(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]\frac{dy}{dx}+\frac{y}{(1+x^2)} = \frac{\arctan x}{(1+x^2)}[/tex] when

[tex]y(\frac{\pi}{4})=0[/tex]

2. Relevant equations

[tex]\frac{dy}{dx}+Q(x)*y=F(x)[/tex]

3. The attempt at a solution

y=u*v

[tex]u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}[/tex]

[tex]\frac{du}{dx}+\frac{u}{(1+x^2)}=0[/tex]

[tex]\int\frac{1}{u}\,du=-\int\frac{1}{(1+x^2)}\,dx [/tex]

[tex]\ln u=-\arctan x [/tex]

[tex]t=\arctan x[/tex]

[tex]t'=\frac{1}{(1+x^2)}[/tex]

[tex] u=\frac{1}{e^t} [/tex]

So let's go back to

[tex]u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}[/tex]

[tex]\frac{1}{e^t}*\frac{dv}{dx}=\frac{t}{(1+x^2)}[/tex]

[tex]v=\int\frac{t*e^t}{(1+x^2)}\,dx[/tex]

Integrating by parts...

[tex]v=t * e^t-e^t*dt+C[/tex]

Going back to y=u*v

[tex]y=t-t'+\frac{C}{e^t}[/tex] - global answer.

The problem starts now... I need to find C when

[tex]y(\frac{\pi}{4})=0[/tex] so i get

[tex]\arctan(\frac{pi}{4})-\frac{1}{(1+\frac{pi^2}{16})}+\frac{C}{e^{\arctan(\frac{pi}{4})}}=0[/tex]

And now i am clueless how to find C. Becouse I cannot find arctan pi/4 and there is a problem with pi^2/16+1

I assume that C=0 becouse graphic of arctan x and 1/(x^2+1) are very close in point 0.79 (pi/4), but i cannot prove it

[tex]C=e^{\arctan(\frac{pi}{4})}*(\frac{1}{(\frac{pi^2}{16}+1)}-\arctan(\frac{pi}{4}))[/tex]

I have to find rational answer...

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# Homework Help: Line integrate find C something wrong. dy/dx+Q(x)*y=F(x)

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