Question

Solve the separable differential equation.
y' = 3 y^2

Answer 1

\(\frac{dy}{dx}=3y^2\\\frac{1}{y^2}dy=3dx\\\int\frac{1}{y^2}dy=\int3dx\\-\frac{1}{y}=3x+c\\y=\frac{-1}{3x+c}\)

Answer 2

make sense?

Answer 3

my answer looks nothing likethat...

Answer 4

Oh wait, forgot this: Using the initial condition: y(3)= 3, find y(1).
y(1) =

Answer 5

but even then, i still don't understand

Answer 6

\(y(3)=3\implies 3=\frac{-1}{3*3+c}\implies27+3c=-1\implies3c=-28 \)
so \(c=\frac{-28}{3}\)
so
\(y(x)=\frac{-1}{3x+\frac{28}{3}}=\frac{-1}{\frac{9x+28}{3}}=\frac{-3}{9x+28}\)
now plug in x=1

Answer 7

what part of the first part do you not get?

Answer 8

Like, what you did follows what I see in the textbook, but I stil can't follow it.

Answer 9

\(\frac{dy}{dx}=3y^2\\\frac{1}{y^2}dy=3dx\)
are you good up till here?

Answer 10

Ok

Answer 11

then we take the integral
\(\int\frac{1}{y^2}dy=\int3dx\)

Answer 12

\[\int\limits_{?}^{?}dyy ^{-1}+c=\int\limits_{?}^{?}3xdx+c\]

Answer 13

?

Answer 14

where did y^(-1) come from, and the + c will come after you take the integral

Answer 15

y^-1 is equiv to 1/y right?

Answer 16

yes but we are looking at 1/y^2

Answer 17

ok i see what you did.

Answer 18

your integral sign shuold be gone, you have taken the intgral, also the dy and dx

Answer 19

why is it -1/y instead of 1/y for your first ever response answer?

Answer 20

you should have y^(-1) + c = 3x+c, and technically these are difference c's but we can just add them together and call them c
so
y^(-1) + c = 3x

Answer 21

go the other way and you will see

take the derivative of -y^(-1)

Answer 22

derive, not integral?

Answer 23

if you want to see where that negative came from...

Answer 24

\(\int x^n dx=\frac{x^{n+1}}{n+1}+c\)

Answer 25

i got y^-2

Answer 26

exactly

Answer 27

thats why the negative is there....

Answer 28

so for you y^(-2) becomes y^(-1) / (-1) = -y^(-1)

Answer 29

I see