1. ## Initial Value Problem

I missed the lecture and the text book goes over this very quickly...

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

This is an autonomous and therefore separable equation so

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

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

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

If I have done this right, what is the next step? If not please correct :$Thank You 2. Originally Posted by Len I missed the lecture and the text book goes over this very quickly...$\displaystyle \frac{dy}{dt}=\frac{1}{2y+3}, y(0)=1$This is an autonomous and therefore separable equation so$\displaystyle dt=(2y+3)dy\displaystyle \int dt= \int (2y+3) dy\displaystyle 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 $\displaystyle y^2+3y=t+C$. Note that if you complete the square, you end up with $\displaystyle \left(y+\tfrac{3}{2}\right)^2=t+C-\tfrac{3}{2}\implies \left(y+\tfrac{3}{2}\right)^2=t+K$, where $\displaystyle K$ is another constant.

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

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

Does this make sense?

3. 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 $\displaystyle y^2+3y=t+C$. Note that if you complete the square, you end up with $\displaystyle \left(y+\tfrac{3}{2}\right)^2=t+C-\tfrac{3}{2}\implies \left(y+\tfrac{3}{2}\right)^2=t+K$, where $\displaystyle K$ is another constant.

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

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

Does this make sense?
I think so, that can be rewrote as

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

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

4. ## Initial Value Problem

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

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

This is an autonomous and therefore separable equation so

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

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

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

If I have done this right, what is the next step? If not please correct :$Thank You Perfect, so far!$\displaystyle y(0)$means the value of$\displaystyle y$when$\displaystyle t = 0$. So you can just plug this into your equation: When$\displaystyle t = 0, y = 1: 0 = 1^2 + 3\times 1 + C.$So$\displaystyle C = ...$? Put this value of$\displaystyle C$back into your equation, and there's your answer for$\displaystyle t$in terms of$\displaystyle y\$.