# Thread: Initial Value Problem

1. ## Initial Value Problem

Hi All,

Can somebody please help me solve this IVP? Or simply explain ivp's to me so I know what I'm supposed to do with this?

(1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0

Thank you

2. Originally Posted by woody198403
Hi All,

Can somebody please help me solve this IVP? Or simply explain ivp's to me so I know what I'm supposed to do with this?

(1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0

Thank you
It is a first order linear equation.
If you put it in standard form and solve for the integrating factor you get the same equation you start with.

so it is the same as

$\frac{d}{dt}\left[ (1+t^2)y\right]=-t\sin(t)$

Now we integrate both sides to get

$(1+t^2)y=t\cos(t)-\sin(t)+C$

You should be able to finish from here.

Good luck.

3. Okay. Here is what ive done, and now Im stuck on a part.

(1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0

dy/dt + (2ty/(1+t^2))*y = (-t sin(t))/(1+t^2)

Then I find the integrating factor, which I think is

=e^(y*ln(1+t^2))

= (1 +t^2)^y

Multiplying this through, you get:

[((1+t^2)^y) * y] ' = [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

Then I think I'm supposed to integrate both sides. I know the l.h.s. is just

((1+t^2)^y)*y

but im having trouble integrating [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

Have I done something wrong here? Why does your answer look so neat compared to mine?

I know that once I have that I can apply the initial condition and workout the particular solution.

4. Originally Posted by woody198403

Multiplying this through, you get:

[((1+t^2)^y) * y] ' = [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

Then I think I'm supposed to integrate both sides. I know the l.h.s. is just

((1+t^2)^y)*y

but im having trouble integrating [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

Have I done something wrong here? Why does your answer look so neat compared to mine?

I know that once I have that I can apply the initial condition and workout the particular solution.

$\frac{-(1+t^2)(t\sin(t)}{1+t^2}=-t\sin(t)$

Edit: You need to use integration by parts twice from here

5. Originally Posted by TheEmptySet
$\frac{-(1+t^2)(t\sin(t)}{1+t^2}=-t\sin(t)$