# Solving differential equations “$c$” value

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.

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

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.
• Do you have any physical interpretation that could limit it? Or some condition on the range/domain of $x/y$ that could influence it? – MRobinson Oct 11 '18 at 15:35
$$dy/dx)=y+y(x^2),$$ $$\implies y'-y(1+x^2)=0$$ 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.