Differential question

• Oct 31st 2009, 04:06 PM
superman69
Differential question
Solve the following initial value problem: http://hosted.webwork.rochester.edu/...068a833fb1.png with http://hosted.webwork.rochester.edu/...1399857701.png
http://hosted.webwork.rochester.edu/...b39caf0291.png

So I isolated the y and it became

8y=(9-dy/dt)t

how would I finish the rest?

• Oct 31st 2009, 04:47 PM
mr fantastic
Quote:

Originally Posted by superman69
Solve the following initial value problem: http://hosted.webwork.rochester.edu/...068a833fb1.png with http://hosted.webwork.rochester.edu/...1399857701.png
http://hosted.webwork.rochester.edu/...b39caf0291.png

So I isolated the y and it became

8y=(9-dy/dt)t

how would I finish the rest?

Divide both sides by t: $\displaystyle \frac{dy}{dt} + \frac{8}{t} y = 9$.

Now use the integrating factor technique.
• Oct 31st 2009, 08:19 PM
HallsofIvy
You can also do this as a "homogeneous" equation. Write it as $\displaystyle \frac{dy}{dt}= \frac{9t- 8y}{t}= 9- 8\frac{y}{t}$.

Let $\displaystyle u= \frac{y}{t}$. Then $\displaystyle y= tu$ so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation.
• Oct 31st 2009, 08:32 PM
superman69
Quote:

Originally Posted by mr fantastic
Divide both sides by t: $\displaystyle \frac{dy}{dt} + \frac{8}{t} y = 9$.

Now use the integrating factor technique.

What is the integrating factor technique? I did not learn that yet.
• Oct 31st 2009, 08:44 PM
superman69
Quote:

Originally Posted by HallsofIvy
You can also do this as a "homogeneous" equation. Write it as $\displaystyle \frac{dy}{dt}= \frac{9t- 8y}{t}= 9- 8\frac{y}{t}$.

Let $\displaystyle u= \frac{y}{t}$. Then $\displaystyle y= tu$ so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation.

So now I use the equation tu'= 9- 9u and plug in u' and u?
• Oct 31st 2009, 09:25 PM
Prove It
Quote:

Originally Posted by superman69
What is the integrating factor technique? I did not learn that yet.

$\displaystyle \frac{dy}{dt} + \frac{8}{t}y = 9$.

The Integrating Factor is $\displaystyle e^{\int{\frac{8}{t}\,dt}} = e^{8\ln{t}}$

$\displaystyle = e^{\ln{t^8}}$

$\displaystyle = t^8$.

So multiply both sides by the integrating factor to get

$\displaystyle t^8\left(\frac{dy}{dt} + \frac{8}{t}y\right) = 9t^8$

$\displaystyle t^8\,\frac{dy}{dt} + 8t^7y = 9t^8$

Can you see that the LHS is a product rule expansion of $\displaystyle \frac{d}{dt}(t^8y)$?

$\displaystyle \frac{d}{dt}(t^8y) = 9t^8$

$\displaystyle t^8y = \int{9t^8\,dt}$

$\displaystyle t^8y = t^9 + C$

$\displaystyle y = t + Ct^{-8}$.

Now use the initial condition to find $\displaystyle C$.
• Oct 31st 2009, 10:18 PM
mr fantastic
Quote:

Originally Posted by superman69
So now I use the equation tu'= 9- 9u and plug in u' and u?

The equation is $\displaystyle t \frac{du}{dt} = 9 - 9u$. It's seperable and your task is to solve for u and so get y from y = ut (and I hope you have learnt about seperable DE's).
• Nov 1st 2009, 03:01 AM
Prove It
Quote:

Originally Posted by mr fantastic
The equation is $\displaystyle t \frac{du}{dt} = 9 - 9u$. It's seperable and your task is to solve for u and so get y from y = ut (and I hope you have learnt about seperable DE's).

It helps if you write the DE as

$\displaystyle \frac{1}{1 - u}\,\frac{du}{dt} = \frac{9}{t}$

so that when you take the integral of both sides with respect to t

$\displaystyle \int{\frac{1}{1 - u}\,\frac{du}{dt}\,dt} = \int{\frac{9}{t}\,dt}$

$\displaystyle \int{\frac{1}{1 - u}\,du} = \int{\frac{9}{t}\,dt}$.

I'm sure you can go from here.