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I very much suspect this but I'm not sure if it's correct:

In solved differential equations, does the constant 'c' always represent the value of the dependent variable when the independent=0 ?

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This is the case in equations such as $\dfrac{dy}{dx}=ay$, where $a$ is a constant. –  Sanath Jun 14 '14 at 16:29
@SanathDevalapurkar ah I see it's in special cases only. Would $\dfrac{dy}{dx}=axy$ also work? –  hb20007 Jun 14 '14 at 16:32
Solving the above given differential equation gives $$\ln |y|=\dfrac{ax^2}{2}+C\implies y=C_1\exp\left(\dfrac{ax^2}{2}\right)$$ When $x=0$, then $C_1=y(0)$, so, yes, it would work. –  Sanath Jun 14 '14 at 16:34
@SanathDevalapurkar Thank you, If this was an answer I'd selected it as best :D –  hb20007 Jun 14 '14 at 16:35
It doesn't exactly answer the question; I provided examples of where it'd work, but, as the counterexamples below show, the statement above is false. –  Sanath Jun 14 '14 at 16:36

2 Answers 2

up vote 11 down vote accepted

Not necessarily: Suppose we have $y = f(x) = e^x + c$

$$f(0) = 1 + c \neq c$$

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In polynomials , we do have that $c = p(0)$ –  amWhy Jun 14 '14 at 16:33
@hb20007 The above comment by amWhy would be a good exercise. –  Sanath Jun 14 '14 at 16:35
@amWhy I am unfamiliar with the notation in your comment. Could you please tell me what 'p' represents and why it's an 'exercise'? –  hb20007 Jun 14 '14 at 16:41
Oh sure, I'm just using $p(x)$ to denote a p olynomial (which we can call $y$, aka, the dependent variable, so $p(0)$ denotes a polynomial $p(x)$ evaluated at $x = 0$. Example: $p(x) = x^3 + 4x^2 + 3x + c \implies p(0) = c.$ –  amWhy Jun 14 '14 at 16:44

No. Suppose $$\frac{dy}{dx}=\sin x$$ then $$y=-\cos x+c$$

At $x=0$ $$y(0)=-1+c\ne c$$

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Dear Edwin, you do not have to use $$<equation>$$ every time. To do inline equations, simply write $<inline equation>$. –  Sanath Jun 14 '14 at 16:34

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