0
$\begingroup$

I have two questions regarding solving differential equations given initial conditions:

1) When do you substitute the initial conditions into the equation to calculate the value of the constant "$c$". Do you substitute it once you integrate both sides of the differential equations and you get a constant "$c$"? Or do you substitute the initial conditions after integrating both sides AS WELL AS rearranging the equations to get $y$ in terms of $x$ and $c$. Using the second method, sometimes you get two values for "$c$" with only one value being correct.

2) When you solve certain differential equations, you get one side written with "$\pm$" in the front. However, only one equation fits the initial conditions even after you solve for the constant "$c$". The one that fits is either the one with the "$+$" or the one with the "$-$" in the front. How do you justify which one is correct without giving geometric representations of both and then saying "according to graph, this one insert equation is correct". Can you somehow solve without getting the "$\pm$" in the front?

Thanks.

$\endgroup$
  • $\begingroup$ Have you forgotten to insert an equation? $\endgroup$ – MRobinson Oct 11 '18 at 15:23
  • 1
    $\begingroup$ The insert equation is just there as a place holder for an example. Not a particular equation $\endgroup$ – Deep Patel Oct 11 '18 at 15:24
  • $\begingroup$ Ah brill, that makes sense! $\endgroup$ – MRobinson Oct 11 '18 at 15:25
1
$\begingroup$

For your first question. Wait until you have got a solution that is dependent on this constant $c$, then plug in your initial conditions to find out it's value. Sometimes you won't have initial conditions so you can just leave $c$ in there.

With regards to the second part, often your answer won't make sense if you pick the $+$ or the $-$. For example if you have $y$ defined as being positive, but taking the $-$ makes it negative. It is often left to you to justify your choice, and if you can't, it can be possible that both hold.

$\endgroup$
  • $\begingroup$ If i solve (dy/dx)=y+y(x^2), the solution is y=+-e^(x+(x^3)/3). The correct solution occurs when y has a positive in the front. However, there are no conditions given for y or x. So how do you justify the positive choice? $\endgroup$ – Deep Patel Oct 11 '18 at 15:31
  • $\begingroup$ Do you have any physical interpretation that could limit it? Or some condition on the range/domain of $x/y$ that could influence it? $\endgroup$ – MRobinson Oct 11 '18 at 15:35
  • $\begingroup$ I am given the initial conditions (a point that the solution curve cuts). But even after I substitute the initial conditions, i get the solution I stated above. However, only one equation fulfils the initial conditions (cuts the given point) even though i got both versions of the solution ("+" and "-") using the initial condition. So, my question is, is there a way to determine which solution is the correct one without saying "according to graph, this one insert equation is correct"? $\endgroup$ – Deep Patel Oct 11 '18 at 15:40
  • $\begingroup$ If only one solution fulfills the initial conditions then this is the correct one $\endgroup$ – MRobinson Oct 11 '18 at 15:49
  • $\begingroup$ thats the only explanation i can give isn't it $\endgroup$ – Deep Patel Oct 11 '18 at 15:50
0
$\begingroup$

$$dy/dx)=y+y(x^2),$$ $$\implies y'-y(1+x^2)=0$$ With integrating factor $$(ye^{-x-x^3/3})'=0$$ Integrate $$y(x)=Ke^{x+x^3/3}$$ You just have a constant K ...and that depends upon initial conditions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.