Find a 3rd degree differential equation

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I'm studying differential equations and in one question on a sheet the teacher gave us for practice I'm asked to find a 3rd degree linear differential equation. I know 2 particular solutions and the solution of the associated homogeneous solution.



My question is: how do I use the particular solutions, I can write a polynome using the homogeneous and another one using the particular ones. I dont know what to do from here. I have a resolution of this question, but none of my colleagues is able to explain it, they just memorized the process for the evaluation, I'm trying to understand it.
Theres a photo of the resolution.



Thank you



The resolution






ordinary-differential-equations







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$endgroup$








  • 1




    $begingroup$
    When you want an answer I suggest you to use mathjax here!
    $endgroup$
    – Fakemistake
    Dec 30 '18 at 19:06










  • $begingroup$
    Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation?
    $endgroup$
    – Arthur
    Dec 30 '18 at 19:16










  • $begingroup$
    You should only need one particular solution. Your solution to the homogeneous equation should have three constants in it that can be used to match the initial conditions. I can't read your photo, so cannot give details.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 19:25
















2












$begingroup$


I'm studying differential equations and in one question on a sheet the teacher gave us for practice I'm asked to find a 3rd degree linear differential equation. I know 2 particular solutions and the solution of the associated homogeneous solution.



My question is: how do I use the particular solutions, I can write a polynome using the homogeneous and another one using the particular ones. I dont know what to do from here. I have a resolution of this question, but none of my colleagues is able to explain it, they just memorized the process for the evaluation, I'm trying to understand it.
Theres a photo of the resolution.



Thank you



The resolution






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    When you want an answer I suggest you to use mathjax here!
    $endgroup$
    – Fakemistake
    Dec 30 '18 at 19:06










  • $begingroup$
    Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation?
    $endgroup$
    – Arthur
    Dec 30 '18 at 19:16










  • $begingroup$
    You should only need one particular solution. Your solution to the homogeneous equation should have three constants in it that can be used to match the initial conditions. I can't read your photo, so cannot give details.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 19:25














2












2








2





$begingroup$


I'm studying differential equations and in one question on a sheet the teacher gave us for practice I'm asked to find a 3rd degree linear differential equation. I know 2 particular solutions and the solution of the associated homogeneous solution.



My question is: how do I use the particular solutions, I can write a polynome using the homogeneous and another one using the particular ones. I dont know what to do from here. I have a resolution of this question, but none of my colleagues is able to explain it, they just memorized the process for the evaluation, I'm trying to understand it.
Theres a photo of the resolution.



Thank you



The resolution






ordinary-differential-equations







share|cite|improve this question











$endgroup$




I'm studying differential equations and in one question on a sheet the teacher gave us for practice I'm asked to find a 3rd degree linear differential equation. I know 2 particular solutions and the solution of the associated homogeneous solution.



My question is: how do I use the particular solutions, I can write a polynome using the homogeneous and another one using the particular ones. I dont know what to do from here. I have a resolution of this question, but none of my colleagues is able to explain it, they just memorized the process for the evaluation, I'm trying to understand it.
Theres a photo of the resolution.



Thank you



The resolution






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 19:13









Arthur

123k7122211




123k7122211










asked Dec 30 '18 at 19:04









fecfec

132




132








  • 1




    $begingroup$
    When you want an answer I suggest you to use mathjax here!
    $endgroup$
    – Fakemistake
    Dec 30 '18 at 19:06










  • $begingroup$
    Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation?
    $endgroup$
    – Arthur
    Dec 30 '18 at 19:16










  • $begingroup$
    You should only need one particular solution. Your solution to the homogeneous equation should have three constants in it that can be used to match the initial conditions. I can't read your photo, so cannot give details.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 19:25














  • 1




    $begingroup$
    When you want an answer I suggest you to use mathjax here!
    $endgroup$
    – Fakemistake
    Dec 30 '18 at 19:06










  • $begingroup$
    Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation?
    $endgroup$
    – Arthur
    Dec 30 '18 at 19:16










  • $begingroup$
    You should only need one particular solution. Your solution to the homogeneous equation should have three constants in it that can be used to match the initial conditions. I can't read your photo, so cannot give details.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 19:25








1




1




$begingroup$
When you want an answer I suggest you to use mathjax here!
$endgroup$
– Fakemistake
Dec 30 '18 at 19:06




$begingroup$
When you want an answer I suggest you to use mathjax here!
$endgroup$
– Fakemistake
Dec 30 '18 at 19:06












$begingroup$
Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation?
$endgroup$
– Arthur
Dec 30 '18 at 19:16




$begingroup$
Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation?
$endgroup$
– Arthur
Dec 30 '18 at 19:16












$begingroup$
You should only need one particular solution. Your solution to the homogeneous equation should have three constants in it that can be used to match the initial conditions. I can't read your photo, so cannot give details.
$endgroup$
– Ross Millikan
Dec 30 '18 at 19:25




$begingroup$
You should only need one particular solution. Your solution to the homogeneous equation should have three constants in it that can be used to match the initial conditions. I can't read your photo, so cannot give details.
$endgroup$
– Ross Millikan
Dec 30 '18 at 19:25










1 Answer
1






active

oldest

votes


















0












$begingroup$

The general $3^{rd}$ order linear differential equation with constant coefficients is $$y'''+ay''+by'+cy=d;a,b,cinBbb R$$



$y_1(x)=x+ln x,y_2(x)=ln x$ are solutions, which gives $frac2{x^3}-frac a{x^2}+frac bx +b+cx+cln x=d=frac2{x^3}-frac a{x^2}+frac bx +cln x$, which gives $cx+b=0forall xtherefore b=c=0$. The associated homogeneous equation is $y'''+ay''=0$ whose solution is $e^{2x} therefore a=-2$. The answer is $$y'''-2y''=frac2{x^3}+frac2{x^2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps
    $endgroup$
    – fec
    Dec 30 '18 at 20:34










  • $begingroup$
    How did you get the $1$ in $P(D)={1,x,e^{2x}}$?
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 4:35










  • $begingroup$
    Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x
    $endgroup$
    – fec
    Jan 2 at 17:53












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The general $3^{rd}$ order linear differential equation with constant coefficients is $$y'''+ay''+by'+cy=d;a,b,cinBbb R$$



$y_1(x)=x+ln x,y_2(x)=ln x$ are solutions, which gives $frac2{x^3}-frac a{x^2}+frac bx +b+cx+cln x=d=frac2{x^3}-frac a{x^2}+frac bx +cln x$, which gives $cx+b=0forall xtherefore b=c=0$. The associated homogeneous equation is $y'''+ay''=0$ whose solution is $e^{2x} therefore a=-2$. The answer is $$y'''-2y''=frac2{x^3}+frac2{x^2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps
    $endgroup$
    – fec
    Dec 30 '18 at 20:34










  • $begingroup$
    How did you get the $1$ in $P(D)={1,x,e^{2x}}$?
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 4:35










  • $begingroup$
    Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x
    $endgroup$
    – fec
    Jan 2 at 17:53
















0












$begingroup$

The general $3^{rd}$ order linear differential equation with constant coefficients is $$y'''+ay''+by'+cy=d;a,b,cinBbb R$$



$y_1(x)=x+ln x,y_2(x)=ln x$ are solutions, which gives $frac2{x^3}-frac a{x^2}+frac bx +b+cx+cln x=d=frac2{x^3}-frac a{x^2}+frac bx +cln x$, which gives $cx+b=0forall xtherefore b=c=0$. The associated homogeneous equation is $y'''+ay''=0$ whose solution is $e^{2x} therefore a=-2$. The answer is $$y'''-2y''=frac2{x^3}+frac2{x^2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps
    $endgroup$
    – fec
    Dec 30 '18 at 20:34










  • $begingroup$
    How did you get the $1$ in $P(D)={1,x,e^{2x}}$?
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 4:35










  • $begingroup$
    Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x
    $endgroup$
    – fec
    Jan 2 at 17:53














0












0








0





$begingroup$

The general $3^{rd}$ order linear differential equation with constant coefficients is $$y'''+ay''+by'+cy=d;a,b,cinBbb R$$



$y_1(x)=x+ln x,y_2(x)=ln x$ are solutions, which gives $frac2{x^3}-frac a{x^2}+frac bx +b+cx+cln x=d=frac2{x^3}-frac a{x^2}+frac bx +cln x$, which gives $cx+b=0forall xtherefore b=c=0$. The associated homogeneous equation is $y'''+ay''=0$ whose solution is $e^{2x} therefore a=-2$. The answer is $$y'''-2y''=frac2{x^3}+frac2{x^2}$$






share|cite|improve this answer











$endgroup$



The general $3^{rd}$ order linear differential equation with constant coefficients is $$y'''+ay''+by'+cy=d;a,b,cinBbb R$$



$y_1(x)=x+ln x,y_2(x)=ln x$ are solutions, which gives $frac2{x^3}-frac a{x^2}+frac bx +b+cx+cln x=d=frac2{x^3}-frac a{x^2}+frac bx +cln x$, which gives $cx+b=0forall xtherefore b=c=0$. The associated homogeneous equation is $y'''+ay''=0$ whose solution is $e^{2x} therefore a=-2$. The answer is $$y'''-2y''=frac2{x^3}+frac2{x^2}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 17:59

























answered Dec 30 '18 at 19:57









Shubham JohriShubham Johri

5,683918




5,683918












  • $begingroup$
    thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps
    $endgroup$
    – fec
    Dec 30 '18 at 20:34










  • $begingroup$
    How did you get the $1$ in $P(D)={1,x,e^{2x}}$?
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 4:35










  • $begingroup$
    Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x
    $endgroup$
    – fec
    Jan 2 at 17:53


















  • $begingroup$
    thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps
    $endgroup$
    – fec
    Dec 30 '18 at 20:34










  • $begingroup$
    How did you get the $1$ in $P(D)={1,x,e^{2x}}$?
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 4:35










  • $begingroup$
    Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x
    $endgroup$
    – fec
    Jan 2 at 17:53
















$begingroup$
thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps
$endgroup$
– fec
Dec 30 '18 at 20:34




$begingroup$
thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps
$endgroup$
– fec
Dec 30 '18 at 20:34












$begingroup$
How did you get the $1$ in $P(D)={1,x,e^{2x}}$?
$endgroup$
– Shubham Johri
Dec 31 '18 at 4:35




$begingroup$
How did you get the $1$ in $P(D)={1,x,e^{2x}}$?
$endgroup$
– Shubham Johri
Dec 31 '18 at 4:35












$begingroup$
Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x
$endgroup$
– fec
Jan 2 at 17:53




$begingroup$
Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x
$endgroup$
– fec
Jan 2 at 17:53


















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