Question

The particular solution of the differential equation dy/dt = y/4 for which y(0) = 20 is

Answer 1

Separation of variables.

Answer 2

(and then using y(0)=20, to find the integration constant C)

Answer 3

these are the choices I have

y = 20e^−0.25t
y = 19 + e^0.25t
y = 20 e^0.25t
y = 20e^4t

Answer 4

Example:
\(\color{#000000}{ \displaystyle \frac{dy}{dt} =\frac{1}{3}y^2~;\quad y(0)=-1 }\)

Solution:

First, rearrange the equation algebraically to "separate
the variables" and integrate. (Like I did right below)
\(\color{#000000}{ \displaystyle \frac{dy}{dt} =\frac{1}{3}y^2\quad \longrightarrow \quad y^{-2}dy=\frac{1}{3}dx\quad \longrightarrow \quad \int y^{-2}dy=\int \frac{1}{3}dx }\)

From this, (after integrating) you get,
\(\color{#000000}{ \displaystyle -y^{-1}=\frac{1}{3}x+C }\)

(I didn't write -C when dividing by -1, since C is an
arbitrary constant and will remain an arbitrary constant
regardless what (non-zero) scalar you multiply it by)
\(\color{#000000}{ \displaystyle y^{-1}=-\frac{1}{3}x+C }\)

Now, just solving for y.
\(\color{#000000}{ \displaystyle y^{-1}=-\frac{x}{3}+C }\)

\(\color{#000000}{ \displaystyle y^{-1}=-\frac{x+C}{3} }\)

\(\color{#000000}{ \displaystyle y=-\frac{3}{x+C} }\)

This is your general solution.

Now, since we are given an initial value (that is y(0)=-1), solve for C.
\(\color{#000000}{ \displaystyle -1=-\frac{3}{0+C}\quad \longrightarrow \quad C=3 }\)

So, your final solution comes out to be
\(\color{#0000ff}{ \displaystyle y=-\frac{3}{x+3} }\)

Answer 5

I came out with a polynomial, but in your case when you integrate you should get ln(y), if you separate the variables correctly.

Answer 6

My example differs, but the procedure is the same.

1) Separate the variables
2) Integrate both sides (don't forget integration constant C)
3) Solve for y
4) Use the initial condition (if given) to find C

(you can do step 4 first and then do step 3, if you like)

Answer 7

would it be a,b,c or d?

Answer 8

Lets solve it to find out.

Answer 9

Oh, I used x instead of t in my example.... switched variables... sorrry, hopefully I didn't mess up too much with this.

Answer 10

In any case, as far as your problem is concerned....

Can you rearrange the equation so that all y's (incl. the differential dy) are on the one side, and all t's (incl the differential dt) are on the other side?

Answer 11

4(dy/y) = dt

Answer 12

Yes, that is good!

Answer 13

Next, we are integrating both sides.

\(\int 4(dy/y)=\int dt\)

Answer 14

now, tell me what do we get when we integrate on both sides?

Answer 15

do we integrate both sides next?

Answer 16

Yes:)

Answer 17

4dx +c=dx^2/2 +c

Answer 18

Redo that integration please.
How do you have differentials in the equation after integrating?

Answer 19

2x^2 + c=x+c

Answer 20

i mean 2y^2 + c = x + c

Answer 21

@SolomonZelman