Solving Second Order Differential Equations: A Guide for Beginners

[-EquinoX-]

Homework Statement

How can I find a general solution to this problem:

5y" + 9y' - 9y = 0

Homework Equations

The Attempt at a Solution

I am confused with second order different equations in it

 

[CompuChip]

Usually with this type of equations (homogeneous ordinary differential equations) the easiest way is to plug in a trial solution $y(t) = e^{\lambda t}$, then solve the resulting polynomial for $\lambda$ and make a linear combination to get the general solution.

(For non-homogeneous equations, add to that a particular solution).

 

[-EquinoX-]

can't I just integrate twice and get the answer? how can I solve this using the method of undeterminant coefficients

 

[Mark44]

can't I just integrate twice and get the answer? how can I solve this using the method of undeterminant coefficients

No. You can integrate y'' twice, but how are you going to integrate y' and y twice?

CompuChip has given you some good advice. Try it out.

 

[-EquinoX-]

I don't get what he meant by make a linear combination

 

[Mark44]

Assume that the solution is of the form y = e^rt. Calculate the first and second derivatives of this function and plug them into your differential equation. After you factor out e^rt, you will have a quadratic that you can solve for r. (CompuChip used lambda, but the idea is the same.) You should get two values for r--r₁ and r₂, meaning that y₁ = e^r₁t and y₂ = e^r₂t are solutions of this DE. The general solution is all possible linear combinations of these two functions. IOW, y = c₁e^r₁t + c₂e^r₂t.

A linear combination of functions is the sum of multiples of those functions.

Clear?

 

[-EquinoX-]

what if it only has one root, say (y-3)^2, so the only root is y = 3

 

[Mark44]

Then your solutions would be e^3t and te^3t. Your general solution would be the linear combinations of these two functions.

IOW, y_g = c₁e^3t + c₂te^3t.

 

[-EquinoX-]

if it's y^2 + 25y , then the two roots are y = 0 and y = -25. so yg = c1 + c2e^(-25t).

 

[CompuChip]

Yes. You can understand why you only get one solution which is non-trivial (i.e. not equal to a constant number everywhere): if r = 0 is a solution to the polynomial, then that means that your equation must have been of the form
a y'' + b y' = 0
so without a term in y.
In that case you can view it as a (simpler) first order equation for v = y':
a v' + b v = 0
which of course only has one independent solution.

In general, if your equation is nth order (i.e. the highest derivative of y that occurs is the nth derivative) you get n independent solutions which you should add (indeed, when you plug in the exponential solution, you get a polynomial of degree n). You should be advised, however, that the roots of the polynomial are not always real, like for example in y'' + y = 0.

 

[-EquinoX-]

well if have initial value y(0) = 2 and y'(0)= 5

then what I get is

y' = -25c2e^(-25t)
5 = -25c2
so c2 = -1/5

and c1 is equal to 11/5

is the true? reason I asked is because the stupid webassign won't accept this answer

 

[Mark44]

well if have initial value y(0) = 2 and y'(0)= 5

then what I get is

y' = -25c2e^(-25t)
5 = -25c2
so c2 = -1/5

and c1 is equal to 11/5

is the true? reason I asked is because the stupid webassign won't accept this answer

Are you talking about this initial value problem? (This is a different problem from the one you first posted in this thread.)
y'' + 25y = 0, y(0) = 2, y'(0) = 5

If so, your solution looks fine to me. Maybe the webassign thing is looking for decimal values instead of fractions.

 

[-EquinoX-]

Are you talking about this initial value problem? (This is a different problem from the one you first posted in this thread.)
y'' + 25y = 0, y(0) = 2, y'(0) = 5

If so, your solution looks fine to me. Maybe the webassign thing is looking for decimal values instead of fractions.

yes this is a different problem, thanks though for confirming and all the help