Question

introduction to differential equations, calc one.

introduction to differential equations, calc one.

@Mathematics

Answer 1

\[dz/dt = t^2z^2\]

Answer 2

seperation of variables

Answer 3

I multiplied both sides by Dtz^3, but now do I need to apply the product rule, and if so, how do I do that backwards?

Answer 4

the z^3 is a little odd

Answer 5

z^2, sorry.

Answer 6

multiply boths sides by dt

dt(dz/dt=z^2 t^2)

dz = z^2 t^2 dt

now divide both sides by z^2

(dz = z^2 t^2 dt)/z^2

dz/z^2 = t^2 dt , and integrate

Answer 7

\[\int \frac{1}{z^2}dz=\int t^2\ dt\]
\[-\frac{1}{z}+c_1=\frac{1}{3} t^3+c_2\]
\[-\frac{1}{z}=\frac{1}{3} t^3+c_2-c_1\]
\[-\frac{1}{z}=\frac{1}{3} t^3+C\]

Answer 8

might have to do some minor mathing to get z in the proper condition yet, but thats the brunt of it

Answer 9

my struggle is intergrating the z, I follow everything you have done with the t side.

Answer 10

the z part then, might be better to see it as z^(-2) instead

Answer 11

\[\int \frac{1}{z^2}dz=\int z^{-2}dz\]

Answer 12

then the power rule for integration is the same as the t side .... add one and divide by it

Answer 13

\[\int z^{-2}dx=\frac{z^{-2+1}}{-2+1}=-z^{-1}=-\frac{1}{z}\]

Answer 14

roger, that's a big help, but let me ask you, why does c2-c1 turn into a singular C?

Answer 15

Its just a convenient way to absorb the arbitrary constants into one mass of arbitrary constant-ness...

Answer 16

when i first learned these i was confused as to why they didnt put a +c on the left side; i eventually found out why

Answer 17

its rough, i thought dx/dy was a symbol until
last week.

Answer 18

yeah, they let you think that as well.

Answer 19

its an operator

Answer 20

like + is the operator for addition; / is the operator for divide, its just an operator

Answer 21

at this level of math you start to see that math isnt as stringent and concrete as you were led to believe

Answer 22

blue pill please.

Answer 23

blue pill?