True or false? If $y_1$ and $y_2$ are solutions to $y^{primeprime} + x^2y = 1$, then $y_1 + y_2$ is also a...












-1












$begingroup$



Prove if this is true or false:
If $y_1$ and $y_2$ are solutions to the differential equation
$$y^{primeprime} + x^2y = 1$$
then $y_1 + y_2$ is also a solution to this equation.




How do I show if this is true or not? Please help me! :)






ordinary-differential-equations







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closed as off-topic by Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers Dec 7 '18 at 0:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Do you know what a linear differential equation is and what the elementary consequences of "linear" for the solution space are?
    $endgroup$
    – LutzL
    Dec 6 '18 at 21:09










  • $begingroup$
    No i dont really know
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:11










  • $begingroup$
    Is it true or not this question?
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:16






  • 1




    $begingroup$
    While $y_1+y_2$ is not a solution, $ty_1+(1-t)y_2$ is a solution for any $t$.
    $endgroup$
    – user614671
    Dec 6 '18 at 21:25
















-1












$begingroup$



Prove if this is true or false:
If $y_1$ and $y_2$ are solutions to the differential equation
$$y^{primeprime} + x^2y = 1$$
then $y_1 + y_2$ is also a solution to this equation.




How do I show if this is true or not? Please help me! :)






ordinary-differential-equations







share|cite|improve this question











$endgroup$



closed as off-topic by Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers Dec 7 '18 at 0:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Do you know what a linear differential equation is and what the elementary consequences of "linear" for the solution space are?
    $endgroup$
    – LutzL
    Dec 6 '18 at 21:09










  • $begingroup$
    No i dont really know
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:11










  • $begingroup$
    Is it true or not this question?
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:16






  • 1




    $begingroup$
    While $y_1+y_2$ is not a solution, $ty_1+(1-t)y_2$ is a solution for any $t$.
    $endgroup$
    – user614671
    Dec 6 '18 at 21:25














-1












-1








-1





$begingroup$



Prove if this is true or false:
If $y_1$ and $y_2$ are solutions to the differential equation
$$y^{primeprime} + x^2y = 1$$
then $y_1 + y_2$ is also a solution to this equation.




How do I show if this is true or not? Please help me! :)






ordinary-differential-equations







share|cite|improve this question











$endgroup$





Prove if this is true or false:
If $y_1$ and $y_2$ are solutions to the differential equation
$$y^{primeprime} + x^2y = 1$$
then $y_1 + y_2$ is also a solution to this equation.




How do I show if this is true or not? Please help me! :)






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 6 '18 at 21:03









Blue

48k870153




48k870153










asked Dec 6 '18 at 20:59









Jacob AndreassonJacob Andreasson

32




32




closed as off-topic by Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers Dec 7 '18 at 0:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers Dec 7 '18 at 0:36


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Scientifica, Lord_Farin, Shailesh, José Carlos Santos, T. Bongers

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Do you know what a linear differential equation is and what the elementary consequences of "linear" for the solution space are?
    $endgroup$
    – LutzL
    Dec 6 '18 at 21:09










  • $begingroup$
    No i dont really know
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:11










  • $begingroup$
    Is it true or not this question?
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:16






  • 1




    $begingroup$
    While $y_1+y_2$ is not a solution, $ty_1+(1-t)y_2$ is a solution for any $t$.
    $endgroup$
    – user614671
    Dec 6 '18 at 21:25


















  • $begingroup$
    Do you know what a linear differential equation is and what the elementary consequences of "linear" for the solution space are?
    $endgroup$
    – LutzL
    Dec 6 '18 at 21:09










  • $begingroup$
    No i dont really know
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:11










  • $begingroup$
    Is it true or not this question?
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:16






  • 1




    $begingroup$
    While $y_1+y_2$ is not a solution, $ty_1+(1-t)y_2$ is a solution for any $t$.
    $endgroup$
    – user614671
    Dec 6 '18 at 21:25
















$begingroup$
Do you know what a linear differential equation is and what the elementary consequences of "linear" for the solution space are?
$endgroup$
– LutzL
Dec 6 '18 at 21:09




$begingroup$
Do you know what a linear differential equation is and what the elementary consequences of "linear" for the solution space are?
$endgroup$
– LutzL
Dec 6 '18 at 21:09












$begingroup$
No i dont really know
$endgroup$
– Jacob Andreasson
Dec 6 '18 at 21:11




$begingroup$
No i dont really know
$endgroup$
– Jacob Andreasson
Dec 6 '18 at 21:11












$begingroup$
Is it true or not this question?
$endgroup$
– Jacob Andreasson
Dec 6 '18 at 21:16




$begingroup$
Is it true or not this question?
$endgroup$
– Jacob Andreasson
Dec 6 '18 at 21:16




1




1




$begingroup$
While $y_1+y_2$ is not a solution, $ty_1+(1-t)y_2$ is a solution for any $t$.
$endgroup$
– user614671
Dec 6 '18 at 21:25




$begingroup$
While $y_1+y_2$ is not a solution, $ty_1+(1-t)y_2$ is a solution for any $t$.
$endgroup$
– user614671
Dec 6 '18 at 21:25










1 Answer
1






active

oldest

votes


















0












$begingroup$

$y_1, y_2$ are solutions of $y^{primeprime} + x^2y = 1$



$implies y_1^{primeprime} + x^2y_1 = 1, y_2^{primeprime} + x^2y_2 = 1$



Now, keep $y=y_1+y_2$,



$implies (y_1+y_2)''+x^2(y_1+y_2)=[y_1^{primeprime} + x^2y_1]+[y_2^{primeprime} + x^2y_2]=1+1=2neq1$.



So, $y_1+y_2$ is not a solution of the given differential equation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 6 '18 at 21:27


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$y_1, y_2$ are solutions of $y^{primeprime} + x^2y = 1$



$implies y_1^{primeprime} + x^2y_1 = 1, y_2^{primeprime} + x^2y_2 = 1$



Now, keep $y=y_1+y_2$,



$implies (y_1+y_2)''+x^2(y_1+y_2)=[y_1^{primeprime} + x^2y_1]+[y_2^{primeprime} + x^2y_2]=1+1=2neq1$.



So, $y_1+y_2$ is not a solution of the given differential equation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 6 '18 at 21:27
















0












$begingroup$

$y_1, y_2$ are solutions of $y^{primeprime} + x^2y = 1$



$implies y_1^{primeprime} + x^2y_1 = 1, y_2^{primeprime} + x^2y_2 = 1$



Now, keep $y=y_1+y_2$,



$implies (y_1+y_2)''+x^2(y_1+y_2)=[y_1^{primeprime} + x^2y_1]+[y_2^{primeprime} + x^2y_2]=1+1=2neq1$.



So, $y_1+y_2$ is not a solution of the given differential equation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 6 '18 at 21:27














0












0








0





$begingroup$

$y_1, y_2$ are solutions of $y^{primeprime} + x^2y = 1$



$implies y_1^{primeprime} + x^2y_1 = 1, y_2^{primeprime} + x^2y_2 = 1$



Now, keep $y=y_1+y_2$,



$implies (y_1+y_2)''+x^2(y_1+y_2)=[y_1^{primeprime} + x^2y_1]+[y_2^{primeprime} + x^2y_2]=1+1=2neq1$.



So, $y_1+y_2$ is not a solution of the given differential equation.






share|cite|improve this answer









$endgroup$



$y_1, y_2$ are solutions of $y^{primeprime} + x^2y = 1$



$implies y_1^{primeprime} + x^2y_1 = 1, y_2^{primeprime} + x^2y_2 = 1$



Now, keep $y=y_1+y_2$,



$implies (y_1+y_2)''+x^2(y_1+y_2)=[y_1^{primeprime} + x^2y_1]+[y_2^{primeprime} + x^2y_2]=1+1=2neq1$.



So, $y_1+y_2$ is not a solution of the given differential equation.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 21:19









Shubham JohriShubham Johri

5,017717




5,017717












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 6 '18 at 21:27


















  • $begingroup$
    Thank you very much!
    $endgroup$
    – Jacob Andreasson
    Dec 6 '18 at 21:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 6 '18 at 21:27
















$begingroup$
Thank you very much!
$endgroup$
– Jacob Andreasson
Dec 6 '18 at 21:25




$begingroup$
Thank you very much!
$endgroup$
– Jacob Andreasson
Dec 6 '18 at 21:25












$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 6 '18 at 21:27




$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 6 '18 at 21:27



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