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o-nauts, before I begin I must stress that I am self-taught and in college I will be taking Calculus I this semester. Nonetheless I have studied on my own up to the Divergence Theorem and Introductory Complex Analysis. In my venture, I now find myself completing an introductory course of Ordinary Differential Equations (which up until now has mostly been Calcu-Algebra to me).

I have now encountered a particular second-order differential equation that can be reduced to a linear, first-order equation of which the solution I have found is not in agreement with what my shows to be the solution, so here goes:

Given $$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$$

Let $p = \frac{dx}{dy}$ and $p\frac{dp}{dy} = \frac{d^2y}{dx^2}$

$$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0 \Rightarrow p\frac{dp}{dy} + p =0$$

Which simplifies to:

$$\frac{dp}{dy} + 1 = 0$$ which when evaluated with separation of variables, we get: $$p + y = 0$$ which is $$\frac{dy}{dx} + y + C_1 = 0$$

I go on to do this:

$$\frac{dy}{dx} + y + C_1 = 0$$ $$\frac{dy}{dx} + y = -C_1$$ which to me is a first order linear equation and can be solved using the canonical integrating factor.

Thus, given the standard for of a first-order linear equation: $$ \frac{dy}{dx} + Py = Q$$ Let: $$P = 1, Q = -C_1 \quad and \quad \rho = e^{\int Pdx}$$

The equation $$\frac{dy}{dx} + y = -C_1$$ evaluates to $$\rho y = \int \rho (-C_1) dx + C_2$$

Which when evaluted (atleast for me), we get:

$$y = C_2e^{-x} - C_1$$

Whereas my textbook's solution is: $$y = C_1e^{-x} + C_2$$

Now, what the !@#$%^& am I doing wrong here?

Oh, and I am using Addison-Wesley's "Calculus and Analytic Geometry" printed in 1968.

Thank You!

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  • $\begingroup$ $-C_1=K_1$ thats it your solution is correct to me Its just a constant after all so writing $-C_1$ or $C_1$ its the same $\endgroup$ – Isham Jul 13 '18 at 14:59
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Your answer is correct, $C_1$ is a constant so writing $-C_1$ or $K_1$ is the same..

Note that you can integrate directly $$y''+y'=0$$ Integrate $$y'+y=K_1$$ Multiply by $e^x$ $$(ye^x)'=K_1e^x$$ Integrate $$ye^x=K_1e^x+K_2$$ Finally $$y(x)=K_1+K_2e^{-x}$$

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Your answer and your book's answer are equivalent because C_1 and C_2 are arbitrary constants to be found by initial information.

Note that your solution could have been shorter as follows:

$$y'' +y' =0$$

$$ u=y' \implies u'+u=0$$ $$ u'=-u \implies u=C_1 e^{-x} $$ $$ y'= C_1 e^{-x} \implies y=-C_1e^{-x}+C_2 $$

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  • $\begingroup$ What I don't understand is that clearly $-C_1& is not $C_1$. I am limited to what I can learn as differential equations are beyond the scope of my bachelor's degree program. Thank you nonetheless!! $\endgroup$ – SSBASE Jul 13 '18 at 15:23
  • $\begingroup$ @SSBASE C_1 is an arbitrary constant, you may call it K or C or -C_1 or whatever else. Once the initial values are given then the constants will be uniquely determined. Sort of like the constant C in an integral. $\endgroup$ – Mohammad Riazi-Kermani Jul 13 '18 at 15:33

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