# Initial Value Problem

• Mar 24th 2009, 06:27 PM
Len
Initial Value Problem
I missed the lecture and the text book goes over this very quickly...

$\frac{dy}{dt}=\frac{1}{2y+3}, y(0)=1$

This is an autonomous and therefore separable equation so

$dt=(2y+3)dy$

$\int dt= \int (2y+3) dy$

$t=y^2+3y+C$

If I have done this right, what is the next step? If not please correct :$Thank You • Mar 24th 2009, 08:37 PM Chris L T521 Quote: Originally Posted by Len I missed the lecture and the text book goes over this very quickly... $\frac{dy}{dt}=\frac{1}{2y+3}, y(0)=1$ This is an autonomous and therefore separable equation so $dt=(2y+3)dy$ $\int dt= \int (2y+3) dy$ $t=y^2+3y+C$ If I have done this right, what is the next step? If not please correct :$

Thank You

It all is correct so far. Just keep in mind that y is the subject of the equation. So we can rewrite what you have as $y^2+3y=t+C$. Note that if you complete the square, you end up with $\left(y+\tfrac{3}{2}\right)^2=t+C-\tfrac{3}{2}\implies \left(y+\tfrac{3}{2}\right)^2=t+K$, where $K$ is another constant.

Now, applying the initial condition $y(0)=1$, we have $\tfrac{25}{4}=K$

Therefore, $\left(y+\tfrac{3}{2}\right)^2=t+\tfrac{25}{4}$

Does this make sense?
• Mar 24th 2009, 08:55 PM
Len
Quote:

Originally Posted by Chris L T521
It all is correct so far. Just keep in mind that y is the subject of the equation. So we can rewrite what you have as $y^2+3y=t+C$. Note that if you complete the square, you end up with $\left(y+\tfrac{3}{2}\right)^2=t+C-\tfrac{3}{2}\implies \left(y+\tfrac{3}{2}\right)^2=t+K$, where $K$ is another constant.

Now, applying the initial condition $y(0)=1$, we have $\tfrac{25}{4}=K$

Therefore, $\left(y+\tfrac{3}{2}\right)^2=t+\tfrac{25}{4}$

Does this make sense?

I think so, that can be rewrote as

$y =\sqrt{t+k}-\frac{3}{2}$
then using initial condition
$(\frac{5}{2})^2=k$

Awesome, thank you very much. :)

Although looking back I didn't need to do that, simply substituting y=1 and t=0 would give me the answer
• Mar 25th 2009, 02:12 AM
Initial Value Problem
Hello Len
Quote:

Originally Posted by Len
I missed the lecture and the text book goes over this very quickly...

$\frac{dy}{dt}=\frac{1}{2y+3}, y(0)=1$

This is an autonomous and therefore separable equation so

$dt=(2y+3)dy$

$\int dt= \int (2y+3) dy$

$t=y^2+3y+C$

If I have done this right, what is the next step? If not please correct :\$

Thank You

Perfect, so far!

$y(0)$ means the value of $y$ when $t = 0$. So you can just plug this into your equation:

When $t = 0, y = 1: 0 = 1^2 + 3\times 1 + C.$

So $C = ...$?

Put this value of $C$ back into your equation, and there's your answer for $t$ in terms of $y$.