Separable First Order Ordinary Differential Equation with Natural Logarithms Please check my work:
$$xy' = 5y$$
$$\int\frac{dy}{y} = 5\int\frac{dx}{x}$$
$$\ln y = 5\ln x + c$$
$$y = 5x + c$$
Is this correct?
 A: In your solution, the fourth line is not a consequence of the third. A correct solution is
$$xy' = 5y$$
$$\int\frac{dy}{y} = 5\int\frac{dx}{x}$$
$$\ln y = 5\ln x + c$$
$$\ln y =\ln x^5+c$$
$$e^{\ln y} = e^{\ln x^5 + c}$$
$$e^{\ln y} = e^{\ln x^5}e^c$$
$$y = x^5e^c$$
$$y = Cx^5$$
where $C=e^c$.
A: $e^{5ln(x)+c}=e^cx^5$ by properties of exponents. Since c was arbritary, you can write this as  $y=cx^5$ where c is another arbitrary constant.
A: On your last step, recognize that $c = \ln c$, because $c$ and $\ln c$ are both arbitrary constants. 
Invoking the rules of logarithms,


*

*$5 \ln x = \ln x^5$ 

*$\ln x^5 + \ln c = \ln cx^5$.


Putting all these together,
$$\ln y = 5 \ln x + c \implies \ln y = \ln x^5 + \ln c \implies \ln y = \ln (cx^5)$$
Now apply the antilog $e$ to both sides:
$$e^{\ln y}=e^{\ln(cx^5)}$$
and we are left with $$y=cx^5$$
A: Actually the solutions are defined on the whole real line and are exactly the functions $y$ such that there exists some constants $(a,b)$ such that
$$
y(x)=\left\{\begin{array}{ccc}ax^5&\text{if}&x\gt0,\\ 0&\text{if}&x=0\\ bx^5&\text{if}&x\lt0.\end{array}\right.
$$
To prove this without dividing (by $y(x)$ when $y(x)=0$?) or taking the logarithm (of $y(x)\lt0$?), note that $xy'=5y$ implies that $y(0)=0$ and that the function $z$ defined by $z(x)=x^{-5}y(x)$ for every $x\ne0$ is such that $z'(x)=0$. Thus, $z$ is constant on $x\gt0$ and on $x\lt0$, QED.
