The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

ODE's problem.

Bit stuck on this question.

Q) Solve the equation,

y′=e2t+1 y'= e^{2t} + 1 subject to y(1)=1.y(1) = 1.


I assume you need to intergrate to get y=12e2t+t+c y = \frac{1}{2} e^{2t} + t + c, then plug in the conditions.


However, here I'm stuck? How do I get the final solution to be,
y=12[e2t−e2]+ty = \frac{1}{2}\left[e^{2t} - e^2 \right] + t
(edited 10 years ago)
Reply 1
Avatar for Notnek
Notnek
21
Original post by BAD AT MATHS
Bit stuck on this question.

Q) Solve the equation,

y′=e2t+1 y'= e^{2t} + 1 subject to y(1)=1.y(1) = 1.


I assume you need to intergrate to get y=12e2t+t+c y = \frac{1}{2} e^{2t} + t + c, then plug in the conditions.


However, here I'm stuck? How do I get the final solution to be,
y=12[e2t−e2]+ty = \frac{1}{2}\left[e^{2t} - e^2 \right] + t

Can you post any working you have where you've tried to plug in the initial conditions?
(edited 10 years ago)
Reply 2
Avatar for BAD AT MATHS
BAD AT MATHS
OP
12
Original post by notnek
Can you post any working you have where you've tried to plug in the initial conditions.


Yeah sure, I got when y(1)=1→y=12e2+1+cy(1) = 1 \rightarrow y = \frac{1}{2} e^2 + 1 + c. However I'm unsure if i'm tackling this in the right way.
Reply 3
Avatar for Notnek
Notnek
21
Original post by BAD AT MATHS
Yeah sure, I got when y(1)=1→y=12e2+1+cy(1) = 1 \rightarrow y = \frac{1}{2} e^2 + 1 + c. However I'm unsure if i'm tackling this in the right way.

y(1)=1y(1)=1 means, when t=1,y=1t=1, y=1. So you can change the left of your equation to 1:

1=12e2+1+c\displaystyle 1 = \frac{1}{2} e^2 + 1 + c

Can you solve this to find c?
Reply 4
Avatar for BAD AT MATHS
BAD AT MATHS
OP
12
Original post by notnek
y(1)=1y(1)=1 means, when t=1,y=1t=1, y=1. So you can change the left of your equation to 1:

1=12e2+1+c\displaystyle 1 = \frac{1}{2} e^2 + 1 + c

Can you solve this to find c?


Oh is it just
1=12e2+1+c⇒1−1=12e2+c\displaystyle 1 = \frac{1}{2} e^2 + 1 + c \Rightarrow 1-1 = \frac{1}{2} e^2 + c

therefore c=−12e2c = - \frac{1}{2} e^2 and hence final solution is
Unparseable latex formula:

y = \frac{1}{2}\left[e^{2t} - e^2] + t

(edited 10 years ago)
Reply 5
Avatar for BabyMaths
BabyMaths
Original post by BAD AT MATHS
Oh is it just
1=12e2+1+c⇒1−1=12e2+c\displaystyle 1 = \frac{1}{2} e^2 + 1 + c \Rightarrow 1-1 = \frac{1}{2} e^2 + c

therefore c=12e2c = \frac{1}{2} e^2 and hence final solution is
Unparseable latex formula:

y = \frac{1}{2}\left[e^{2t} - e^2] + t



You have c+e^2/2=0 so c=?
Reply 6
Avatar for BAD AT MATHS
BAD AT MATHS
OP
12
Original post by BabyMaths
You have c+e^2/2=0 so c=?


c=−12e2 c = -\frac{1}{2}e^2

Please say I'm correct. :redface:
(edited 10 years ago)
Reply 7
Avatar for BabyMaths
BabyMaths
Original post by BAD AT MATHS
c=−12e2 c = -\frac{1}{2}e^2

Please say I'm correct. :redface:


correct

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you