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!
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 $$y(x)=K_1+K_2e^{-x}$$