# Differential Equations, solve the following: y^(4) - y'' - 2y' +2y = 0

• komarxian
I said that it's not useful, in general, to approach a differential equation by factoring its associated polynomial. This is because finding the roots of a polynomial is not always feasible, and even when it is, the solutions may not be easy to deal with in practice. In these cases, other methods such as numerical solutions are often preferred.

## Homework Statement

Solve the following differential equations/initial value problem:

y^(4) - y'' - 2y' +2y = 0 Hint: e^-x sinx is a solution

## Homework Equations

I was attempting to solve this problem by using a characteristic equation.

## The Attempt at a Solution

y'''' -y'' -2y' + 2y = 0 --> r^4 -r^2 -2r +2 = 0

r^2(r^2 - 1) -2(r - 1) = 0

r^2 (r+1)(r-1) -2(r-1) = 0

(r^2(r+1) -2)(r-1) = 0

from this I get r = 1, but I know there are more solutions; How do I solve for the other solutions? And then write it as the solution?

komarxian said:

## Homework Statement

Solve the following differential equations/initial value problem:

y^(4) - y'' - 2y' +2y = 0 Hint: e^-x sinx is a solution

## Homework Equations

I was attempting to solve this problem by using a characteristic equation.

## The Attempt at a Solution

y'''' -y'' -2y' + 2y = 0 --> r^4 -r^2 -2r +2 = 0

r^2(r^2 - 1) -2(r - 1) = 0
This isn't a useful approach, IMO, unless you are able to factor the cubic polynomial.
komarxian said:
r^2 (r+1)(r-1) -2(r-1) = 0

(r^2(r+1) -2)(r-1) = 0

from this I get r = 1, but I know there are more solutions; How do I solve for the other solutions? And then write it as the solution?
Use the given hint. If ##y = e^{-x}\sin(x)## is a solution, then another solution is ##y = e^{-x}\cos(x)## is also a solution. This means that ##r_1 = -1 + i## and ##r_2 = -1 - i## are roots of some quadratic characteristic equation.

komarxian
komarxian said:

## Homework Statement

Solve the following differential equations/initial value problem:

y^(4) - y'' - 2y' +2y = 0 Hint: e^-x sinx is a solution

## Homework Equations

I was attempting to solve this problem by using a characteristic equation.

## The Attempt at a Solution

y'''' -y'' -2y' + 2y = 0 --> r^4 -r^2 -2r +2 = 0

r^2(r^2 - 1) -2(r - 1) = 0

r^2 (r+1)(r-1) -2(r-1) = 0

(r^2(r+1) -2)(r-1) = 0

from this I get r = 1, but I know there are more solutions; How do I solve for the other solutions? And then write it as the solution?

You have ##p(r) \equiv r^2 - r^2 -2r+2 = (r-1)(r^3+r^2-2).## Notice that the polynomial ##q(r) = r^3+r^2 -2## has a root ##r = 1##, so ##(r-1)## is a factor. That gives ##q(r) = (r-1)(r^2+2r+2) ##, hence ##p(r) = (r-1)^2 (r^2+2r+2).## Thus, ##r=1## is a double root of ##p(r)## and the other roots are found by solving a simple quadratic equation.

komarxian
Mark44 said:
This isn't a useful approach, IMO, unless you are able to factor the cubic polynomial.

Use the given hint. If ##y = e^{-x}\sin(x)## is a solution, then another solution is ##y = e^{-x}\cos(x)## is also a solution. This means that ##r_1 = -1 + i## and ##r_2 = -1 - i## are roots of some quadratic characteristic equation.
Thank you!

Ray Vickson said:
You have ##p(r) \equiv r^2 - r^2 -2r+2 = (r-1)(r^3+r^2-2).## Notice that the polynomial ##q(r) = r^3+r^2 -2## has a root ##r = 1##, so ##(r-1)## is a factor. That gives ##q(r) = (r-1)(r^2+2r+2) ##, hence ##p(r) = (r-1)^2 (r^2+2r+2).## Thus, ##r=1## is a double root of ##p(r)## and the other roots are found by solving a simple quadratic equation.
Thank you! I get it now; got to work on the algebra XD
I've got 99 problems and it's all algebra pretty much.

Mark44 said:
This isn't a useful approach, IMO, unless you are able to factor the cubic polynomial.

Use the given hint. If ##y = e^{-x}\sin(x)## is a solution, then another solution is ##y = e^{-x}\cos(x)## is also a solution. This means that ##r_1 = -1 + i## and ##r_2 = -1 - i## are roots of some quadratic characteristic equation.

You say that solving the characteristic equation may not be useful. However, sometimes that's all you can do. For example, the differential equation ##y''' - 2 y''
+3 y' - 4 = 0## has characteristic equation ##r^3-2 r^2+3 r - 4 = 0##. This can be solve usingCardano's formulas (for example), but that is really not a useful way to go: numerical solutions are far preferable. That gives the solutions as ##y_1 = e^{rx},## ## y_2 = e^{\alpha x} \cos(\beta x)## and ##y_3 = e^{\alpha x} \sin(\beta x),## where ##r \doteq 1.650629192,## ##\alpha \doteq 0.1746854042## and ##\beta \doteq 1.546868888.##

jim mcnamara
Ray Vickson said:
You say that solving the characteristic equation may not be useful. However, sometimes that's all you can do.
I didn't say that solving the char. equation wasn't useful in this case, if the OP wasn't able to get other solutions out of the remaining cubic polynomial.