# Integrating by parts

• Apr 6th 2011, 08:12 AM
hansbahia
Integrating by parts
Solve the differential equations specified using an integrating factor.
Solve the initial-value problem
dy/dt = 1/(t+1)y + 4t^2 + 4t, y(1) = 10

so dy/dt = 1/(t+1)y + 4t^2 + 4t, y(1) = 10

by using the integrating factor(IF)

IF= e^(integral of P(x))
P(x)= 1/(t+1), so IF=e^(1/(t+1))=e^(ln(t+1))=t+1

now that I have the integrating factor I multiply everything by t+1

dy/dt = 1/(t+1)y + 4t^2 + 4t
(t+1) dy/dt = (t+1)(1/(t+1))y + (t+1)4t^2 + (t+1)4t
multiplying...
(t+1)dy/dt = y + 4t^3+4t^2 + 4t^2+4t
(t+1) dy/dt = y + 4t^3 + 8t^2+4t, 4t^3 + 8t^2+4t= (t+1)(4t^2+4t)
dividing both sides by t+1

(t+1)/(t+1) dy/dt = (y + (t+1)(4t^2+4t))(t+1)

simplifying...

dy/dt= y/(t+1)+4t^2+4t

So far is the integrating by factors right?
And I'm getting stuck on dy/dt= y/(t+1)+4t^2+4t

Now how can I differentiate this first-order linear ordinary equation {dy/dt= y/(t+1)+4t^2+4t} ?
• Apr 6th 2011, 08:16 AM
Ackbeet
You should write your IVP this way:

$\displaystyle \dfrac{dy}{dt}-\dfrac{y}{t+1}=4t^{2}+4t,\quad y(1)=0.$

Then you can see that your IF is off by a sign. Once you get the IF, what you do is multiply the entire DE by the IF (makes sense, since it is called an integrating factor, after all!). The LHS should then be the derivative of a product, which you can integrate directly. The RHS will still just have t's in it - no y's. So you can integrate that side as well.

Make sense?
• Apr 6th 2011, 02:52 PM
hansbahia
Ohhhh...
Now I see! I forgot about that. That makes everything easier

since
dy/dt-1(t+1)y=4t^2+4t

then the Integrating factor will be 1/(t+1)

If I multiply everything by the IF, then i would get...
1/(t+1)dy/dt-(1/(t+1)^2)y=1/(t+1)(4t^2+4t)
since the RHS is a derivative of a product...integral of -1/(t+1)^2 is equal 1/(t+1). Plus if I simplify the RHS I would get
d/dt[1/(t+1)y]=4t
then integrate
1/(t+1)y=2t^2+C
solve for y...
y=2t^2(t+1)+C(t+1)

Then to solve for C I plug 1 to t, then I get C=3

So y=2t^3+2t^2+3t+3
• Apr 7th 2011, 03:35 AM
Ackbeet
That looks right to me. You can always check your solution against the original DE (plug it in: do you get equality?), and against the initial condition.