Therefore, y(t) = k1y1(t) + k2y2(t) is a solution to the given DE.

In summary, the question is asking to verify that a linear combination of two solutions of a second order linear equation is also a solution. With the given information, we can use the Linearity Principle to show that inserting the linear combination into the equation results in an equality. This can be done by solving each solution for y'' and plugging it into the other equation, resulting in 0=0.
  • #1
KevinL
37
0
I understand how to solve a normal second order linear equation, but this question in the homework is a bit more theoretical and I'm a bit confused.

"Suppose y1(t) and y2(t) are solutions of y'' + py' + qy = 0

Verify that y(t) = k1y1(t) + k2y2(t) is also a solution for any choice of constants k1 and k2."

If we were given actual functions (i.e. e^-t) this would be very simple. Just plug it into the DE and see that you have an equality. But when given arbitrary functions I don't know how to verify that its an equality. I am guessing it has something to do with the Linearity Principle, but I don't understand how to actually verify it.
 
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  • #2
Since it's homework, I probably shouldn't give the answer. But a hint...

Since y_1 and y_2 are solutions, then

y1'' + py1' + qy1 = 0
y2'' + py2' + qy2 = 0

How about plugging (k1y1 + k2y2) into the original equation and rearranging a bit?
 
  • #3
That got me started on the right track, I think. With the two equations you have, solve each one for y''. Then plug that into the other equation and everything nicely cancels so that 0=0.
 
  • #4
Sounds like you have it. You just need to show that inserting y into the equation will equal zero as well.

(k1y1 + k2y2)'' + p(k1y1 + k2y2)' + q(k1y1 + k2y2) = 0

k1y1'' + k2y2'' + pk1y1' + pk2y2' + qk1y1 + qk2y2 = 0

k1(y1'' + py1' + qy1) + k2(y2'' + py2' + qy2) = 0

k1*0 + k2*0 = 0 ==> 0 = 0
 

Related to Therefore, y(t) = k1y1(t) + k2y2(t) is a solution to the given DE.

1. What is a second order linear equation?

A second order linear equation is a mathematical equation that contains a second derivative term (y'') and can be written in the form of ay'' + by' + cy = 0. It is called a linear equation because the highest degree of the dependent variable is 1, and it is called a second order equation because it contains a second derivative term.

2. What is the general solution to a second order linear equation?

The general solution to a second order linear equation is a function that satisfies the equation for any values of the independent variable. It can be written in the form of y(x) = c1y1(x) + c2y2(x), where y1(x) and y2(x) are two linearly independent solutions and c1 and c2 are arbitrary constants.

3. What is the difference between homogeneous and non-homogeneous second order linear equations?

A homogeneous second order linear equation is one where the constant term is equal to 0, while a non-homogeneous equation has a non-zero constant term. The general solution for a homogeneous equation only contains the arbitrary constants c1 and c2, while the solution for a non-homogeneous equation also contains a particular solution that satisfies the non-zero constant term.

4. What is the characteristic equation and how is it used in solving second order linear equations?

The characteristic equation for a second order linear equation is obtained by setting the equation equal to 0 and solving for the roots of the polynomial. These roots are used to find the two linearly independent solutions, y1(x) and y2(x), which are then used in the general solution. The characteristic equation is also used to determine whether the solutions are real, complex, or repeated.

5. Can a second order linear equation have an infinite number of solutions?

No, a second order linear equation can only have two linearly independent solutions. However, if the equation is non-homogeneous, the general solution will also contain a particular solution, leading to an infinite number of solutions. Additionally, a second order linear equation can have a repeated solution, which is counted as one solution but can take on different forms.

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