1. ## Differential question

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?

2. Originally Posted by superman69
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.

3. 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.

4. Originally Posted by mr fantastic
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.

5. Originally Posted by HallsofIvy
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?

6. Originally Posted by superman69
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 $C$.

7. Originally Posted by superman69
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).

8. Originally Posted by mr fantastic
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.