Is this true about differential equations?

[gikiian]

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

 

[pasmith]

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

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

 

[gikiian]

But what if [itex]y_{p1}[/itex] and [itex]y_{p2}[/itex] are linearly dependent in the considered vector space?

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

 

[HallsofIvy]

Did you check for yourself that [itex]y_{P1}+ y_{P2}[/itex] satisfies the equation?

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

If [itex]y_{P1}[/itex] and [itex]y_{P2}[/itex] are NOT independent, then [itex]y_{p1}+ y_{P2}[/itex] would NOT be independent of either [itex]y_{P1}[/itex] or [itex]y_{P2}[/itex] (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 [itex]r^2- 2r+ 1= (r- 1)^2= 0[/itex] which has the single root r= 1. [itex]y= e^x[/itex] is a solution. [itex]y= 3e^x[/itex] is also a solution- though NOT an independent solution. But still [itex]e^x+ 3e^x= 4e^x[/itex] is a solution to the equation.)

 

[gikiian]

Did you check for yourself that [itex]y_{P1}+ y_{P2}[/itex] satisfies the equation?

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

If [itex]y_{P1}[/itex] and [itex]y_{P2}[/itex] are NOT independent, then [itex]y_{p1}+ y_{P2}[/itex] would NOT be independent of either [itex]y_{P1}[/itex] or [itex]y_{P2}[/itex] (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 [itex]y_p[/itex], involved in the general solution just by looking at [itex]y_c[/itex],[itex]y_{p1}[/itex] and [itex]y_{p2}[/itex]?