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

[komarxian]

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?

 

[Mark44]

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.

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.

 

[Ray Vickson]

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]

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!

 

[komarxian]

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.

 

[Ray Vickson]

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 $\beta \doteq 1.546868888.$

 

[Mark44]

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.