I have a problem with this question:
Solve the differential equation $ \sqrt{1-x^2}\frac {dy}{dx} = -x(1+y) $, writing the general solution y as an explicit function of x.
This is my answer:
$$ \sqrt{1-x^2}\frac {dy}{dx} = -x(1+y) $$ $$ \frac 1{1+y}dy = -\frac x{\sqrt{1-x^2}} dx $$ $$ \int\frac 1{1+y}dy = \int-\frac x{\sqrt{1-x^2}} dx $$ $$ ln|1+y| = \sqrt{1-x^2} $$ $$ 1+y=e^{\sqrt{\ 1-x^2}} $$ $$ y=e^{\sqrt{\ 1-x^2}}-1 $$
Ok so I cross checked with an online calculator(eMathHelp - I find it to be very reliable) but the answer was $ y=C_1e^{\sqrt{1-x^2}}-1 $. I tried figuring out where the $C_1$ came from and saw that it's indefinite integral but my answer is $ y=e^{\sqrt{1-x^2}}-1+C_1 $. I don't know why the eMath calculator is multiplying. I also checked using the WolframAlpha online calculator and it gave the same answer as eMathHelp so I'm probably wrong but how did they get that? Can anyone please shed some light on this?
Thanks.
$\endgroup$ 61 Answer
$\begingroup$You should write $ln|1+y| = \sqrt{1-x^2}+C$ where $C$ is a constant of integration. From here we get $|1+y| = e^{\sqrt{1-x^2}+C}=e^Ce^{\sqrt{1-x^2}}$ or $y=C_1e^{\sqrt{1-x^2}}-1$. $C_1=e^C$ is also going to be a constant.
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