why two solutions to DE are contradictory? 
Solution to the given differential equation $$\frac{dx}{dt}=4.9-0.196x$$
  is given by 
a) $x=25+ke^{-0.196t}\qquad$ b) $x=50+ke^{-0.196t}\qquad$ c) $x=50-ke^{-0.196t}\qquad$ 
d) $x=25-ke^{-0.196t}\qquad$ where k is some constant

my try:
method(1)
$$\frac{dx}{4.9-0.196x}=dt$$
$$\int \frac{dx}{4.9-0.196x}=t+c$$
substituting $4.9-0.196 x=z$, $dx=-\frac{dz}{0.196}$, i get
$$\int \frac{dz}{z}=-0.196(t+c)$$
$$\ln z=\ln(4.9-0.196x)=-0.196(t+c) \tag{*}$$
$$0.196x=4.9-e^{-0.196(t+c)}=4.9-e^{-0.196c}\cdot e^{-0.196t}$$
$$x=25-ke^{-0.196t}\tag 1$$
method(2) $$\frac{dx}{dt}=4.9-0.196x$$
$$\frac{dx}{dt}+0.196x=4.9$$
now, use integration factor $I.F.=e^{\int 0.196\ dt}=e^{0.196 t}$
so the solution is 
$$x\cdot (I.F.)=\int (I.F.)\cdot 4.9\ dt+c$$
$$x\cdot e^{0.196 t}=\int e^{0.196 t}\cdot 4.9\ dt+c$$
$$x\cdot e^{0.196 t}=\frac{4.9}{0.196}e^{0.196 t}+c=25e^{0.196 t}+c$$
$$x=25+ke^{-0.196 t}\tag 2$$
now, you see i am getting two different solutions (1) & (2) to the same D.E., but i don't know which one is correct & why.
please explain me where i am wrong or which is the correct option & why?   

When $\text{a}\space\wedge\space\text{b}\in\mathbb{R}$:
$$x'(t)=\text{a}-\text{b}x(t)\Longleftrightarrow\int\frac{x'(t)}{\text{a}-\text{b}x(t)}\space\text{d}t=\int1\space\text{d}t$$
Use:


*

*Substitute $u=\text{a}-\text{b}x(t)$ and $\text{d}u=-\text{b}x'(t)\space\text{d}t$:
$$\int\frac{x'(t)}{\text{a}-\text{b}x(t)}\space\text{d}t=-\frac{1}{\text{b}}\int\frac{1}{u}\space\text{d}u=-\frac{\ln\left|u\right|}{\text{b}}+\text{C}=\text{C}-\frac{\ln\left|\text{a}-\text{b}x(t)\right|}{\text{b}}$$

*$$\int1\space\text{d}t=t+\text{C}$$


So, we get:
$$-\frac{\ln\left|\text{a}-\text{b}x(t)\right|}{\text{b}}=t+\text{C}$$

Another way, is using Laplace transform:
$$x'(t)=\text{a}-\text{b}x(t)\to\text{s}\text{X}(\text{s})-x(0)=\frac{\text{a}}{s}-\text{b}\text{X}(\text{s})\Longleftrightarrow\text{X}(\text{s})=\frac{\text{a}+\text{s}x(0)}{\text{s}(\text{b}+\text{s})}$$
With inverse Laplace transform you'll find:
$$x(t)=\frac{e^{-\text{b}t}\left(\text{a}\left(e^{\text{b}t}-1\right)+\text{b}x(0)\right)}{\text{b}}=\frac{\text{a}}{\text{b}}+\frac{e^{-\text{b}t}\left(\text{b}x(0)-\text{a}\right)}{\text{b}}$$
