Question

Solve the initial value problem: y'=sqrt(xy), y(1)=4.

Answer 1

integrate y'

Answer 2

I know it'd be y, but how about the other side?

Answer 3

\[y'=\frac{dy}{dx}\]\[\frac{dy}{dx}=\sqrt{xy}\]

Answer 4

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

Answer 5

Now integrate both sides

Answer 6

Ahh I see....Forgot this haha.

Answer 7

Just a separable differential equation

Answer 8

But how do I integrate dy/sqrt(y)?

Answer 9

\[y^{-\frac{1}{2}}dy\]

Answer 10

Do it the way you'd integrate any other expression (x^(n+1)/(n+1) + C)

Answer 11

\[\large \int\limits \frac{1}{\sqrt{y}}=\int\limits\sqrt{x}dx\]\[\large\int\limits y^{-1/2}dy=\int\limits x^{1/2}dx\]

Answer 12

are u sure?

Answer 13

I got c=10/3, 2sqrt(y)=(2/3)x^3/2+10/3, right?

Answer 14

I think you're both off by a sign.

Answer 15

I might be wrong, but jhanny definitely is off by a sign.

Answer 16

\[\large\int\limits y^{-1/2}dy=\int\limits x^{1/2}dx\] \[\large \frac{y^{1/2}}{1/2} = \frac{x^{3/2}}{3/2}+c \]\[\large 2\sqrt{y} = \frac23x^{3/2}+c\]

Answer 17

i was off by an entire power. lol

Answer 18

Thank you guys so much.

Answer 19

Np