I am entirely new to Differential Equations and am stuck with the following:
$(y')^2=|4y|$
$y(1)=4$
Which of the following are solutions to the problem?
a. $y=(x+1)^2$
b. $y=(3-x)^2$
c. $y=(x+1)·|x+1|$
d. $y=0$
e. $y=(3-x)·|3-x|$
I would say all of them, except for d. However, I am not sure, and c. and e. in particular are throwing me off, and I am not sure how to approach them. Any help would be tremendously appreciated!
All solutions except d) are of the form
$$y=\pm(x-a)^2$$ where the sign only depends on the sign of $x-a$.
But in all cases
$$(y')^2=(\pm2(x-a))^2=4(x-a)^2=4|y|.$$
We also have
$$(0')^2=4|0|.$$
Remains to check the initial condition.