Question

Separable Equations
\[\frac{dy}{dx}=\frac{\sqrt{x}}{e^{y}}\]
\[\int e^{y}dy=\int \sqrt{x}dx \]
\[\frac{e^{y}}{2x}=y\]
Yes?

Answer 1

Nope.

\[\int e^{y}dy \neq \int \sqrt{x}dx\]
\[\int e^{y}dy = e^y + C\]
\[\int \sqrt{x} dx = \int x^{1/2} dx = \frac{x^{1/2+1}}{1/2+1} + C = \frac{2}{3}x^{3/2} + C\]

Answer 2

Your second line of logic is false.

Answer 3

wait? Why is the second line that he has is bad?

Answer 4

Disagree?

Answer 5

To solve for Y, you need to equate both answers and take the natural log of both sides.

Answer 6

Well we want to separate the variables \[\frac{dy}{dx}=\frac{\sqrt{x}}{e^y}\]
Multiply e^y on both sides
Multiply dx on both sides
\[e^y dy =\sqrt{x} dx\]
Now integrate both sides

Answer 7

\[\int\limits_{}^{}e^y dy= \int\limits_{}^{}\sqrt{x} dx\]

Answer 8

exactly

Answer 9

is my integration correction though?

Answer 10

\[\int\limits_{}^{}e^y dy = \int\limits_{}^{}x^\frac{1}{2} dx\]

Answer 11

Hey @MathSofiya agent showed you how to integrate the right side

Answer 12

and the left side

Answer 13

ok...so I did it wrong, lolz

Answer 14

After integrating, your answer should look like:
e^y = 2/3x^3/2 + c. Then, take the natural log of both sides to get Y.

Answer 15

I see

Answer 16

thanks!

Answer 17

\[\ln|e^y| = ln|(2/3)x^{3/2} + C|\]
\[y = ln|(2/3)x^{3/2}| + C\]

Although why we don't have to rename C is a technical mystery I've never quite understood.

I stand corrected @myininaya , it's the 3\(^{rd}\) line, not the 2\(^{nd}\). :-)

Answer 18

thanks agentx5

Answer 19

why the absolute value?

Answer 20

@agentx5 lol okay

Answer 21

Because you can't rule out if the result of what's inside is negative for some reason, the domain of logarithms between this: 0<x<\(\infty\)

I'll leave you to think about why ;-)

PS: If there's some sort of conditions we can use to know (2/3)x^(3/2) is always positive then we can drop the absolute value bars

Answer 22

fair enough

Answer 23

:)