# General solution to 2nd order DE

Hi, im just looking for clarification whether or not im doing these problems correctly, heres an example:

## Homework Statement

Find the general solution to
x'' -2x' + 5x = 0

## Homework Equations

Charasteristic polynomial.
General solution form for complex roots.

## The Attempt at a Solution

form the char polynomial:
m^2 -2m +5 = 0

Solve for roots:
(-2 (+-)SQRT(2^2 - 4(1)(5)))/2
-1+SQRT(-16)
-1-SQRT(-16)
m1 = -1+4i
m2 = -1-4i

Insert into general solution form:

y = (e^ax)(Acoswx + Bsinwx)
y = (e^-1x)(Acos4ix + Bsin4ix)

General solution: (?)
y = (e^-x)(Acos4ix + Bsin4ix)

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Mark44
Mentor
Hi, im just looking for clarification whether or not im doing these problems correctly, heres an example:

## Homework Statement

Find the general solution to
x'' -2x' + 5x = 0

## Homework Equations

Charasteristic polynomial.
General solution form for complex roots.

## The Attempt at a Solution

form the char polynomial:
m^2 -2m +5 = 0

Solve for roots:
(-2 (+-)SQRT(2^2 - 4(1)(5)))/2
-1+SQRT(-16)
-1-SQRT(-16)
m1 = -1+4i
m2 = -1-4i

Insert into general solution form:

y = (e^ax)(Acoswx + Bsinwx)
y = (e^-1x)(Acos4ix + Bsin4ix)

General solution: (?)
y = (e^-x)(Acos4ix + Bsin4ix)
Almost!
y = e-x(Acos(4x) + Bsin(4x))

You can verify that this is the solution by calculating y' and y'' and substituting them into your differential equation.

Nice, thanks!

What do I do if im asked to write a solution that satisfies the conditions
x(0) = 1
x´(0) =0

Mark44
Mentor
Your solution is actually x(t) = e-t(Acos(4t) + Bsin(4t)). I hadn't noticed earlier that you had switched your solution to be y as a function of x.

Use your initial conditions to substitute x = 1 and t = 0 into your solution, and x' = 0 and t = 1 into the derivative of your solution. That will give you two equations in two unknowns, so you should be able to solve for the constants A and B.

Cool, I got it now, thx.