Differential Equation Question

[Calu]

I have a differential equation

y'' + y' -2y = 3e^-2x + 5cosx

y = y_c + y_p

I found

y_c = Ae^-2x + Be^x

for A, B arb. Const.

Then when selecting a trial function to find the particular integral, y_p I came up with:

y_p = ae^-2x + bcosx + csinx

However the correct trial function to choose is:

y_p = axe^-2x + bcosx + csinx

Where the ae^-2x has been multiplied by x. I was wondering in what instance I would have to multiply the term in the y_p equation by x or x².

 

[LCKurtz]

I have a differential equation

y'' + y' -2y = 3e^-2x + 5cosx

y = y_c + y_p

I found

y_c = Ae^-2x + Be^x

for A, B arb. Const.

Then when selecting a trial function to find the particular integral, y_p I came up with:

y_p = ae^-2x + bcosx + csinx

However the correct trial function to choose is:

y_p = axe^-2x + bcosx + csinx

Where the ae^-2x has been multiplied by x. I was wondering in what instance I would have to multiply the term in the y_p equation by x or x².

When you plug in your expression for $y_p$, you know you need to get out a $3e^{-2x}$ term. But if you plug $ae^{-2x}$ into the equation, you know you will get zero because it is in $y_c$. That's when you want to try a term $y_p=axe^{-2x}$. There's more to that story but that's the short answer.

 

[STEMucator]

You want the terms in $y_p$ to be linearly independent of the terms in $y_c$. Note that $f(x) = 3e^{-2x} +5cos(x)$.

Your original guess for $y_p$ includes a multiple of one of your homogeneous solutions, namely the $Ae^{-2x}$ term, so it cannot be linearly independent of $y_c$. Multiplying by $x$ is a quick shortcut to obtaining a linearly independent $y_p$ guess.

This method was developed by D'alembert, you can find out more by checking out "reduction order".

 

[Calu]

When you plug in your expression for $y_p$, you know you need to get out a $3e^{-2x}$ term. But if you plug $ae^{-2x}$ into the equation, you know you will get zero because it is in $y_c$. That's when you want to try a term $y_p=axe^{-2x}$. There's more to that story but that's the short answer.

You want the terms in $y_p$ to be linearly independent of the terms in $y_c$. Note that $f(x) = 3e^{-2x} +5cos(x)$.

Your original guess for $y_p$ includes a multiple of one of your homogeneous solutions, namely the $Ae^{-2x}$ term, so it cannot be linearly independent of $y_c$. Multiplying by $x$ is a quick shortcut to obtaining a linearly independent $y_p$ guess.

This method was developed by D'alembert, you can find out more by checking out "reduction order".

These were both great answers! Thanks a lot!