Find the specified particular solution of ($x^2+2y')y''+2xy'=0$ while $y(0)=1$ and $y'(0)=0$
up vote
0
down vote
favorite
In this problem I am able to understand that it must be solved using reduction of order. But I am not able to proceed with the sum & also I am confused of where to substitute the given values of $y$ and $y'$.
differential-equations
asked Nov 25 at 9:32
Renuka
74
add a comment |
up vote
0
down vote
favorite
In this problem I am able to understand that it must be solved using reduction of order. But I am not able to proceed with the sum & also I am confused of where to substitute the given values of $y$ and $y'$.
differential-equations
asked Nov 25 at 9:32
Renuka
74
I got stuck up again. I proceeded as you told but I am getting it as (x^2 + 2p)^2=x^4+C. Now how to separate the variables??. Sorry for the trouble.
– Renuka
Nov 25 at 10:06
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In this problem I am able to understand that it must be solved using reduction of order. But I am not able to proceed with the sum & also I am confused of where to substitute the given values of $y$ and $y'$.
differential-equations
asked Nov 25 at 9:32
Renuka
74
In this problem I am able to understand that it must be solved using reduction of order. But I am not able to proceed with the sum & also I am confused of where to substitute the given values of $y$ and $y'$.
differential-equations
differential-equations
asked Nov 25 at 9:32
Renuka
74
asked Nov 25 at 9:32
Renuka
74
edited Nov 25 at 9:45
asked Nov 25 at 9:32
Renuka
74
asked Nov 25 at 9:32
Renuka
74
asked Nov 25 at 9:32
Renuka
74
74
I got stuck up again. I proceeded as you told but I am getting it as (x^2 + 2p)^2=x^4+C. Now how to separate the variables??. Sorry for the trouble.
– Renuka
Nov 25 at 10:06
add a comment |
I got stuck up again. I proceeded as you told but I am getting it as (x^2 + 2p)^2=x^4+C. Now how to separate the variables??. Sorry for the trouble.
– Renuka
Nov 25 at 10:06
I got stuck up again. I proceeded as you told but I am getting it as (x^2 + 2p)^2=x^4+C. Now how to separate the variables??. Sorry for the trouble.
– Renuka
Nov 25 at 10:06
I got stuck up again. I proceeded as you told but I am getting it as (x^2 + 2p)^2=x^4+C. Now how to separate the variables??. Sorry for the trouble.
– Renuka
Nov 25 at 10:06
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
We have$$2y'y''+x^2y''+2xy'=0$$by integrating we obtain$$(y')^2+(x^2y')+C=0$$since $y'(0)=0$ we have $C=0$ which yields to $$y'=0\text{and/or}\ y'+x^2=0$$therefore$$y=D\text{and/or}\ y'=-{x^3over 3}+D$$once more we have $y(0)=1$ therefore $$y=1\text{and/or }\y=1-{x^3over 3}$$
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
add a comment |
up vote
1
down vote
Start with $p=y'$ to get
$$left(2 p+x^2right) p'+2 x p=0qquad text{with}qquad p(0)=0$$
Let now $$u=2p+x^2implies p=frac {u-x^2}2implies p'=frac{u'-2x}2$$ Replace to get
$$frac{1}{2} u u'-x^3=0$$ which seems to be easy to integrate.
When done, go back to $p$ and fix the constant of integration to get the condition.
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 at 13:59
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
We have$$2y'y''+x^2y''+2xy'=0$$by integrating we obtain$$(y')^2+(x^2y')+C=0$$since $y'(0)=0$ we have $C=0$ which yields to $$y'=0\text{and/or}\ y'+x^2=0$$therefore$$y=D\text{and/or}\ y'=-{x^3over 3}+D$$once more we have $y(0)=1$ therefore $$y=1\text{and/or }\y=1-{x^3over 3}$$
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
add a comment |
up vote
3
down vote
We have$$2y'y''+x^2y''+2xy'=0$$by integrating we obtain$$(y')^2+(x^2y')+C=0$$since $y'(0)=0$ we have $C=0$ which yields to $$y'=0\text{and/or}\ y'+x^2=0$$therefore$$y=D\text{and/or}\ y'=-{x^3over 3}+D$$once more we have $y(0)=1$ therefore $$y=1\text{and/or }\y=1-{x^3over 3}$$
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
add a comment |
up vote
3
down vote
up vote
3
down vote
We have$$2y'y''+x^2y''+2xy'=0$$by integrating we obtain$$(y')^2+(x^2y')+C=0$$since $y'(0)=0$ we have $C=0$ which yields to $$y'=0\text{and/or}\ y'+x^2=0$$therefore$$y=D\text{and/or}\ y'=-{x^3over 3}+D$$once more we have $y(0)=1$ therefore $$y=1\text{and/or }\y=1-{x^3over 3}$$
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
We have$$2y'y''+x^2y''+2xy'=0$$by integrating we obtain$$(y')^2+(x^2y')+C=0$$since $y'(0)=0$ we have $C=0$ which yields to $$y'=0\text{and/or}\ y'+x^2=0$$therefore$$y=D\text{and/or}\ y'=-{x^3over 3}+D$$once more we have $y(0)=1$ therefore $$y=1\text{and/or }\y=1-{x^3over 3}$$
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
answered Nov 25 at 10:14
Mostafa Ayaz
13.4k3836
13.4k3836
add a comment |
add a comment |
up vote
1
down vote
Start with $p=y'$ to get
$$left(2 p+x^2right) p'+2 x p=0qquad text{with}qquad p(0)=0$$
Let now $$u=2p+x^2implies p=frac {u-x^2}2implies p'=frac{u'-2x}2$$ Replace to get
$$frac{1}{2} u u'-x^3=0$$ which seems to be easy to integrate.
When done, go back to $p$ and fix the constant of integration to get the condition.
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 at 13:59
add a comment |
up vote
1
down vote
Start with $p=y'$ to get
$$left(2 p+x^2right) p'+2 x p=0qquad text{with}qquad p(0)=0$$
Let now $$u=2p+x^2implies p=frac {u-x^2}2implies p'=frac{u'-2x}2$$ Replace to get
$$frac{1}{2} u u'-x^3=0$$ which seems to be easy to integrate.
When done, go back to $p$ and fix the constant of integration to get the condition.
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 at 13:59
add a comment |
up vote
1
down vote
up vote
1
down vote
Start with $p=y'$ to get
$$left(2 p+x^2right) p'+2 x p=0qquad text{with}qquad p(0)=0$$
Let now $$u=2p+x^2implies p=frac {u-x^2}2implies p'=frac{u'-2x}2$$ Replace to get
$$frac{1}{2} u u'-x^3=0$$ which seems to be easy to integrate.
When done, go back to $p$ and fix the constant of integration to get the condition.
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
Start with $p=y'$ to get
$$left(2 p+x^2right) p'+2 x p=0qquad text{with}qquad p(0)=0$$
Let now $$u=2p+x^2implies p=frac {u-x^2}2implies p'=frac{u'-2x}2$$ Replace to get
$$frac{1}{2} u u'-x^3=0$$ which seems to be easy to integrate.
When done, go back to $p$ and fix the constant of integration to get the condition.
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
edited Nov 25 at 13:58
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
answered Nov 25 at 9:44
Claude Leibovici
117k1156131
117k1156131
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 at 13:59
add a comment |
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 at 13:59
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 at 10:12
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 at 13:59
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 at 13:59
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012627%2ffind-the-specified-particular-solution-of-x22yy2xy-0-while-y0-1-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I got stuck up again. I proceeded as you told but I am getting it as (x^2 + 2p)^2=x^4+C. Now how to separate the variables??. Sorry for the trouble.
– Renuka
Nov 25 at 10:06