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?   
 A: Because $k$ is an arbitrary constant, the two expressions describe the same set of possible solutions, so they are equivalent. 
However, with these integrations involving logarithms, there are some subtleties to be careful about with minus signs and absolute values. Many students learn these in a slightly incorrect way in calc II and then they get confused when they get to differential equations and these subtleties start to matter.
For example, if we solve $y'=y$ by separation of variables, we find
$$\int \frac{dy}{y} = \int dt \\
\ln(|y|) = t+C_1 \\
|y|=e^{t+C_1}=e^t e^{C_1} = C_2 e^t.$$
Here $C_1$ was arbitrary but $C_2$ is not: it must be positive. On the other hand, when I go to remove the absolute value to solve for $y$, I get a $\pm$:
$$y=\pm C_2 e^t$$
so now I can think of $\pm C_2$ as $C_3$ which could be any nonzero number. Moreover $y=0$ is a "nonseparable" solution, so in fact we could have $y=C e^t$ for any real number $C$. Essentially the same thing happens in your problem.
You can avoid this by solving the problem by definite integration. Specifically you can solve the problem by writing
$$\int_{y_0}^y \frac{dz}{z} = \int_{t_0}^t ds$$
so that you have $\ln(y/y_0)=t-t_0$ and $y=y_0 e^{t-t_0}$. The trick here is that $\int_a^b \frac{dx}{x} = \ln(b/a)$ regardless of the sign of $a,b$...assuming of course that they have the same sign so that the integration makes sense.
A: They are both correct.
$k$ is any real number.
Observe that the set $$\{5k | k \in \mathbb R\}$$ is the same as $$\{-5k | k \in \mathbb R\}$$
A: 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:
$$x(t)=\frac{e^{-\text{b}t}\left(\text{a}\left(e^{\text{b}t}-1\right)+\text{b}x(0)\right)}{\text{b}}=\frac{\text{a}}{\text{b}}+\frac{e^{-\text{b}t}\left(\text{b}x(0)-\text{a}\right)}{\text{b}}$$
