Mixing
Differential equation $2x^4yy'+y^4 = 4x^6$
$begingroup$
I have differential equation
$2x^4yy'+y^4 = 4x^6$
How to find real parameter $m$ for which, when we introduce substitution $y=z^m$, given equation becomes first order homogeneous differential equation?
real-analysis ordinary-differential-equations integro-differential-equations
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
$endgroup$
add a comment |
$begingroup$
I have differential equation
$2x^4yy'+y^4 = 4x^6$
How to find real parameter $m$ for which, when we introduce substitution $y=z^m$, given equation becomes first order homogeneous differential equation?
real-analysis ordinary-differential-equations integro-differential-equations
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
$endgroup$
add a comment |
$begingroup$
I have differential equation
$2x^4yy'+y^4 = 4x^6$
How to find real parameter $m$ for which, when we introduce substitution $y=z^m$, given equation becomes first order homogeneous differential equation?
real-analysis ordinary-differential-equations integro-differential-equations
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
$endgroup$
I have differential equation
$2x^4yy'+y^4 = 4x^6$
How to find real parameter $m$ for which, when we introduce substitution $y=z^m$, given equation becomes first order homogeneous differential equation?
real-analysis ordinary-differential-equations integro-differential-equations
real-analysis ordinary-differential-equations integro-differential-equations
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
asked Jan 26 at 0:30
Nikola MijuškovićNikola Mijušković
274
274
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$
2x^4y y'+y^4=4x^6tag 1
$$
We write
$$
x^4(y^2)'+(y^2)^2=4x^6
$$
Set $w=y^2$. Hence
$$
x^4w'+w^2=4x^6tag 2
$$
This equation has partial solution $w=-4x^3$. So we set $w=-4x^3z$ and (2) becomes
$$
x(1+3z-4z^2+xz')=0
$$
Hence
$$
z'=(4z^2-3z-1)/x
$$
or
$$
frac{dz}{4z^2-3z-1}=frac{dx}{x}
$$
or integrating
$$
frac{1}{5}log(1-z)-frac{1}{5}log(1+4z)=c+log x
$$
or
$$
z=frac{1-Ax^5}{1+4Ax^5}textrm{, }A=const
$$
Hence
$$
w=-4x^3frac{1-Ax^5}{1+4Ax^5}
$$
and finaly
$$
y=y(x)=2sqrt{x^3frac{Ax^5-1}{4x^5+1}}textrm{, }A=const
$$
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
add a comment |
$begingroup$
Ans: $frac{3}{2}$
we have,
$2x^4yy'+y^4 = 4x^6$
rearranging,
$y'=frac{4x^6-y^4}{2x^4y}$
now
substituting $$y=z^m$$
$$dy=mz^{m-1}dz$$
$$z'=frac{4x^6-z^{4m}}{mz^{2m-1}x^4}$$
Now for $m=frac32$ equation is homogeneous.
answered Jan 26 at 1:17
TanmayTanmay
444
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
2x^4y y'+y^4=4x^6tag 1
$$
We write
$$
x^4(y^2)'+(y^2)^2=4x^6
$$
Set $w=y^2$. Hence
$$
x^4w'+w^2=4x^6tag 2
$$
This equation has partial solution $w=-4x^3$. So we set $w=-4x^3z$ and (2) becomes
$$
x(1+3z-4z^2+xz')=0
$$
Hence
$$
z'=(4z^2-3z-1)/x
$$
or
$$
frac{dz}{4z^2-3z-1}=frac{dx}{x}
$$
or integrating
$$
frac{1}{5}log(1-z)-frac{1}{5}log(1+4z)=c+log x
$$
or
$$
z=frac{1-Ax^5}{1+4Ax^5}textrm{, }A=const
$$
Hence
$$
w=-4x^3frac{1-Ax^5}{1+4Ax^5}
$$
and finaly
$$
y=y(x)=2sqrt{x^3frac{Ax^5-1}{4x^5+1}}textrm{, }A=const
$$
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
add a comment |
$begingroup$
$$
2x^4y y'+y^4=4x^6tag 1
$$
We write
$$
x^4(y^2)'+(y^2)^2=4x^6
$$
Set $w=y^2$. Hence
$$
x^4w'+w^2=4x^6tag 2
$$
This equation has partial solution $w=-4x^3$. So we set $w=-4x^3z$ and (2) becomes
$$
x(1+3z-4z^2+xz')=0
$$
Hence
$$
z'=(4z^2-3z-1)/x
$$
or
$$
frac{dz}{4z^2-3z-1}=frac{dx}{x}
$$
or integrating
$$
frac{1}{5}log(1-z)-frac{1}{5}log(1+4z)=c+log x
$$
or
$$
z=frac{1-Ax^5}{1+4Ax^5}textrm{, }A=const
$$
Hence
$$
w=-4x^3frac{1-Ax^5}{1+4Ax^5}
$$
and finaly
$$
y=y(x)=2sqrt{x^3frac{Ax^5-1}{4x^5+1}}textrm{, }A=const
$$
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
add a comment |
$begingroup$
$$
2x^4y y'+y^4=4x^6tag 1
$$
We write
$$
x^4(y^2)'+(y^2)^2=4x^6
$$
Set $w=y^2$. Hence
$$
x^4w'+w^2=4x^6tag 2
$$
This equation has partial solution $w=-4x^3$. So we set $w=-4x^3z$ and (2) becomes
$$
x(1+3z-4z^2+xz')=0
$$
Hence
$$
z'=(4z^2-3z-1)/x
$$
or
$$
frac{dz}{4z^2-3z-1}=frac{dx}{x}
$$
or integrating
$$
frac{1}{5}log(1-z)-frac{1}{5}log(1+4z)=c+log x
$$
or
$$
z=frac{1-Ax^5}{1+4Ax^5}textrm{, }A=const
$$
Hence
$$
w=-4x^3frac{1-Ax^5}{1+4Ax^5}
$$
and finaly
$$
y=y(x)=2sqrt{x^3frac{Ax^5-1}{4x^5+1}}textrm{, }A=const
$$
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
$endgroup$
$$
2x^4y y'+y^4=4x^6tag 1
$$
We write
$$
x^4(y^2)'+(y^2)^2=4x^6
$$
Set $w=y^2$. Hence
$$
x^4w'+w^2=4x^6tag 2
$$
This equation has partial solution $w=-4x^3$. So we set $w=-4x^3z$ and (2) becomes
$$
x(1+3z-4z^2+xz')=0
$$
Hence
$$
z'=(4z^2-3z-1)/x
$$
or
$$
frac{dz}{4z^2-3z-1}=frac{dx}{x}
$$
or integrating
$$
frac{1}{5}log(1-z)-frac{1}{5}log(1+4z)=c+log x
$$
or
$$
z=frac{1-Ax^5}{1+4Ax^5}textrm{, }A=const
$$
Hence
$$
w=-4x^3frac{1-Ax^5}{1+4Ax^5}
$$
and finaly
$$
y=y(x)=2sqrt{x^3frac{Ax^5-1}{4x^5+1}}textrm{, }A=const
$$
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
answered Jan 26 at 1:27
Nikos Bagis Nikos Bagis
2,482616
2,482616
add a comment |
add a comment |
$begingroup$
Ans: $frac{3}{2}$
we have,
$2x^4yy'+y^4 = 4x^6$
rearranging,
$y'=frac{4x^6-y^4}{2x^4y}$
now
substituting $$y=z^m$$
$$dy=mz^{m-1}dz$$
$$z'=frac{4x^6-z^{4m}}{mz^{2m-1}x^4}$$
Now for $m=frac32$ equation is homogeneous.
answered Jan 26 at 1:17
TanmayTanmay
444
$endgroup$
add a comment |
$begingroup$
Ans: $frac{3}{2}$
we have,
$2x^4yy'+y^4 = 4x^6$
rearranging,
$y'=frac{4x^6-y^4}{2x^4y}$
now
substituting $$y=z^m$$
$$dy=mz^{m-1}dz$$
$$z'=frac{4x^6-z^{4m}}{mz^{2m-1}x^4}$$
Now for $m=frac32$ equation is homogeneous.
answered Jan 26 at 1:17
TanmayTanmay
444
$endgroup$
add a comment |
$begingroup$
Ans: $frac{3}{2}$
we have,
$2x^4yy'+y^4 = 4x^6$
rearranging,
$y'=frac{4x^6-y^4}{2x^4y}$
now
substituting $$y=z^m$$
$$dy=mz^{m-1}dz$$
$$z'=frac{4x^6-z^{4m}}{mz^{2m-1}x^4}$$
Now for $m=frac32$ equation is homogeneous.
answered Jan 26 at 1:17
TanmayTanmay
444
$endgroup$
Ans: $frac{3}{2}$
we have,
$2x^4yy'+y^4 = 4x^6$
rearranging,
$y'=frac{4x^6-y^4}{2x^4y}$
now
substituting $$y=z^m$$
$$dy=mz^{m-1}dz$$
$$z'=frac{4x^6-z^{4m}}{mz^{2m-1}x^4}$$
Now for $m=frac32$ equation is homogeneous.
answered Jan 26 at 1:17
TanmayTanmay
444
answered Jan 26 at 1:17
TanmayTanmay
444
answered Jan 26 at 1:17
TanmayTanmay
444
answered Jan 26 at 1:17
TanmayTanmay
444
444
add a comment |
add a comment |
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