# Question about solving linear first order non-homogeneous ODEs

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• CGandC
CGandC
TL;DR Summary
Why solution of ## y'+2xy = 8x ## is ## y=4+C e^{-x^2} ## and not ## y=e^{-C_1}(4+C_2 e^{-x^2}) ##?
A general equation for linear first order non-homogeneous ODE is: ## y' + a(x)y = b(x) ##.
The procedure to solve ( assuming ## a(x) , b(x) ## are continuous so that the fundamental theorem of calculus could be used ) it is to multiply it by ## e^{A(x)} ## ( here ## A'(x) = a(x) ## ) s.t. ## A(x) = \int^x { a(t)dt } ## and we get ## (ye^{A(x)})'=b(x)e^{A(x)} ##.
Hence ## ye^{A(x)}=\int^x {b(t)e^{A(t)}dt + C} ##, hence ## y(x) = e^{-A(x)}(\int^x {b(t)e^{A(t)}dt } + C ) ##.

The question is: why when we consider## A(x) = \int^x { a(t)dt } ## we do not add another constant? , i.e., why are we not writing ## A(x) = \int^x { a(t)dt } + \tilde{C} ## where ## \tilde{C} ## is another constant?.

Example: Consider ## y'+2xy = 8x ##. I tried solving it as follows:
Here ## A'(x) = 2x ## and so ## A(x) = x^2 + C_1 ##. Multiply both sides of ODE by ## e^{A(x)} = e^{x^2 + C_1 } ## and so ## (ye^{x^2 +C_1} )' = 8xe^{x^2 } ##, hence ## ye^{x^2 +C_1} = \int^x 8te^{t^2 }dt = 4e^{x^2} + C_2 ##; from this we have ## y=e^{-C_1}(4+C_2 e^{-x^2}) ##.
However, the "true" solution according to the procedure above is ## y=4+C e^{-x^2} ## ( only one constant ). Why is it so? It seems to me like the constant part ## e^{-C_1} ## cannot be "intermingled" into the other constant ## C_2 ## and that we must have ## y=e^{-C_1}(4+C_2 e^{-x^2}) ##.

Thanks for any help!______________________

Edit: as I was typing this I just noticed that when I multiplied both side and that I've written ## (ye^{x^2 +C_1} )' = 8xe^{x^2 } ## - this is wrong since it must be ## (ye^{x^2 +C_1} )' = 8xe^{x^2 + C_1 } ## , this was my mistake ( and then it turns out the constant ## C_1 ## doesn't come up in the solution so it doesn't matter if we write ## A(x) = \int^{x} a(t)dt ## or ## A(x) = \int^{x} a(t)dt + C_1 ## )

Last edited:

CGandC
CGandC said:
Example: Consider y′+2xy=8x. I tried solving it as follows:
$$\frac{dy}{8-2y}=xdx$$
$$\ln |y-4|=-x^2+C_1$$
$$y=4+C_2e^{-x^2}$$

Last edited:
CGandC
Mark44 said:
Yes,

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