Initial value problem $dy/dx =\sqrt{y^2}$

Question

how does one go about solving the IVP

$dy/dx = \sqrt{y^2}$ ; with $y(0) = C$ in the area of $x = 0$?

My solution:

$dy/dx = y$

$\implies$ $(1/y) dy = 1 dx$

$\implies$ $\int (1/y) dy = \int 1 dx$

$\implies$ $\ln |y| = x + c$

$\implies$ $y = e^{x+c}$ .

However, checking the answer sheet to the problem, it says that the correct answer is $y=c\times e^x$.

Thanks for any help in advance!

Answer 1

Remember $C$ is just a constant. Use properties of exponentiation $$ y= e^{x+c}= e^x \cdot e^c= C e^x $$ where $C= e^c$. So although the constants $c,C$ are different, the solutions are the same, e.g. $$ 2e^x= e^{\ln 2} e^x= e^{x+\ln 2} $$ In this case, $C=2$ and $c= \ln 2$.