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
<br /> a_3(x) (y_{p1} + y_{p2})&#039;&#039;&#039; + a_2(x) (y_{p1} + y_{p2})&#039;&#039; + a_1(x) (y_{p1} + y_{p2})&#039;<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}?