Question

initial value problem:

2y' * sqroot(x) = y
y(0) = 1

Answer 1

\[2y'\sqrt{x} = y\]
y(0) = 1

Answer 2

\[2dy/dx \sqrt{x}=y,\]
dy/y=dx/(2(sqrt(x),
\[\ln \left| y \right|=\sqrt{x}+c\]
subst. (0,1) to get
ln(1)=sqrt(0) +c
so c=0

Answer 3

Im hoping you know how to integrate , good luck!

Answer 4

thank you lbt =3