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I have the differential equation:

$\dfrac{dy}{dt} = \dfrac{1}{2y+3}$

with the initial value of $y(0) = 1$

So I solve the diffeq via separation of variables:

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

But from here, how do I solve for the function? Am I "allowed" to use the initial value information with an implicit function such as this, and give an implicit function as my answer, as below?

$y^2 + 3y = t + 4$

As far as I can tell, it should be perfectly fine to leave my answer as an implicit function. And I don't see a way to solve for $y$ here. TO be honest, I used a diffeq solver to check my work, and that tool did come up with an explicit solution, though when I tried to see where it came from I had no idea.

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  • $\begingroup$ This is a quadratic equaion for $y$. $\endgroup$ – kmitov Jan 13 '16 at 3:25
  • $\begingroup$ @kmitov But i do not want to solve for a particular value of $y$, I want to solve for a particular solution $y(t)$. $\endgroup$ – user278703 Jan 13 '16 at 3:30
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Yes, you should use your known initial condition to solve for $c$.

Proceeding to solve for $y(t)$, we write $y^2+3y+\frac{9}{4}-\frac{9}{4}=\left(y+\frac{3}{2}\right)^2-\frac{9}{4}$ by completing the square.

Now, we write $\left(y+\frac{3}{2}\right)^2-\frac{9}{4}=t+4.$ and solve for $y$ in terms of $t$.

Adding $\frac{9}{4}$, taking the square root, and subtracting $\frac{9}{2}$, we get: $$y(t)=\sqrt{t+\frac{25}{4}}-\frac{3}{2}$$

EDIT: I had misplaced a $3$.

Edit: To address your comment: Note that we can write $y(t)=\sqrt{t+\frac{25}{4}}-\frac{3}{2}=\sqrt{\frac{1}{4}(4t+25)}-\frac{3}{2}=\frac{1}{2}\sqrt{(4t+25)}-\frac{3}{2}=\frac{1}{2}(\sqrt{(4t+25)}-3).$

The reason that $c$ is different is because $c$ is some arbitrary constant. In this case, we have already solved for it.

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  • $\begingroup$ But $\frac{9}{2} + \frac{9}{2}$ doesn't equal 3?? And $\frac{9}{2} * \frac{9}{2}$ doesn't equal $\frac{9}{4}$?? How can you factor $y^2 + 3y + \frac{9}{4}$ and get $(y+ \frac{9}{2})^2$? Shouldn't the correct factoring have $\frac{3}{2}$ instead of $\frac{9}{2}$? $\endgroup$ – user278703 Jan 13 '16 at 3:46
  • $\begingroup$ I'm not factoring. I'm completing the square. For $x^2+bx$, we can add and subtract $(\frac{b}{2})^2$ to get $x^2+bx+(\frac{b}{2})^2-(\frac{b}{2})^2$, which is equal to $(x+\frac{b}{2})^2-(\frac{b}{2})^2$. Now, let $b=3$. $\endgroup$ – zz20s Jan 13 '16 at 3:51
  • $\begingroup$ How? If you factor $y^2 + 3y + \frac{9}{4}$ it equals $(y+\frac{3}{2})^2$. If we remove the $-\frac{9}{4}$ from each side of the equation you wrote, it is $y^2 + 3y + \frac{9}{4} = (y+\frac{9}{2})^2$, which is wrong. Edit: I have seen your edit, thanks $\endgroup$ – user278703 Jan 13 '16 at 3:56
  • $\begingroup$ No problem. Do you understand the solution now? $\endgroup$ – zz20s Jan 13 '16 at 4:02
  • $\begingroup$ I understand, When I solve the diffeq on an online solver it gives $y(t) = \frac{1}{2}(-\sqrt{4t+c} - 3)$. I'm not sure where they got the $4$, or the $-3$, but I guess I'll play with it and see. $\endgroup$ – user278703 Jan 13 '16 at 4:04
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From where you arrived,

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

Substituting $y(0)=1$, we get $c=4$

Hence, we get $y^2+3y-4=t$

Proceed by completing the square for $y^2+3y-4=(y+\frac{3}{2})^2-\frac{7}{4}$

$(y+\frac{3}{2})^2-\frac{7}{4}=t$

Then we get $y=\frac{1}{2}(3±\sqrt{4t+7})$

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