Question

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. Explain your reasoning. Compare your answers to the previous problem. Recall that the general solution has the form y(t) = yh(t) + yp(t).

(A) y' - 2y = 0
(B) y' - 2y = 1
(C) y" - 4y = 0
(D) y" - 4y = e^(3t)

Answer 1

(A)
`y=ke^(2t)` with
`k\in\mathbb{R}`.

(B)
`y=ke^(2t)/2-1/2` with
`k\in\mathbb{R}`

(C)
`y=k_1e^(2t)+k_2e^(-2t)` with
`k_1,k_2\in\mathbb{R}`

(D)
`y=k_1e^(2t)+k_2e^(-2t)+e^(3t)/5` with
`k_1,k_2\in\mathbb{R}`,

Explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then
`0=y'-2y=(dy)/(dt)-2y\Rightarrow (dy)/(dt)=2y \Rightarrow  (1)/(2y)dy=dt \ \Rightarrow \int (1)/(2y)dy=\int dt \Rightarrow \ln |y|^(1/2)=t+C \Rightarrow |y|^(1/2)=e^y=e^(t+C)=e^(C)e^t} \Rightarrow y=ke^(2t)`. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form
`e^(2t)` with
`t` real.

(B) Proceeding and the previous item, we obtain
`1=y'-2y=(dy)/(dt)-2y\Rightarrow (dy)/(dt)=2y+1 \Rightarrow  (1)/(2y+1)dy=dt \ \Rightarrow \int (1)/(2y+1)dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^(1/2)=e^^(1/2)=e^(t+C)=e^(C)e^t \Rightarrow y=ke^(2t)/2-1/2`. Which is not a vector space with the usual operations (this is because
`-1/2`), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set
`y=e^(mt) \Rightarrow y''=m^2e^(mt)` and we obtain the characteristic equation
`0=y''-4y=m^2e^(mt)-4e^(mt)=(m^2-4)e^(mt)\Rightarrow m^(2)-4=0\Rightarrow m=\pm 2` and then the general solution is
`y=k_1e^(2t)+k_2e^(-2t)` with
`k_1,k_2\in\mathbb{R}`, and as in the first items the set of solutions form a vector space.

(D) Using C, let be
`y=me^(3t)` we obtain that it must satisfies
`3^2m-4m=1\Rightarrow m=1/5` and then the general solution is
`y=k_1e^(2t)+k_2e^(-2t)+e^(3t)/5` with
`k_1,k_2\in\mathbb{R}`, and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).