Determining “trial” solution for particular integral












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In the DE $x^2y''+xy'-4y=15x^3$, supposedly the particular integral can be found by attempting $y=Cx^3$ and then substituting. However it's not clear to me why this is. I would have thought the most logical starting point would be an arbitrary polynomial of degree $3$, i.e. $Ax^3+Bx^2+Cx+D$. How does one know that the true solution will only be of the form $Cx^3$?










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  • $begingroup$
    Will $A$ have a chance at influencing $B$? Are any of $A,B,C,D$ coupled after inserting the full polynomial?
    $endgroup$
    – LutzL
    Jan 8 at 21:59










  • $begingroup$
    May I confess that I would have done the same as you ?
    $endgroup$
    – Claude Leibovici
    Jan 9 at 6:20
















0












$begingroup$


In the DE $x^2y''+xy'-4y=15x^3$, supposedly the particular integral can be found by attempting $y=Cx^3$ and then substituting. However it's not clear to me why this is. I would have thought the most logical starting point would be an arbitrary polynomial of degree $3$, i.e. $Ax^3+Bx^2+Cx+D$. How does one know that the true solution will only be of the form $Cx^3$?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Will $A$ have a chance at influencing $B$? Are any of $A,B,C,D$ coupled after inserting the full polynomial?
    $endgroup$
    – LutzL
    Jan 8 at 21:59










  • $begingroup$
    May I confess that I would have done the same as you ?
    $endgroup$
    – Claude Leibovici
    Jan 9 at 6:20














0












0








0





$begingroup$


In the DE $x^2y''+xy'-4y=15x^3$, supposedly the particular integral can be found by attempting $y=Cx^3$ and then substituting. However it's not clear to me why this is. I would have thought the most logical starting point would be an arbitrary polynomial of degree $3$, i.e. $Ax^3+Bx^2+Cx+D$. How does one know that the true solution will only be of the form $Cx^3$?










share|cite|improve this question









$endgroup$




In the DE $x^2y''+xy'-4y=15x^3$, supposedly the particular integral can be found by attempting $y=Cx^3$ and then substituting. However it's not clear to me why this is. I would have thought the most logical starting point would be an arbitrary polynomial of degree $3$, i.e. $Ax^3+Bx^2+Cx+D$. How does one know that the true solution will only be of the form $Cx^3$?







ordinary-differential-equations






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asked Jan 8 at 21:53









cb7cb7

1036




1036












  • $begingroup$
    Will $A$ have a chance at influencing $B$? Are any of $A,B,C,D$ coupled after inserting the full polynomial?
    $endgroup$
    – LutzL
    Jan 8 at 21:59










  • $begingroup$
    May I confess that I would have done the same as you ?
    $endgroup$
    – Claude Leibovici
    Jan 9 at 6:20


















  • $begingroup$
    Will $A$ have a chance at influencing $B$? Are any of $A,B,C,D$ coupled after inserting the full polynomial?
    $endgroup$
    – LutzL
    Jan 8 at 21:59










  • $begingroup$
    May I confess that I would have done the same as you ?
    $endgroup$
    – Claude Leibovici
    Jan 9 at 6:20
















$begingroup$
Will $A$ have a chance at influencing $B$? Are any of $A,B,C,D$ coupled after inserting the full polynomial?
$endgroup$
– LutzL
Jan 8 at 21:59




$begingroup$
Will $A$ have a chance at influencing $B$? Are any of $A,B,C,D$ coupled after inserting the full polynomial?
$endgroup$
– LutzL
Jan 8 at 21:59












$begingroup$
May I confess that I would have done the same as you ?
$endgroup$
– Claude Leibovici
Jan 9 at 6:20




$begingroup$
May I confess that I would have done the same as you ?
$endgroup$
– Claude Leibovici
Jan 9 at 6:20










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