Find the specified particular solution of ($x^2+2y')y''+2xy'=0$ while $y(0)=1$ and $y'(0)=0$
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 '18 at 9:32
Renuka
74
add a comment |
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 '18 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 '18 at 10:06
add a comment |
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 '18 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 '18 at 9:32
Renuka
74
asked Nov 25 '18 at 9:32
Renuka
74
edited Nov 25 '18 at 9:45
asked Nov 25 '18 at 9:32
Renuka
74
asked Nov 25 '18 at 9:32
Renuka
74
asked Nov 25 '18 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 '18 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 '18 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 '18 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 '18 at 10:06
add a comment |
2 Answers
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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 '18 at 10:14
Mostafa Ayaz
13.6k3836
add a comment |
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 '18 at 9:44
Claude Leibovici
119k1157132
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 '18 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 '18 at 13:59
add a comment |
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2 Answers
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2 Answers
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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 '18 at 10:14
Mostafa Ayaz
13.6k3836
add a comment |
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 '18 at 10:14
Mostafa Ayaz
13.6k3836
add a comment |
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 '18 at 10:14
Mostafa Ayaz
13.6k3836
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 '18 at 10:14
Mostafa Ayaz
13.6k3836
answered Nov 25 '18 at 10:14
Mostafa Ayaz
13.6k3836
answered Nov 25 '18 at 10:14
Mostafa Ayaz
13.6k3836
answered Nov 25 '18 at 10:14
Mostafa Ayaz
13.6k3836
13.6k3836
add a comment |
add a comment |
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 '18 at 9:44
Claude Leibovici
119k1157132
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 '18 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 '18 at 13:59
add a comment |
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 '18 at 9:44
Claude Leibovici
119k1157132
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 '18 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 '18 at 13:59
add a comment |
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 '18 at 9:44
Claude Leibovici
119k1157132
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 '18 at 9:44
Claude Leibovici
119k1157132
edited Nov 25 '18 at 13:58
answered Nov 25 '18 at 9:44
Claude Leibovici
119k1157132
answered Nov 25 '18 at 9:44
Claude Leibovici
119k1157132
answered Nov 25 '18 at 9:44
Claude Leibovici
119k1157132
119k1157132
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 '18 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 '18 at 13:59
add a comment |
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 '18 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 '18 at 13:59
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 '18 at 10:12
I think you made a mistake: $p'=dfrac{u'-2x}{2}$.
– Batominovski
Nov 25 '18 at 10:12
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 '18 at 13:59
@Batominovski. Thanks for pointing the typo.
– Claude Leibovici
Nov 25 '18 at 13:59
add a comment |
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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 '18 at 10:06