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I am learning simple differential equations, and I know that the way they are solved is basically some magic with $dy$ and $dx$ behaving like normal fractions but then suddenly being integrated and I find this whole thing a bit weird as I can't explain to myself how come this works. However, while trying to solve a simple equation $y' = y + a$, I came up with this: $$ \frac{dy}{dx} = y + a \\ \frac{dy}{y+a} = dx \\ \int \frac{1}{y+a} dy = \int 1 dx \\ \text{substitute u = y + a} \\ \text{therefore dy = d(u - a) = du - da} \\ \int \frac{1}{u} du - da = \int 1 dx \\ \int \frac{1}{u} du - \int 1 da = \int 1 dx \\ \ln u - a = x \\ \ln (y+a) = x + a \\ y = e^{x+a} - a $$ I checked the solution and it seems correct but the whole substitution thing seems a bit... sketchy to me. Therefore I ask, is this a valid way of solving differential equations?

Note: I omitted the constants for simplification

lav_shaun
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1 Answers1

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You mention that you have omitted the constants for simplification, however it appears that $a$ is a constant in this equation, as I can't decipher what the independent variable of $a$ is. Assuming that $a$ is constant, a solution to this ODE is $$y(x)=ce^x-a$$ for $c$ constant.

The part where you're going awry is where you $u$-substitute.

Your choice of $u$ is good, however it's not valid to say $dy = d(u-a)$ because $a$ is constant. Your $du$ for this substitution will just be $dy$ since $u$ is a function of $y$. That is, $$u(y)=y+a\\ \frac{du}{dy}=1\\du=dy$$

Alex D
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  • Yes, $a$ is just an arbitary number. Also, I found that another solution to this particular equation is $y = 1 - a$. Any way to get to this solution from the equation? – lav_shaun Apr 03 '20 at 08:49
  • $y=1-a$ is not a solution to the original ODE. You can see this if you try substituting it into the original equation, $\frac{dy}{dx}=y(x)+a$. If, $y=1-a$, then $\frac{dy}{dx}=0$. Thus you get $0=(1-a)+a=1$ and $0=1$ is a contradiction. – Alex D Apr 03 '20 at 09:26