Question

Question regarding first-order linear ordinary differential equation. How do i proceed? for the equation 3y'-2y = e^((-pi*t)/2) where y(0)=a find the critical value a0 where a0 is the value for which the transition fro one type of behavior to another occurs. The book doesn't explain it at all.

Answer 1

Here's how it's written in the book.

a) Draw a direction field for the given differential equation. How do the solutions appear to behave as t becomes large? Does the behavior depend on the choice of initial value a? Let a0 be the value of a for which the transition from one type of behavior to another occurs. Estimate the value of a0. (I got this partially done except the a0 part)
(b) Solve the initial value problem (I am pretty sure i got this part right) and find the critical value of a0 exactly (now here's the problem, the book has never explained how to do this)
(c) Describe the behavior of the solution corresponding to the initial value a0

for the equation: \[3y'-2y = e^{-\frac{ -\pi*t }{ 2 }}\]

now, the general solution is \[y=\frac{ 2e ^{\frac{ \pi*t }{ 2 }} }{ 3\pi-4 }+c* e ^{\frac{ 2t }{ 3 }}\] (i figured that out myself)

from there we can see that for y(0) \[y(0)=c+\frac{ 2 }{ 3\pi-4 }\]

the answer in the textbook states \[a0=\frac{ -2 }{ 3\pi+4 }\]

they also have a different general solution\[y=-3e ^{\frac{ t }{ 3 }}+\left( a+3 \right)e ^{\frac{ t }{ 2 }}\]

now how did they arrive at these conclussions?

Answer 2

oops\[\frac{ 2+a \left( 3\pi+4 \right) e ^{\frac{ 2t }{ 3 }}-2e ^{\frac{ -\pi*t }{ 2 }}}{ 3\pi+4 }\] the last equation of my previous post is incomplete it should be

Answer 3

I got something different from your general solution.

Answer 4

Your general solution has a few wrong signs I believe.

Answer 5

I solved it i screwed up with a - somewhere apperantly. Thanks

Answer 6

\[y'-\frac{2}{3}y=\frac{1}{3}e^{ -\pi t /2}\]
Multiply both sides be the integrating factor
The integrating factor is
\[e^{\int\limits\limits \frac{-2}{3} dt}=e^{\frac{-2}{3} t+k} (note: \text{ pick k=0} ) = e^{\frac{-2}{3}t} \]
\[e^{\frac{-2}{3}t}y'-e^{\frac{-2}{3}t}y=\frac{1}{3}e^{\frac{-2}{3}t}e^{\frac{- \pi t}{2}} \]
\[(e^{\frac{-2}{3}t} y)'=\frac{1}{3}e^{(\frac{-2}{3}-\frac{\pi}{2})t} \]

Answer 7

But like I said okay. Good job you. :p

Answer 8

it's fine to leave the replies for future reference and others who might get the same or similar problem. none of my classmates could help me either! (and they are supposed to know it by now, we have a test in less than a week).

Good to know there is at least 1 person on this website who has a good understanding of differential equations so that when i am stuck i can ask someone to give me a hint in the right direction.

Answer 9

There are others on here who have great knowledge in differential equations.

Answer 10

Also I will try my best. My memory in differential equations is best in solving differential equations be separating variables or using in integrating factor when it comes to solving linear differential equations.