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
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.