Solving differential equations "$c$" value

Question

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.

Answer 1

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.

Answer 2

$$dy/dx)=y+y(x^2),$$ $$\implies y'-y(1+x^2)=0$$ 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.

Answer 3

Substituting Initial Conditions: Typically, you substitute the initial conditions after integrating both sides of the differential equation and obtaining the general solution with an arbitrary constant "C". This general solution will have the form of an equation relating y and x. Then, you apply the initial conditions to determine the specific value of "C" that satisfies the given conditions. This allows you to find the particular solution that fits the initial conditions. Sometimes, rearranging the equation to solve for "y" explicitly in terms of "x" and "C" can yield multiple possible values for "C". In such cases, you must evaluate which value(s) of "C" satisfy the given initial conditions. It's crucial to ensure consistency with the initial conditions when determining the appropriate value(s) for "C".

Dealing with "±" in the Solution: In some cases, after integrating and simplifying the differential equation, you might end up with an equation that contains "±" in the solution. This "±" indicates that there are two possible solutions, one with the positive sign and another with the negative sign. However, only one of these solutions will satisfy the given initial conditions. To determine the correct solution, substitute the initial conditions into both possible solutions separately. Evaluate which one satisfies the initial conditions and matches the given values. This will allow you to select the appropriate solution that fits the initial conditions.

It's important to note that for some differential equations, the "±" arises due to the nature of the equation itself, and it may not be possible to avoid it. In such cases, evaluating the solutions using the initial conditions becomes necessary to determine the correct one.

By applying these methods, you can ensure that you find the particular solution that satisfies the given initial conditions and accurately represents the behavior of the system described by the differential equation.

I hope this clarifies the process for you! Feel free to ask any additional questions.