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

[DragonIce]

Homework Statement

\frac{dy}{dx}+\frac{y}{(1+x^2)} = \frac{\arctan x}{(1+x^2)} when
y(\frac{\pi}{4})=0

Homework Equations

\frac{dy}{dx}+Q(x)*y=F(x)

The Attempt at a Solution

y=u*v
u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}
\frac{du}{dx}+\frac{u}{(1+x^2)}=0
\int\frac{1}{u}\,du=-\int\frac{1}{(1+x^2)}\,dx
\ln u=-\arctan x
t=\arctan x
t'=\frac{1}{(1+x^2)}
u=\frac{1}{e^t}
So let's go back to
u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}
\frac{1}{e^t}*\frac{dv}{dx}=\frac{t}{(1+x^2)}
v=\int\frac{t*e^t}{(1+x^2)}\,dx
Integrating by parts...
v=t * e^t-e^t*dt+C
Going back to y=u*v
y=t-t'+\frac{C}{e^t} - global answer.

The problem starts now... I need to find C when
y(\frac{\pi}{4})=0 so i get

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

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
C=e^{\arctan(\frac{pi}{4})}*(\frac{1}{(\frac{pi^2}{16}+1)}-\arctan(\frac{pi}{4}))
I have to find rational answer...

 

[vela]

Homework Statement

\frac{dy}{dx}+\frac{y}{(1+x^2)} = \frac{\arctan x}{(1+x^2)} when
y(\frac{\pi}{4})=0

Homework Equations

\frac{dy}{dx}+Q(x)*y=F(x)

The Attempt at a Solution

y=u*v
u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}
\frac{du}{dx}+\frac{u}{(1+x^2)}=0
\int\frac{1}{u}\,du=-\int\frac{1}{(1+x^2)}\,dx
\ln u=-\arctan x
t=\arctan x
dt=\frac{1}{(1+x^2)}

That should be
$$dt = \frac{1}{(1+x^2)}\,dx$$

u=\frac{1}{e^t}
So let's go back to
u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}
\frac{1}{e^t}*\frac{dv}{dx}=\frac{t}{(1+x^2)}
v=\int\frac{t*e^t}{(1+x^2)}\,dx
Integrating by parts...
v=t * e^t-e^t*dt+C

That's wrong. You can't have a lone dt sitting around.

Going back to y=u*v
y=t-dt+\frac{C}{e^t} - global answer.

The problem starts now... I need to find C when
y(\frac{\pi}{4})=0 so i get

arctan pi/4-1/(1+pi^2/16)+C/e^arctan pi/4=0

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

 

[DragonIce]

That's wrong. You can't have a lone dt sitting around.

(u*v)'=u'*v+u*v'
I meant it like dt=t'
I should write ' instead of d, but i am right here i think.
If u=f(x) and v=e^q(x)
Then (u*v)'=f'(x)*e^q(x)+f(x)*q'(x)*e^q(x)
It doesn't solve my problem with Constant

 

[vela]

No, that's not right.

 

[DragonIce]

Anyway if you are right i still don't know what to do with
arctan pi/4-1+C/e^arctan pi/4=0
C=e^arctan x - arctan x * e^arctan x
I have to transform it somehow to get rational answer

 

[jackmell]

You're all over the place with that. Try and treat it more carefully. First write it as:

y'+\frac{1}{1+x^2}y=\frac{\arctan(x)}{1+x^2}

so the integrating factor is:

\mu=e^{\arctan(x)}

apply that to both sides and integrate to get:

\int d\left(y e^{\arctan(x)}\right)=\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx

or:

ye^{\arctan(x)}=\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx

so now you just have to integrate

\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx

You can do that right?

 

[DragonIce]

You're all over the place with that. Try and treat it more carefully. First write it as:

y'+\frac{1}{1+x^2}y=\frac{\arctan(x)}{1+x^2}

so the integrating factor is:

\mu=e^{\arctan(x)}

apply that to both sides and integrate to get:

\int d\left(y e^{\arctan(x)}\right)=\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx

or:

ye^{\arctan(x)}=\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx

so now you just have to integrate

\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx

You can do that right?

well i did that... if you look carefully you can notice that i treat t=arctan(x)
so basicaly \int e^t\frac{t}{1+x^2}dx equal to \int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx
and my answer is v=\arctan(x)*e^{\arctan(x)}-\frac{e^{\arctan(x)}}{x^2+1}+C and then i turn back to y=u*v and gety=\arctan(x)-\frac{1}{x^2+1}+\frac{C}{e^{\arctan(x)}} i need to aply that y(\frac{pi}{4})=0 and find C, but i am stack.

 

[jackmell]

You shouldn't have t's and x's in the same integrand. Start with:

\int e^{\arctan(x)} \frac{\arctan(x)}{1+x^2}dx

and let:

u=e^{\arctan(x)}

then:

du=\frac{e^{\arctan(x)}}{1+x^2}dx

now substitute all that in the integral to get:

\int \ln(u)du

see, all u's now. Now integrate that and don't forget the constant of integration.

 

[vela]

well i did that... if you look carefully you can notice that i treat t=arctan(x) so basically
\int e^t\frac{t}{1+x^2}dx equal to \int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx
and my answer is v=\arctan(x)*e^{\arctan(x)}-\frac{e^{\arctan(x)}}{x^2+1}+C

Which isn't correct. If you differentiate your answer, you do not get the integrand back. From here, it appears your mistake is resulting from your sloppy notation.