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

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

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! :)

edited Dec 6 '18 at 21:03

Blue

48k870153

asked Dec 6 '18 at 20:59

Jacob Andreasson

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

add a comment |

-1

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! :)

edited Dec 6 '18 at 21:03

Blue

48k870153

asked Dec 6 '18 at 20:59

Jacob Andreasson

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

add a comment |

-1

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! :)

edited Dec 6 '18 at 21:03

Blue

48k870153

asked Dec 6 '18 at 20:59

Jacob Andreasson

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

edited Dec 6 '18 at 21:03

Blue

48k870153

asked Dec 6 '18 at 20:59

Jacob Andreasson

edited Dec 6 '18 at 21:03

Blue

48k870153

asked Dec 6 '18 at 20:59

Jacob Andreasson

edited Dec 6 '18 at 21:03

Blue

48k870153

edited Dec 6 '18 at 21:03

Blue

48k870153

edited Dec 6 '18 at 21:03

Blue

48k870153

asked Dec 6 '18 at 20:59

Jacob Andreasson

asked Dec 6 '18 at 20:59

Jacob Andreasson

asked Dec 6 '18 at 20:59

Jacob Andreasson

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

add a comment |

$begingroup$
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

Do you know what a linear differential equation is and what the elementary consequences of "linear" for the solution space are?

– LutzL
Dec 6 '18 at 21:09

No i dont really know

– Jacob Andreasson
Dec 6 '18 at 21:11

Is it true or not this question?

– Jacob Andreasson
Dec 6 '18 at 21:16

While $y_1+y_2$ is not a solution, $ty_1+(1-t)y_2$ is a solution for any $t$.

– user614671
Dec 6 '18 at 21:25

add a comment |

1 Answer
1

active

oldest

votes

$y_1, y_2$ 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.

answered Dec 6 '18 at 21:19

Shubham Johri

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

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

$y_1, y_2$ 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.

answered Dec 6 '18 at 21:19

Shubham Johri

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

add a comment |

$y_1, y_2$ 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.

answered Dec 6 '18 at 21:19

Shubham Johri

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

add a comment |

$y_1, y_2$ 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.

answered Dec 6 '18 at 21:19

Shubham Johri

5,017717

$y_1, y_2$ 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.

answered Dec 6 '18 at 21:19

Shubham Johri

5,017717

answered Dec 6 '18 at 21:19

Shubham Johri

5,017717

answered Dec 6 '18 at 21:19

Shubham Johri

5,017717

answered Dec 6 '18 at 21:19

Shubham Johri

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

add a comment |

$begingroup$
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

Thank you very much!

– Jacob Andreasson
Dec 6 '18 at 21:25

You're welcome.

– Shubham Johri
Dec 6 '18 at 21:27

add a comment |

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