Question

differential equations by separation of variables
dy/dx= 4y
(help with how to solve the differential equations)

Answer 1

i know how to do them, but i dont know how to check on the calculator and im not sure if i am doing them correctly.

Answer 2

\[\int\limits_{}^{}(\frac{1}{4y})dy=\int\limits_{}^{}dx\]

Answer 3

To check, substitute the function you got, differentiate with respect to x and see if you get 4y.

Answer 4

if you want to verify your result, you don't use a calculator. You use pencil and paper. The very best way to verify you have the right result to a differential equation is to substitute your solution into the differential equation and show that it satisfies it.

Answer 5

james check out my prob too http://openstudy.com/study#/updates/4f0d9f5de4b084a815fcdc36

Answer 6

btw, what solution did you find to this equation?

Answer 7

dy/dx=4y
1/4y dy=dx
∫▒〖1/4y dy=∫▒1 dx〗
1/4 lny=x+c
lny^(1/4)=x+c

Answer 8

\[ \frac {dy}{dx}=4y\]
\[\frac {dy}{4y}=dx\]
\[\int\limits \frac {1}{4y}dy=\int\limits dx\]
\[\frac {1}{4}\ln y=x+C\]
\[\ln y=4x+C _{1}\]
\[y=e ^{4x+4C _{1}}\]
\[y=e ^{4x}e ^{4C _{1}}=e ^{4x}C\]

Answer 9

4 of 1/4 cannot be taken to the RHS it will become the power of ln's variable

Answer 10

I didn`t get what your saying

Answer 11

1/4lny becomes lny^(1/4)

Answer 12

Cinar's solution is correct.

Answer 13

but as far as i have studied the co-effecient of lny becomes the power of y.can u explain?

Answer 14

Yes, that's right.
\[ a \ln b = \ln b^a \]
But because we want to solve for y, we don't take that step, or if we do, it's just an intermediate of one, until we get to y is a function of x. That is what cinar has done.

Answer 15

okay.got it

Answer 16

tmrk, you can go on to solve problem like this,

\[\ln y ^{1/4}=x+c => y ^{1/4}=e ^{x+c} => ( y ^{1/4})^{4}=(e ^{x+c})^{4}\]
\[y=e ^{4x+4c} => y=e ^{4x}e ^{4c} =>y=e ^{4x}C\]

Answer 17

thanks i got it.