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My differential equations book shows that the solution of $\frac{dy}{dt}+\frac{t}{2}y=4t$

is $y=\frac{\int_{}^{}e^{t^2}dt}{e^{0.25t^2}} + \frac{c}{e^{0.25t^2}}$

However, it then goes on by saying that we can represent the integral(by changing the variable of integration to s) by:

$y=\frac{\int_{0}^{t}e^{s^2}ds}{e^{0.25t^2}} +\frac{c}{e^{0.25t^2}}$

So how is it possible to represent an indefinite integral with a definite integral like the one above.

I tried using the same method with the integral of $e^t$ but the define integral ends up with a -1 while the indefinite one ends up with an arbitrary constant.

The problem is from Elementary Differential Equations and Boundary Value Problems 10th edition by Boyce the integral can be found in top of page 39

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    Try doing both integrals and seeing what happens. I know it sounds insane, but actually write both down carefully and formally. I think you'll find it very shocking. It sounds like I'm kidding, but I promise I'm not. I remember being very befuddled by similar things in the past. Trust me. Write them out. – 2017-01-26
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    I tried it with $e^t$ as I mentioned above and it gave me a different answer than that of the indefinite integral ( have you read the question !!) – 2017-01-26
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    Yes, BACK TACK, I read the question. The point I was making is that if you do what I said correctly, your answers will differ by a constant. – 2017-01-26
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    So what you are saying is that we can always rewrite an indefinite as a definite one in the same manner above. So my guess is that differing by a constant is not an issue since they will be lumped together and be called and arbitrary constant ? – 2017-01-26
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    Yes, or there is a subtle copying error in the question. In my experience, they may also just be acting lazy with their notation. I would ask the instructor, or perhaps e-mail the author directly. But My hunch was exactly what you wrote. – 2017-01-26
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    You're asking us a question about your differential equations book, but you won't tell us what book it is? What the hell? And there's something wrong with those integrals, no dt or ds in them? – 2017-01-26
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    The book is Elementary Differential Equations and-Boundary Value Problems 10th edition by Boyce the integral can be found in top of page 39 – 2017-01-27

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May be the confusion comes from some ambiguities in the writing of your equations.

In $\quad\frac{dy}{dx}+\frac{t}{2}y=4t\quad$ what is the variable ? Is it $x$, or $t$ , or is the unknown function of two variables $y(x,t)$ ?

You wrote the solution $\quad y=\frac{\int e^{t^2}}{e^{0.25t^2}} + \frac{c}{e^{0.25t^2}}\:.\quad$ This is ambiguous because in $\int e^{t^2}$ it is not specified which is the variable of integration.

Supposing that the ODE be correctly written as : $$\frac{dy}{dt}+\frac{t}{2}y=4t$$

Then, your solution $\quad y=\frac{\int e^{t^2}dt}{e^{0.25t^2}} + \frac{c}{e^{0.25t^2}}\quad$ is false.

The correct result is : $$y=\frac{\int 4te^{0.25t^2}dt}{e^{0.25t^2}}=8+\frac{c}{e^{0.25t^2}}$$ An arbitrary constant $c$ appears when the indefinite integral is written on the form of a definite integral plus a constant : $$\int 4te^{0.25t^2}dt=\int_0^t 4se^{0.25s^2}ds+C=\left[8e^{0.25s^2} \right]_{s=0}^{s=t}+C = 8e^{0.25t^2}-8+C$$ Let $c=-8+C$ $$\int 4te^{0.25t^2}dt=8e^{0.25t^2}+c$$ $$y=\frac{\int 4te^{0.25t^2}dt}{e^{0.25t^2}}=8+\frac{c}{e^{0.25t^2}}$$ Since you don't show all details of your calculus, we cannot point out where exactly is your mistake.

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    Sorry, I was in a hurry when I wrote the post. it is supposed to be $dy/dt$ not $dy/dx$. I also corrected the post with the appropriate variables of integration and added the book and the page number where this can be found – 2017-01-27