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?   

They are both correct.
$k$ is any real number.
Observe that the set $$\{5k | k \in \mathbb R\}$$ is the same as $$\{-5k | k \in \mathbb R\}$$
