Question

Help with this differential equation please, it won't take my answer in Webwork.

Answer 1

hang on

Answer 2

did you try the preview answer button?
to see if it looks as expected?
unless you've used e^ before on ww, you should try exp( )

Answer 3

Is the answer right though? I'll try typing in preview

Answer 4

actually it isn't. looked like it was.

Answer 5

Help me fix it?

Answer 6

general solution:
\[y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}\]
you found λ 1 and 2 which were 1/5 and 4/5

so we have:
\[y(t) = c_1 e^{t/5} + c_2 e^{4t/5}\]
the boundary conditions are used to define c and c2.

y(0) = 1, when plugged in, leads to c1 + c2 = 1
to use y'(0), we find derivative:\[y'(t) = \frac{ 1 }{ 5 }e^{t/5} + \frac{ 4 }{ 5 }e^{4t/5}\]
\[y'(0) = \frac{ 1 }{ 5 }c_1 + \frac{ 4 }{ 5} c_2 = 0 = \frac{ 1 }{ 5 }(c_1 + 4c_2)\]
this leads to c1 + 4c2 = 0

combined with c1 + c2 = 1, we can solve for c1 and c2.

turns out they are:\[c_1 = -\frac{ 4 }{ 3 }\]\[c_2 = -\frac{ 1 }{ 3 }\]
so\[y(t) = \frac{ 4 }{ 3 }e^{t/5} - \frac{ 1 }{ 3 }e^{4t/5}\]

Answer 7

since y(t) = 4/3e^(t/5) - 1/3e^(4t/5) do we plug in values?

Answer 8

i forgot to include c1 and c2 in the derivative the first time

Answer 9

so its y'(t) = 4/3e^(t/5) - 1/3e^(4t/5) or no

Answer 10

\[y_1 = \frac{ 4 }{ 3 }e^{t/5}\]\[y_2 =- \frac{ 1 }{ 3 }e^{4t/5}\]

Answer 11

it is, but with c1 and c2, so 4/3 c1 e^(t/5) etc

Answer 12

when I type those into webwork it says incorrect

Answer 13

oh wow my bad

Answer 14

y(t) is the whole thing. y_1 + y_2

the other question has different boundary conditions. just noticed that

Answer 15

y(t) being y1 on webwork

Answer 16

I need y1 and y2

Answer 17

so it should be your screenshot but replace c1 and c2

Answer 18

I have W(t)

Answer 19

I never got y1

Answer 20

y1 isnt:

\[\frac{ 4 }{ 3 }e^{t/5} - \frac{ 1 }{ 3 }e^{4t/5}\]?

Answer 21

it is whats y2 now?

Answer 22

try solving for c1 and c2 the way i did while i try it too

Answer 23

using the initial/boundary conditions

Answer 24

ok but I might get it wrong :(

Answer 25

i got it. ill give u chance to try though :)

Answer 26

so whats the first step?

Answer 27

y(0) = 0
\[y(0) = c_1 e^{0/5} + c_2 e^{4(0)/5} = 0 = c_1 + c_2\]
y'(0) = 1\[y'(0) = \frac{ 1 }{ 5 }c_1e^{0/5} + \frac{ 4 }{ 5 }c_2e^{4(0)/5} = 1 = \frac{ 1 }{ 5 }c_1 + \frac{ 4 }{ 5 }c_2\]

now that you have 2 equations, you can solve for 1 and 2

Answer 28

for c1 and c2**

Answer 29

y2(0) = 0 though

Answer 30

that's the first line. y(0) = 0

Answer 31

y'(0) = 1 do you plug the equations back in?

Answer 32

you plug it in y'(t) which is the derivative of y(t)

Answer 33

so you said 1/5 c1 + 4/5 c2?

Answer 34

= 1

Answer 35

c1 is -4/3 and c2 is -1/3

Answer 36

1/5e^(t/5)+4/5e^(4t/5) ?

Answer 37

not in this one.

we have\[c_1 + c_2 = 0\]so\[c_1 = -c_2\]
then we have\[\frac{ 1 }{ 5 }c_1 + \frac{ 4 }{ 5 }c_2 = 1\]
plugging in c1 = -c2:\[-\frac{ 1 }{ 5 }c_2 + \frac{ 4 }{ 5 }c_2 = 1\]leads to\[c_2 = \frac{ 5 }{ 3 }\]plug in c1 = -c2:\[c_1 = -\frac{ 5 }{ 3 }\]

Answer 38

so whats the final solution?

Answer 39

\[y_2 = -\frac{ 5 }{ 3 }e^{t/5} +\frac{ 5 }{ 3 }e^{4t/5}\]

Answer 40

ok I see, some work I just did not understand

Answer 41

\[y(t) = c_1e^{t/5} + c_2e^{4t/5}\]\[y'(t) = \frac{ 1 }{ 5 }c_1e^{t/5} + \frac{ 4 }{ 5 }c_2e^{4t/5}\]

we plugged in the boundary conditions y(0) and y'(0)

for the second one y(0) = 0

we plug that in y(t):
\[y(0) = 0 = c_1e^{0/5} + c_2e^{4(0)/5}\]since e^0 = 1,\[_1 + c_2 = 0\] same deal for y'(0) = 1

Answer 42

second to last line:
\[c_1 + c_2 = 0\]

Answer 43

thank you :)

Answer 44

glad i could help ^_^