# Thread: First order differential equations

1. ## First order differential equations

Consider the first order differential equation:

d^2y/dx^2 = -((lambda + 1)^2)y

Write down the general solution.
Find the particular solution satisfying dy/dx = 0, y = 1 when x = 0.

I didnt think this was first order, but maybe if you integrate first. I couldnt get an answer that looked right.

2. Originally Posted by BogStandard
Consider the first order differential equation:

d^2y/dx^2 = -((lambda + 1)^2)y

Write down the general solution.
Find the particular solution satisfying dy/dx = 0, y = 1 when x = 0.

I didnt think this was first order, but maybe if you integrate first. I couldnt get an answer that looked right.
this is a second order differential equation with constant coefficients. we proceed by assuming a solution of $\displaystyle y = e^{rx}$ and then forming the characteristic equation. which in this case is:

$\displaystyle r^2 + ( \lambda + 1)^2 = 0$ (do you see why?)

now solve for the two roots of this equation, $\displaystyle r_1$ and $\displaystyle r_2$. your general solution is of the form $\displaystyle y(x) = Ae^{r_1t} + Be^{r_2t}$

for the particular solution (we want to know what A and B are): you know that $\displaystyle y'(0) = 0$ and $\displaystyle y(0) = 1$, we can use these equations to solve for the (constant) unknowns $\displaystyle A$ and $\displaystyle B$ to obtain our particular solution

can you continue?

3. I would normally be able to solve this for a first order, but for second order im not sure.

This is an exam question, and the paper says its first order, which has completely thrown me off.

So im afraid I dont understand your method!

4. Originally Posted by BogStandard
I would normally be able to solve this for a first order, but for second order im not sure.

This is an exam question, and the paper says its first order, which has completely thrown me off.

So im afraid I dont understand your method!
there is something amiss here. $\displaystyle \frac {d^2y}{dx^2}$ means it is second order, because that is the second derivative. please type the question exactly as stated

5. Originally Posted by Jhevon
there is something amiss here. $\displaystyle \frac {d^2y}{dx^2}$ means it is second order, because that is the second derivative. please type the question exactly as stated
Yeah I know what you're thinking, I thought it too, but that's literally what the question says.
It's most probably a typo or something, or some really unobvious trick question. Pretty bad for an exam paper.

So could you explain to me the same method but for second order differential equations?

6. Originally Posted by BogStandard
Yeah I know what you're thinking, I thought it too, but that's literally what the question says.
It's most probably a typo or something, or some really unobvious trick question. Pretty bad for an exam paper.

So could you explain to me the same method but for second order differential equations?
my explanation was for second order. first-orders are different. to be first order, you would have to have dy/dx not d^2y/dx^2. if that was what they meant though, they would not have given us y'(0) and y(0). only y(k) is required for first order. so something is diffinately off here. i suspect the only typo is "first order", maybe it should really be a second order DE, in which case you follow the method i gave you.

if you have a first order DE of the form $\displaystyle y' = ky$, where $\displaystyle k$ is a constant, you proceed as follows

divide through by $\displaystyle y$:
$\displaystyle \frac {y'}y = k$

integrate both sides:
$\displaystyle \ln y = kx + C$

solve for y:
$\displaystyle y = e^{kx + C}$

$\displaystyle \Rightarrow y = Ae^{kx}$ ....$\displaystyle A$ is a constant, you can find it if you are given intial data, say y(0) = ...