Setting up particular solution for nonhomogenous diff eq'n

[Rosebud]

Given the nonhomogenous differential equation y'' + 3y' + 2y = -10e^(3t), the roots are r = -2 & -1, & the characteristic eq'n is yc(x) = c1e^(-2t) + c2e^(-t)

How do we go about setting up the particular solution?

There is no repetition between terms so I know that we do not add a variable to the particular solution. Since we have -10e^3t do we set up the particular solution as yp(t) = -Ae^3t? OR just yp(t) = Ae^3t? I know that A just represents a constant but do I need to include the negative sign or not?

Thank you.

 

[LCKurtz]

Given the nonhomogenous differential equation y'' + 3y' + 2y = -10e^(3t), the roots are r = -2 & -1, & the characteristic eq'n is yc(x) = c1e^(-2t) + c2e^(-t)

How do we go about setting up the particular solution?

There is no repetition between terms so I know that we do not add a variable to the particular solution. Since we have -10e^3t do we set up the particular solution as yp(t) = -Ae^3t? OR just yp(t) = Ae^3t? I know that A just represents a constant but do I need to include the negative sign or not?

Thank you.

It doesn't matter whether you use the $-A$ or $A$. Try them.

 

[Mark44]

Given the nonhomogenous differential equation y'' + 3y' + 2y = -10e^(3t), the roots are r = -2 & -1, & the characteristic eq'n is yc(x) = c1e^(-2t) + c2e^(-t)

How do we go about setting up the particular solution?

There is no repetition between terms so I know that we do not add a variable to the particular solution. Since we have -10e^3t do we set up the particular solution as yp(t) = -Ae^3t? OR just yp(t) = Ae^3t? I know that A just represents a constant but do I need to include the negative sign or not?

Thank you.

In future posts, please do not delete the homework template.

 

[Rosebud]

It doesn't matter whether you use the $-A$ or $A$. Try them.

When setting up the particular solution as yp(t) = Ae^(3t), yp'(t) = 3Ae^(3t) and yp''(t) = 9Ae^(3t). After substituting these back into the original diff eq'n we get A = -1/2.

However, when setting up the particular sol'n as yp(t) = -Ae^(3t), yp'(t) = -3Ae^(3t) and yp''(t) = -9Ae^(3t). After substituting these back into the original diff eq'n we get A = 1/2.

So, I do believe there is a difference because -1/2 is not equal to 1/2.

Which substitution should I use, A or -A?

 

[Rosebud]

In future posts, please do not delete the homework template.

You mean those 1.

2.

3.

things?

 

[Mark44]

You mean those 1.

2.

3.

things?

Yes, and the sections for Problem Statement, Relevant Equations, and Attempt. Per forum rules, the homework template is required for homework problems.

 

[Rosebud]

It doesn't matter whether you use the $-A$ or $A$. Try them.

Oh, I get it now. -A = 1/2 and A = 1/2 so it doesn't matter... derp.

Thank you so much!

 

[Rosebud]

Yes, and the sections for Problem Statement, Relevant Equations, and Attempt. Per forum rules, the homework template is required for homework problems.

Sounds good, I will use them from now on.

 

[Mark44]

When setting up the particular solution as yp(t) = Ae^(3t), yp'(t) = 3Ae^(3t) and yp''(t) = 9Ae^(3t). After substituting these back into the original diff eq'n we get A = -1/2.

However, when setting up the particular sol'n as yp(t) = -Ae^(3t), yp'(t) = -3Ae^(3t) and yp''(t) = -9Ae^(3t). After substituting these back into the original diff eq'n we get A = 1/2.

So, I do believe there is a difference because -1/2 is not equal to 1/2.

Which substitution should I use, A or -A?

Sounds good, I will use them from now on.

OK, good. We appreciate it.

I should mention that there's no point in writing -Ae^3t, as A could be positive or negative. All that will come out in the wash when you solve for A.