solve `dy/dt= -8t-8ty` when with y(0)=14

Calculus 1 and 2 questions with single variable are posted here.

Moderator: ChangBroot

Forum rules
To type int use \`int\` and the same way to type sqrt simply type \`sqrt\` or Please view the examples forum.

As you type, live preview is displayed so you know exactly what are you typing.

solve `dy/dt= -8t-8ty` when with y(0)=14

Postby barnamah » Thu 31 March, 2011 11:50

please solve for y in `dy/dt= -8t-8ty` when with y(0)=14
Explore and know. That is asked.
User avatar
barnamah
 
Posts: 102
Joined: Mon 14 March, 2011 08:37
Location: Canada

Re: solve `dy/dt= -8t-8ty` when with y(0)=14

Postby Integr88 » Thu 31 March, 2011 12:12

we have `dy/dt= -8t-8ty`

factor out the t from the right side and we have:
`dy/dt= -8t(1+y)`

Now multiply both side by `1/(-8t)` and we have:
`1/(1+y)*dy/dt= -8t(1+y)*1/(1+y)`

the 1+y one right cancels out and we have:

`1/(1+y)*dy/dt= -8t` which can be written as `dy/(1+y)*1/dt= -8t`

Now our mission is to move `1/dt` to his friend -8t on the right and we can accomplish that by multiplying both side by dt and have:
`dy/(1+y)*1/dt*dt= -8t*dt`

the dt on the left cancels out and we have:
`dy/(1+y)= -8t*dt`

Now we can integrate :
`intdy/(1+y)=int -8t*dt`

1-The integrate the right side `int -8t*dt`and we have
`-8*t^(1+1)/(1+1)=(-8t^2)/2=-4t^2` plus a constant

2-The integrate the left side `intdy/(1+y)` we can rewrite it as:
`intdy/(1+y)=ln(1+y)` plus a constant

Putting them together and we have:
`ln(1+y)=-4t^2+C` where C is a constant

our aim is to find y so we write:
`e^(ln(1+y))=e^(-4t^2+C)` one the left `e^ln(1+y)` cancel each other and we left with 1+y. so we write it as:

`|1+y|=e^(-4t^2+C)` we rewrite the right side as
`|1+y|=e^(-4t^2)e^C`
We wrote `e^(-4t^2+C)` as `e^(-4t^2)e^C` is due to the fact that `a^5*a^6=a^(5+6)` here we just did the reverse.


`|1+y|=e^(-4t^2)e^C` to get rid of the absolute sign we can add `+-` to the right and we have:

`1+y=+-e^(-4t^2)e^C`
Because have `(+-)(e^C)e^(-4t^2)` and all of them are multiplied to a `e^C` which is constant we can simplify the `+-e^5=` an arbitrary constant.
we have now:
`1+y=me^(-4t^2)`

subtract 1 from both side and have:
`y=me^(-4t^2)-1`

At this point we got the y.
the question say solve with initial value y(0)=14 which means when t=0, y=14 so we sub 0 and 14 in our equation:
`14=me^(-40^2)-1=me^0-1=m(1)-1` so
14=m-1 `:.` m=15.

Now can write our full solution as `y=(15)e^(-4t^2)=15e^(-4t^2)`

`:.` the final solution`y=15e^(-4t^2)`
User avatar
Integr88
 
Posts: 71
Joined: Tue 15 March, 2011 13:27


Return to Calculus single variable

Who is online

Users browsing this forum: No registered users and 1 guest

cron