= \int_{}^{} \frac{1}{x ^{2}} dx}\)
\(\displaystyle{ -ln\left| a-y\right| = - \frac{1}{x} +C _{1}}\)
\(\displaystyle{ ln\left| a-y\right| = ln(e ^{\frac{1}{x} -C _{1}})}\)
\(\displaystyle{ \left|a-y \right|=e ^{ \frac{1}{x}} \cdot \frac{1}{e ^{C _{1} } } \Rightarrow a-y=e ^{ \frac{1}{x}} \cdot C \Rightarrow y=-e ^{ \frac{1}{x}} \cdot C+a}\)
Czy ja tu coś źle robię?
\(\displaystyle{ x ^{2} \frac{dy}{dx} +y -a= 0}\)
\(\displaystyle{ x ^{2} \frac{dy}{dx}= a -y}\)
\(\displaystyle{ \int_{}^{} \frac{1}{a -y} dy= \int_{}^{} \frac{1}{x ^{2}} dx}\)
\(\displaystyle{ -ln\left| a-y\right| = - \frac{1}{x} +C _{1}}\)
\(\displaystyle{ ln\left| a-y\right| = ln(e ^{\frac{1}{x} -C _{1}})}\)
\(\displaystyle{ \left|a-y \right|=e ^{ \frac{1}{x}} \cdot \frac{1}{e ^{C _{1} } } \Rightarrow a-y=e ^{ \frac{1}{x}} \cdot C \Rightarrow y=-e ^{ \frac{1}{x}} \cdot C+a}\)