Question

@study100 Help me on these two questions please

Answer 1

I'll try ;u;

Answer 2

I did what we did in the previous ones right and took "y" out of the equation and took the integral based on x

Answer 3

\[\large \large \frac{ dy }{ \sqrt{y} } = \frac{ dx }{ \sqrt{x}}\]

Answer 4

and just intergrate both sides : DDD

Answer 5

\[\frac{ 2 }{\sqrt{\frac{1}{x}}}+ C\]

Answer 6

\[\large 2y^{1/2} = 2 x ^{1/2} + C\]

Answer 7

the other side is \[\frac{ y^2}{2} +C\]

Answer 8

lol im wrong then

Answer 9

f(x) =\[2\sqrt{x}\]

Answer 10

well integrate we have
\[\large 2 y ^{-1/2 +1}\] = \[2 x ^ {-1/2 + 1} + C???\]

Answer 11

simplify I got
\[\sqrt{y} = \sqrt{x} +C\]

Answer 12

nice i simplified and got the same thing

Answer 13

one more question and thats it

Answer 14

Ok! xD I almost have to eat dinner too!
\[\large \frac{ dy }{ y } = 3x dx\]

Answer 15

\[\large \ln (y) =\frac{ 3x^{2} }{ 2 } + C\]

Answer 16

ok

Answer 17

should I exponentiate it? o.o like e^lny to get y alone?

Answer 18

\[\large e^{lny} = e^{3x^{2}/2+C}\]

Answer 19

got it

Answer 20

\[y = Ce^{3x^{2}/2}\]

Answer 21

the answer is