The differential equation is $$y'=\arctan(x+y)$$$$y(0)=1$$ I tried to resolve it by change of variable $z=x+y$ and I got the following $$\int \frac{dz}{\arctan(z)+1} = x+c$$ with $c \in \mathbb{R}$. However I cannot resolve the integral, I also tried to put $$x=\arctan(x+y)$$ But it didn't lead me anywhere. Is there a better way to do this?
How to solve $y'=\arctan(x+y)$
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ordinary-differential-equations
1 Answers
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setting $$x+y=u$$ then we have $$1+y'=u'$$ and you have to solve $$u'=\arctan(u)+1$$ which is easy to solve and then i would solve the equation $$x=\arctan(x+y)$$ for $$y$$
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2This is what OP did. The question is about making further progress. – 2017-01-07
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1Asked for $\int\frac{dx}{1+\arctan x}$, WolframAlpha says that no integral is known with elementary functions (and it considers as elementary many more functions than usual). – 2017-01-07
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0yes i think it is true, there is no elementary solution – 2017-01-07
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0@Dr. Sonnhard Graubner This is exactly what he said :) You have used $u$ instead of $z$. – 2017-01-07
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0thank you for your help, but the real problem is the integral i can't find a solution for it. Is the condition y(0)=1 of any help? – 2017-01-07
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0i will try it give me time – 2017-01-07