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I began with a linear, first-order differential equation and an initial condition, and then I simplified the equation to the following:

$$y = \frac{1}{2}\int x^\frac{3}{2} dx $$

Which simplifies further to:

$$y = \frac{1}{2} * (\frac{2}{5}x^\frac{5}{2} + C)$$

Is it necessary to distribute the $\frac{1}{2}$ to the constant of integration $C$, since I am now going to try and solve for an exact value of $C$ with the initial condition?

If I didn't need to solve for $C$, I imagine that it wouldn't matter either way, since $C$ just represents an arbitrary number. But whether I do or don't in this case seems to affect the outcome. For the initial condition $y(0) = 1$, it seems that I could either get $C=1$ or $C=2$.

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    It doesn't matter either way you do it here, as distributing or not, one still has to solve for *some constant* through the initial condition. (personally, reduce/simplify where possible first) – 2017-01-26
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    You can write your constants however you want. You can write $f(x)+c$, where $c=5$, or $f(x)+5c$, where $c=1$. In either case, the textbook will probably say that the constant term is $5$. – 2017-01-26
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    The point is that the constant of integration is a constant - the value is unknown unless there is some condition to determine it. When there are limits the same constant appears twice and cancels. Often integrals may look different but differ by a constant e.g. http://math.stackexchange.com/questions/2113950/how-come-frac11-x-and-fracx1-x-have-the-same-derivative . A scalar multiple of a constant is still a constant. – 2017-01-26

2 Answers 2

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There is no difference whether you distribute the $\frac{1}{2}$ or not. The reason you seem to be getting $C=1$ or $C=2$ is because you should technically be using different variable symbols, since both $C$'s are different. (Which is why I substitute $\frac{1}{2}C=k$).

Let's first do this problem without distributing $C$: $$y=\frac{1}{2}\left(\frac{2}{5}x^{\frac{5}{2}}+C\right)$$ Substituting your initial condition gives: $$1=\frac{1}{2}(\frac{2}{5}\cdot 0 + C)$$ $$1=\frac{1}{2}C$$ $$C=2$$ Therefore, your solution is: $$y=\frac{1}{2}\left(\frac{2}{5}x^{\frac{5}{2}}+2\right)=\frac{1}{5}x^{\frac{5}{2}}+1 \tag{1}$$ Now, let's do this problem while distributing $C$: $$y=\frac{1}{2}\left(\frac{2}{5}x^{\frac{5}{2}}+C\right)=\frac{1}{5}x^\frac{5}{2}+\frac{1}{2}C$$ Note that since $C$ is an arbitrary constant, we can substitute $\frac{1}{2}C=k$. $$y=\frac{1}{5}x^\frac{5}{2}+k \tag{2}$$ Substituting initial conditions: $$1=\frac{1}{5}\cdot 0 + k$$ $$k=1$$ Which still gives you the same solution as in $(1)$ if the initial condition is substituted into $(2)$.

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It is often easier to treat formulas mechanically. But if you have difficulty with the understanding , you should write all the details of what is asserted or implied: $$y(x)=(1/2)\int x^{3/2}dx \quad \text { for all } x$$ IFF there exists a constant $C$ such that $$y(x)=(1/2)(C+(2/5)x^{5/2} \quad \text {for all } x$$ IFF there exists constant $C$ such that $$y(x)=C/2+x^{5/2}/5 \quad \text { for all } x.$$ At this point we can say that this is equivalent to: There exists a constant $D=2C$ such that $$y(x)=D+x^{5/2}/5 \quad \text { for all } x$$ which is equivalent to: There exists a constant $D$ such that $$y(x)=D+x^{5/2} \quad \text { for all } x.$$