Differential question

superman69 · #1 $Old$ October 31st, 2009, 06:06 PM

Solve the following initial value problem:

with

So I isolated the y and it became

8y=(9-dy/dt)t

how would I finish the rest?

mr fantastic · #2 $Old$ October 31st, 2009, 06:47 PM

Quote:

Originally Posted by superman69 $View Post$

Solve the following initial value problem:

with

So I isolated the y and it became

8y=(9-dy/dt)t

how would I finish the rest?

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

Now use the integrating factor technique.

HallsofIvy · #3 $Old$ October 31st, 2009, 10:19 PM

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

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

superman69 · #4 $Old$ October 31st, 2009, 10:32 PM

Quote:

Originally Posted by mr fantastic $View Post$

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

Now use the integrating factor technique.

What is the integrating factor technique? I did not learn that yet.

superman69 · #5 $Old$ October 31st, 2009, 10:44 PM

Quote:

Originally Posted by HallsofIvy $View Post$

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

Let $u= \frac{y}{t}$ . Then $y= tu$ 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?

Prove It · #6 $Old$ October 31st, 2009, 11:25 PM

Quote:

Originally Posted by superman69 $View Post$

What is the integrating factor technique? I did not learn that yet.

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

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

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

$= t^8$ .

So multiply both sides by the integrating factor to get

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

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

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

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

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

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

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

Now use the initial condition to find $C$ .

mr fantastic · #7 $Old$ November 1st, 2009, 12:18 AM

Quote:

Originally Posted by superman69 $View Post$

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

The equation is $t \frac{du}{dt} = 9 - 9u$ . 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).

Prove It · #8 $Old$ November 1st, 2009, 05:01 AM

Quote:

Originally Posted by mr fantastic $View Post$

The equation is $t \frac{du}{dt} = 9 - 9u$ . 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

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

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

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

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

I'm sure you can go from here.