rates of decay: how do i do dy/dx =x+y ?
- Thread starter ezra
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I would write the ODE in standard linear form:
\(\displaystyle \frac{dy}{dx}-y=x\)
Now, it appears you have correctly computed your integrating factor \(\displaystyle \mu(x)=e^{-x}\) and so we get:
\(\displaystyle e^{-x}\frac{dy}{dx}-e^{-x}y=xe^{-x}\)
\(\displaystyle \frac{d}{dx}\left(e^{-x}y\right)=xe^{-x}\)
Integrating with respect to \(\displaystyle x\), we obtain:
\(\displaystyle e^{-x}y=-e^{-x}(x+1)+c_1\)
It appears you made a minor error integrating. If you can post your work, perhaps we can figure out where.
Hence:
\(\displaystyle y(x)=c_1e^x-(x+1)\)
\(\displaystyle \frac{dy}{dx}-y=x\)
Now, it appears you have correctly computed your integrating factor \(\displaystyle \mu(x)=e^{-x}\) and so we get:
\(\displaystyle e^{-x}\frac{dy}{dx}-e^{-x}y=xe^{-x}\)
\(\displaystyle \frac{d}{dx}\left(e^{-x}y\right)=xe^{-x}\)
Integrating with respect to \(\displaystyle x\), we obtain:
\(\displaystyle e^{-x}y=-e^{-x}(x+1)+c_1\)
It appears you made a minor error integrating. If you can post your work, perhaps we can figure out where.
Hence:
\(\displaystyle y(x)=c_1e^x-(x+1)\)
Last edited:
HallsofIvy
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You can do a direct integration, letting u= y+ x. Then y= u- x and \(\displaystyle \frac{dy}{dx}= \frac{du}{dx}- 1= u\).
\(\displaystyle \frac{du}{dx}= u+ 1\)
\(\displaystyle \frac{du}{u+ 1}= dx\)
Integrating, \(\displaystyle ln(u+ 1)= x+ C\) so that \(\displaystyle u+ 1= e^{x+ C}= C'e^x\). Then \(\displaystyle u= y+ x= C'e^x- 1\) so \(\displaystyle y= C'e^x- x- 1\).
\(\displaystyle \frac{du}{dx}= u+ 1\)
\(\displaystyle \frac{du}{u+ 1}= dx\)
Integrating, \(\displaystyle ln(u+ 1)= x+ C\) so that \(\displaystyle u+ 1= e^{x+ C}= C'e^x\). Then \(\displaystyle u= y+ x= C'e^x- 1\) so \(\displaystyle y= C'e^x- x- 1\).