How to solve implicit solution of IVP The given question is:
$$
 dy/dt + 2y = 1\ ;\qquad    y(0)= 5/2
$$
when i solve this i get $\ln(-4)=c$
now the problem is how to solve $\ln(-4)$?
You can get rid of the "arbitrary constant" as follows.

Cauchy's problem:

$$\tag{1} \begin{cases} y^\prime (t) +2y(t)=1 \\ y(0)=5/2 \end{cases}$$

has unique global solution (because the ODE is linear with constant coefficients); this solution is continuous in $\mathbb{R}$, hence $y^\prime (t)=1-2y(t)$ is continuous in $\mathbb{R}$ and $y(t)$ is of class $C^1$ in $\mathbb{R}$... Bootstrapping one gets $y(t)$ of class $C^\infty$ in $\mathbb{R}$.

Moreover, the ODE in (1) has only one constant solution, namely $\bar{y}(t):=1/2$, and it forces its nonconstant global solutions to be either $&gt;1/2$ or $&lt;1/2$ everywhere in $\mathbb{R}$: in fact, if a solution $\phi(t)$ has $\phi(t_1)&lt;1/2&lt;\phi(t_2)$, then (by continuity) there exists $T$ s.t. $\phi(T)=1/2$ therefore $\phi(t)=\bar{y}(t)$ everywhere (by uniqueness), a contradiction. In the present case, since $y(0)=5/2&gt;1/2$, one has $y(t)&gt;1/2$ everywhere in $\mathbb{R}$; hence $y^\prime (t)=1-2y(t)&lt;0$ in $\mathbb{R}$, so that $y(t)$ is strictly decreasing.

Now, let $t&gt;0$; one can divide both sides of $y^\prime (t)=1-2y(t)$ by $1-2y(t)\neq 0$ and integrate over $[0,t]$ to get:

$$\int_0^t \frac{y^\prime (\tau)}{1-2y(\tau)}\ \text{d} \tau = \int_0^t \text{d} \tau \; ;$$

since $y(\tau)$ is of class $C^\infty$ and strictly decreasing, one can make the change of variable $\eta = y(\tau)$ in the LHside to get:

$$\int_{5/2}^{y(t)} \frac{1}{1-2\eta}\ \text{d} \eta =t\; ,$$

hence:

$$-\frac{1}{2}\log |1-2\eta|\Big|_{5/2}^{y(t)}=t \; ;$$

keeping in mind that $1-2y(t)$ is negative in $\mathbb{R}$, one finds:

$$\log \frac{4}{2y(t)-1} =2t$$

and finally:

$$\tag{2} y(t)=2e^{-2t}+\tfrac{1}{2} \; .$$

On the other hand, if $t&lt;0$ one has to integrate over $[t,0]$ to recover the same elementary expression (2) for the solution. Therefore (2) gives the unique global solution to problem (1).
this is a linear differential equation and the solution of this equation is as follow:

IF= = e^∫▒〖2 dt〗
= e^2t
Now solution of equation is:
y e^2t = ∫▒〖1 .e^2t  dt〗
y e^2t = e^2t/2 + c  ……………………….eq(1)
Now put t=0 and y =  5/2
5/2 = 1/2+ c
C = 2
Putting this in eq(1), we get
y e^2t = e^2t/2+ 2
y=1/2+  1/2 e^(-2t)