Is this true about differential equations?

[gikiian]

If $a_3(x)y'''+a_2(x) y''+a_1(x) y'+a_0(x)y=f(x)$ is an ODE with particular solution $y_{p1}$
and $a_3(x)y'''+a_2(x) y''+a_1(x) y'+a_0(x)y=g(x)$ is an ODE with particular solution $y_{p2}$,
then the ODE $a_3(x)y'''+a_2(x) y''+a_1(x) y'+a_0(x)y=f(x)+g(x)$ has the particular solution $y_{p1}+y_{p2}$.

 

[pasmith]

If $a_3(x)y'''+a_2(x) y''+a_1(x) y'+a_0(x)y=f(x)$ is an ODE with particular solution $y_1$
and $a_3(x)y'''+a_2(x) y''+a_1(x) y'+a_0(x)y=g(x)$ is an ODE with particular solution $y_2$,
then the ODE $a_3(x)y'''+a_2(x) y''+a_1(x) y'+a_0(x)y=f(x)+g(x)$ has the particular solution $y_{p1}+y_{p2}$.

If that were true, then you must have
$$a_3(x) (y_{p1} + y_{p2})''' + a_2(x) (y_{p1} + y_{p2})'' + a_1(x) (y_{p1} + y_{p2})'<br /> + a_0(x) (y_{p1} + y_{p2}) = f(x) + g(x).$$
Is that the case? Check for yourself.

 

[gikiian]

But what if $y_{p1}$ and $y_{p2}$ are linearly dependent in the considered vector space?

Will the particular solution to the third equation still be $y_{p1}+y_{p2}$, or will it more be like $y_{p1}+xy_{p2}$?

 

[HallsofIvy]

Did you check for yourself that $y_{P1}+ y_{P2}$ satisfies the equation?

You are confusing "satisfies the equation" with "is an independent solution to the equation".

If $y_{P1}$ and $y_{P2}$ are NOT independent, then $y_{p1}+ y_{P2}$ would NOT be independent of either $y_{P1}$ or $y_{P2}$ (so we could not use it to construct a "general solution") but it would be a solution.

(There is nothing special about fact that the given example is non-homogeneous. The characteristic equation of the differential y''- 2y'+ y= 0 is $r^2- 2r+ 1= (r- 1)^2= 0$ which has the single root r= 1. $y= e^x$ is a solution. $y= 3e^x$ is also a solution- though NOT an independent solution. But still $e^x+ 3e^x= 4e^x$ is a solution to the equation.)

 

[gikiian]

Did you check for yourself that $y_{P1}+ y_{P2}$ satisfies the equation?

You are confusing "satisfies the equation" with "is an independent solution to the equation".

If $y_{P1}$ and $y_{P2}$ are NOT independent, then $y_{p1}+ y_{P2}$ would NOT be independent of either $y_{P1}$ or $y_{P2}$ (so we could not use it to construct a "general solution") but it would be a solution.

I get the point! But can we predict the particular solution, say $y_p$, involved in the general solution just by looking at $y_c$,$y_{p1}$ and $y_{p2}$?