Mixing
First order non linear ODE with Bernoulli
Clash Royale CLAN TAG#URR8PPP
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4
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I have a problem with this equation: $ y'(x)-xy(x)=-xy^4(x) $ with initial condition $ y(x_0)=y_0$.
I'm arrived to prove that $ y_0= (Ce^-frac32x_0^2+1)^-3 $ but now i can't move on. Moreover, WolframAlpha's solution is next to impossible…
Thanks for any help!
differential-equations
edited 1 hour ago
Harry49
5,0972827
asked 1 hour ago
Marco Pittella
524
add a comment |Â
up vote
4
down vote
favorite
I have a problem with this equation: $ y'(x)-xy(x)=-xy^4(x) $ with initial condition $ y(x_0)=y_0$.
I'm arrived to prove that $ y_0= (Ce^-frac32x_0^2+1)^-3 $ but now i can't move on. Moreover, WolframAlpha's solution is next to impossible…
Thanks for any help!
differential-equations
edited 1 hour ago
Harry49
5,0972827
asked 1 hour ago
Marco Pittella
524
You may set $z=y^-3$.
– dmtri
1 hour ago
Do you mean you cannot solve for $C$ ?
– dmtri
1 hour ago
it's exponent $-1/3$ not $-3$ since we have that$$y(x)=frac 1 (Ke^-3x^2/2+1)^1/3$$
– Isham
1 hour ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have a problem with this equation: $ y'(x)-xy(x)=-xy^4(x) $ with initial condition $ y(x_0)=y_0$.
I'm arrived to prove that $ y_0= (Ce^-frac32x_0^2+1)^-3 $ but now i can't move on. Moreover, WolframAlpha's solution is next to impossible…
Thanks for any help!
differential-equations
edited 1 hour ago
Harry49
5,0972827
asked 1 hour ago
Marco Pittella
524
I have a problem with this equation: $ y'(x)-xy(x)=-xy^4(x) $ with initial condition $ y(x_0)=y_0$.
I'm arrived to prove that $ y_0= (Ce^-frac32x_0^2+1)^-3 $ but now i can't move on. Moreover, WolframAlpha's solution is next to impossible…
Thanks for any help!
differential-equations
differential-equations
edited 1 hour ago
Harry49
5,0972827
asked 1 hour ago
Marco Pittella
524
edited 1 hour ago
Harry49
5,0972827
asked 1 hour ago
Marco Pittella
524
edited 1 hour ago
Harry49
5,0972827
edited 1 hour ago
Harry49
5,0972827
edited 1 hour ago
Harry49
5,0972827
5,0972827
asked 1 hour ago
Marco Pittella
524
asked 1 hour ago
Marco Pittella
524
asked 1 hour ago
Marco Pittella
524
524
You may set $z=y^-3$.
– dmtri
1 hour ago
Do you mean you cannot solve for $C$ ?
– dmtri
1 hour ago
it's exponent $-1/3$ not $-3$ since we have that$$y(x)=frac 1 (Ke^-3x^2/2+1)^1/3$$
– Isham
1 hour ago
add a comment |Â
You may set $z=y^-3$.
– dmtri
1 hour ago
Do you mean you cannot solve for $C$ ?
– dmtri
1 hour ago
it's exponent $-1/3$ not $-3$ since we have that$$y(x)=frac 1 (Ke^-3x^2/2+1)^1/3$$
– Isham
1 hour ago
You may set $z=y^-3$.
– dmtri
1 hour ago
You may set $z=y^-3$.
– dmtri
1 hour ago
Do you mean you cannot solve for $C$ ?
– dmtri
1 hour ago
Do you mean you cannot solve for $C$ ?
– dmtri
1 hour ago
it's exponent $-1/3$ not $-3$ since we have that$$y(x)=frac 1 (Ke^-3x^2/2+1)^1/3$$
– Isham
1 hour ago
it's exponent $-1/3$ not $-3$ since we have that$$y(x)=frac 1 (Ke^-3x^2/2+1)^1/3$$
– Isham
1 hour ago
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
1
down vote
Hint: Write this as $$fracy'(x)y(x)-y(x)^4=x$$
edited 45 mins ago
amWhy
190k27221433
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
add a comment |Â
up vote
0
down vote
Substitute $z=1/y^3$
$$z'+3xz=3x$$
Tis equation is separable
$$z'=3x(1-z)$$
$$int frac dz1-z=frac 32x^2+C$$
$$-ln (z-1)=frac 32x^2+C$$
$$implies y^3(x)=frac 1 Ke^-3x^2/2+1$$
Therefore
$$y_0^3=frac 1 Ke^-3x_0^2/2+1$$
$$Ke^-3x_0^2/2=frac 1 y_0^3-1$$
$$K=left (frac 1 y_0^3-1 right)e^3x_0^2/2$$
answered 50 mins ago
Isham
11.2k3929
So the final result is $y=frac1[(frac1y_0^3-1)e^-frac32x^2]e^-frac32x^2+1$?
– Marco Pittella
35 mins ago
@marco dont forget the exponent $1/3$ and you can also write $e^-3/2(x^2+x_0^2)$
– Isham
6 mins ago
add a comment |Â
up vote
0
down vote
We have $$C=e^frac3x_0^22(y_0^-1/3-1)$$
edited 41 mins ago
amWhy
190k27221433
answered 1 hour ago
dmtri
956518
add a comment |Â
up vote
0
down vote
Let us solve the Bernoulli differential equation $$y'(x)-xy(x)=-xy(x)^4$$ with initial condition $y (x_0)=y_0$. Dividing by $y^4$ and setting $u=y^-3$, we have the linear ODE
$$
frac13 u'(x)+xu(x) = x
$$
with initial condition $u (x_0)=y_0^-3$. The solution obtained by integrating factor reads
beginaligned
u(x) &= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + 3 int_x_0^x t e^3t^2/2 ,text d t right) \
&= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + e^3x^2/2 - e^3x_0^2/2 right) ,
endaligned
from which one deduces $y=u^-1/3$.
answered 57 mins ago
Harry49
5,0972827
add a comment |Â
up vote
0
down vote
To reach a better understanding of my problem, i will write all the passages.
$fracy^'y^4=fracxyy^3-xrightarrow fracy^'y^4=xy^-3-x$
Now i put $z=y^-3rightarrow z^'=-3xz+3x$.
$y_0(x)=Ce^A(x)rightarrow A(x)=int -3xdx=-frac32x^2rightarrow y_0(x)=Ce^-frac32x^2$
$y_p(x)=e^A(x)B(x)rightarrow B(x)=int -3xcdot e^A(x)dx=int -3xcdot e^frac32x^2dx$
Now i put $frac32x^2=trightarrow dt=3xdxrightarrow dx=fracdx3x$
Since $int -3xe^tfracdt3x=-e^frac32x^2rightarrow y_px=1$, i obtain $y(x)=Ce^-frac32x^2+1$. But since $y^-3=z$...the result that i wrote.
answered 48 mins ago
Marco Pittella
524
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Write this as $$fracy'(x)y(x)-y(x)^4=x$$
edited 45 mins ago
amWhy
190k27221433
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
add a comment |Â
up vote
1
down vote
Hint: Write this as $$fracy'(x)y(x)-y(x)^4=x$$
edited 45 mins ago
amWhy
190k27221433
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Write this as $$fracy'(x)y(x)-y(x)^4=x$$
edited 45 mins ago
amWhy
190k27221433
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
Hint: Write this as $$fracy'(x)y(x)-y(x)^4=x$$
edited 45 mins ago
amWhy
190k27221433
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
edited 45 mins ago
amWhy
190k27221433
edited 45 mins ago
amWhy
190k27221433
edited 45 mins ago
amWhy
190k27221433
190k27221433
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
answered 1 hour ago
Dr. Sonnhard Graubner
69.1k32761
69.1k32761
add a comment |Â
add a comment |Â
up vote
0
down vote
Substitute $z=1/y^3$
$$z'+3xz=3x$$
Tis equation is separable
$$z'=3x(1-z)$$
$$int frac dz1-z=frac 32x^2+C$$
$$-ln (z-1)=frac 32x^2+C$$
$$implies y^3(x)=frac 1 Ke^-3x^2/2+1$$
Therefore
$$y_0^3=frac 1 Ke^-3x_0^2/2+1$$
$$Ke^-3x_0^2/2=frac 1 y_0^3-1$$
$$K=left (frac 1 y_0^3-1 right)e^3x_0^2/2$$
answered 50 mins ago
Isham
11.2k3929
So the final result is $y=frac1[(frac1y_0^3-1)e^-frac32x^2]e^-frac32x^2+1$?
– Marco Pittella
35 mins ago
@marco dont forget the exponent $1/3$ and you can also write $e^-3/2(x^2+x_0^2)$
– Isham
6 mins ago
add a comment |Â
up vote
0
down vote
Substitute $z=1/y^3$
$$z'+3xz=3x$$
Tis equation is separable
$$z'=3x(1-z)$$
$$int frac dz1-z=frac 32x^2+C$$
$$-ln (z-1)=frac 32x^2+C$$
$$implies y^3(x)=frac 1 Ke^-3x^2/2+1$$
Therefore
$$y_0^3=frac 1 Ke^-3x_0^2/2+1$$
$$Ke^-3x_0^2/2=frac 1 y_0^3-1$$
$$K=left (frac 1 y_0^3-1 right)e^3x_0^2/2$$
answered 50 mins ago
Isham
11.2k3929
So the final result is $y=frac1[(frac1y_0^3-1)e^-frac32x^2]e^-frac32x^2+1$?
– Marco Pittella
35 mins ago
@marco dont forget the exponent $1/3$ and you can also write $e^-3/2(x^2+x_0^2)$
– Isham
6 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Substitute $z=1/y^3$
$$z'+3xz=3x$$
Tis equation is separable
$$z'=3x(1-z)$$
$$int frac dz1-z=frac 32x^2+C$$
$$-ln (z-1)=frac 32x^2+C$$
$$implies y^3(x)=frac 1 Ke^-3x^2/2+1$$
Therefore
$$y_0^3=frac 1 Ke^-3x_0^2/2+1$$
$$Ke^-3x_0^2/2=frac 1 y_0^3-1$$
$$K=left (frac 1 y_0^3-1 right)e^3x_0^2/2$$
answered 50 mins ago
Isham
11.2k3929
Substitute $z=1/y^3$
$$z'+3xz=3x$$
Tis equation is separable
$$z'=3x(1-z)$$
$$int frac dz1-z=frac 32x^2+C$$
$$-ln (z-1)=frac 32x^2+C$$
$$implies y^3(x)=frac 1 Ke^-3x^2/2+1$$
Therefore
$$y_0^3=frac 1 Ke^-3x_0^2/2+1$$
$$Ke^-3x_0^2/2=frac 1 y_0^3-1$$
$$K=left (frac 1 y_0^3-1 right)e^3x_0^2/2$$
answered 50 mins ago
Isham
11.2k3929
edited 44 mins ago
answered 50 mins ago
Isham
11.2k3929
answered 50 mins ago
Isham
11.2k3929
answered 50 mins ago
Isham
11.2k3929
11.2k3929
So the final result is $y=frac1[(frac1y_0^3-1)e^-frac32x^2]e^-frac32x^2+1$?
– Marco Pittella
35 mins ago
@marco dont forget the exponent $1/3$ and you can also write $e^-3/2(x^2+x_0^2)$
– Isham
6 mins ago
add a comment |Â
So the final result is $y=frac1[(frac1y_0^3-1)e^-frac32x^2]e^-frac32x^2+1$?
– Marco Pittella
35 mins ago
@marco dont forget the exponent $1/3$ and you can also write $e^-3/2(x^2+x_0^2)$
– Isham
6 mins ago
So the final result is $y=frac1[(frac1y_0^3-1)e^-frac32x^2]e^-frac32x^2+1$?
– Marco Pittella
35 mins ago
So the final result is $y=frac1[(frac1y_0^3-1)e^-frac32x^2]e^-frac32x^2+1$?
– Marco Pittella
35 mins ago
@marco dont forget the exponent $1/3$ and you can also write $e^-3/2(x^2+x_0^2)$
– Isham
6 mins ago
@marco dont forget the exponent $1/3$ and you can also write $e^-3/2(x^2+x_0^2)$
– Isham
6 mins ago
add a comment |Â
up vote
0
down vote
We have $$C=e^frac3x_0^22(y_0^-1/3-1)$$
edited 41 mins ago
amWhy
190k27221433
answered 1 hour ago
dmtri
956518
add a comment |Â
up vote
0
down vote
We have $$C=e^frac3x_0^22(y_0^-1/3-1)$$
edited 41 mins ago
amWhy
190k27221433
answered 1 hour ago
dmtri
956518
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have $$C=e^frac3x_0^22(y_0^-1/3-1)$$
edited 41 mins ago
amWhy
190k27221433
answered 1 hour ago
dmtri
956518
We have $$C=e^frac3x_0^22(y_0^-1/3-1)$$
edited 41 mins ago
amWhy
190k27221433
answered 1 hour ago
dmtri
956518
edited 41 mins ago
amWhy
190k27221433
edited 41 mins ago
amWhy
190k27221433
edited 41 mins ago
amWhy
190k27221433
190k27221433
answered 1 hour ago
dmtri
956518
answered 1 hour ago
dmtri
956518
answered 1 hour ago
dmtri
956518
956518
add a comment |Â
add a comment |Â
up vote
0
down vote
Let us solve the Bernoulli differential equation $$y'(x)-xy(x)=-xy(x)^4$$ with initial condition $y (x_0)=y_0$. Dividing by $y^4$ and setting $u=y^-3$, we have the linear ODE
$$
frac13 u'(x)+xu(x) = x
$$
with initial condition $u (x_0)=y_0^-3$. The solution obtained by integrating factor reads
beginaligned
u(x) &= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + 3 int_x_0^x t e^3t^2/2 ,text d t right) \
&= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + e^3x^2/2 - e^3x_0^2/2 right) ,
endaligned
from which one deduces $y=u^-1/3$.
answered 57 mins ago
Harry49
5,0972827
add a comment |Â
up vote
0
down vote
Let us solve the Bernoulli differential equation $$y'(x)-xy(x)=-xy(x)^4$$ with initial condition $y (x_0)=y_0$. Dividing by $y^4$ and setting $u=y^-3$, we have the linear ODE
$$
frac13 u'(x)+xu(x) = x
$$
with initial condition $u (x_0)=y_0^-3$. The solution obtained by integrating factor reads
beginaligned
u(x) &= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + 3 int_x_0^x t e^3t^2/2 ,text d t right) \
&= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + e^3x^2/2 - e^3x_0^2/2 right) ,
endaligned
from which one deduces $y=u^-1/3$.
answered 57 mins ago
Harry49
5,0972827
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let us solve the Bernoulli differential equation $$y'(x)-xy(x)=-xy(x)^4$$ with initial condition $y (x_0)=y_0$. Dividing by $y^4$ and setting $u=y^-3$, we have the linear ODE
$$
frac13 u'(x)+xu(x) = x
$$
with initial condition $u (x_0)=y_0^-3$. The solution obtained by integrating factor reads
beginaligned
u(x) &= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + 3 int_x_0^x t e^3t^2/2 ,text d t right) \
&= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + e^3x^2/2 - e^3x_0^2/2 right) ,
endaligned
from which one deduces $y=u^-1/3$.
answered 57 mins ago
Harry49
5,0972827
Let us solve the Bernoulli differential equation $$y'(x)-xy(x)=-xy(x)^4$$ with initial condition $y (x_0)=y_0$. Dividing by $y^4$ and setting $u=y^-3$, we have the linear ODE
$$
frac13 u'(x)+xu(x) = x
$$
with initial condition $u (x_0)=y_0^-3$. The solution obtained by integrating factor reads
beginaligned
u(x) &= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + 3 int_x_0^x t e^3t^2/2 ,text d t right) \
&= e^-3 (x^2-x_0^2)/2 left(y_0^-3 + e^3x^2/2 - e^3x_0^2/2 right) ,
endaligned
from which one deduces $y=u^-1/3$.
answered 57 mins ago
Harry49
5,0972827
edited 35 mins ago
answered 57 mins ago
Harry49
5,0972827
answered 57 mins ago
Harry49
5,0972827
answered 57 mins ago
Harry49
5,0972827
5,0972827
add a comment |Â
add a comment |Â
up vote
0
down vote
To reach a better understanding of my problem, i will write all the passages.
$fracy^'y^4=fracxyy^3-xrightarrow fracy^'y^4=xy^-3-x$
Now i put $z=y^-3rightarrow z^'=-3xz+3x$.
$y_0(x)=Ce^A(x)rightarrow A(x)=int -3xdx=-frac32x^2rightarrow y_0(x)=Ce^-frac32x^2$
$y_p(x)=e^A(x)B(x)rightarrow B(x)=int -3xcdot e^A(x)dx=int -3xcdot e^frac32x^2dx$
Now i put $frac32x^2=trightarrow dt=3xdxrightarrow dx=fracdx3x$
Since $int -3xe^tfracdt3x=-e^frac32x^2rightarrow y_px=1$, i obtain $y(x)=Ce^-frac32x^2+1$. But since $y^-3=z$...the result that i wrote.
answered 48 mins ago
Marco Pittella
524
add a comment |Â
up vote
0
down vote
To reach a better understanding of my problem, i will write all the passages.
$fracy^'y^4=fracxyy^3-xrightarrow fracy^'y^4=xy^-3-x$
Now i put $z=y^-3rightarrow z^'=-3xz+3x$.
$y_0(x)=Ce^A(x)rightarrow A(x)=int -3xdx=-frac32x^2rightarrow y_0(x)=Ce^-frac32x^2$
$y_p(x)=e^A(x)B(x)rightarrow B(x)=int -3xcdot e^A(x)dx=int -3xcdot e^frac32x^2dx$
Now i put $frac32x^2=trightarrow dt=3xdxrightarrow dx=fracdx3x$
Since $int -3xe^tfracdt3x=-e^frac32x^2rightarrow y_px=1$, i obtain $y(x)=Ce^-frac32x^2+1$. But since $y^-3=z$...the result that i wrote.
answered 48 mins ago
Marco Pittella
524
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To reach a better understanding of my problem, i will write all the passages.
$fracy^'y^4=fracxyy^3-xrightarrow fracy^'y^4=xy^-3-x$
Now i put $z=y^-3rightarrow z^'=-3xz+3x$.
$y_0(x)=Ce^A(x)rightarrow A(x)=int -3xdx=-frac32x^2rightarrow y_0(x)=Ce^-frac32x^2$
$y_p(x)=e^A(x)B(x)rightarrow B(x)=int -3xcdot e^A(x)dx=int -3xcdot e^frac32x^2dx$
Now i put $frac32x^2=trightarrow dt=3xdxrightarrow dx=fracdx3x$
Since $int -3xe^tfracdt3x=-e^frac32x^2rightarrow y_px=1$, i obtain $y(x)=Ce^-frac32x^2+1$. But since $y^-3=z$...the result that i wrote.
answered 48 mins ago
Marco Pittella
524
To reach a better understanding of my problem, i will write all the passages.
$fracy^'y^4=fracxyy^3-xrightarrow fracy^'y^4=xy^-3-x$
Now i put $z=y^-3rightarrow z^'=-3xz+3x$.
$y_0(x)=Ce^A(x)rightarrow A(x)=int -3xdx=-frac32x^2rightarrow y_0(x)=Ce^-frac32x^2$
$y_p(x)=e^A(x)B(x)rightarrow B(x)=int -3xcdot e^A(x)dx=int -3xcdot e^frac32x^2dx$
Now i put $frac32x^2=trightarrow dt=3xdxrightarrow dx=fracdx3x$
Since $int -3xe^tfracdt3x=-e^frac32x^2rightarrow y_px=1$, i obtain $y(x)=Ce^-frac32x^2+1$. But since $y^-3=z$...the result that i wrote.
answered 48 mins ago
Marco Pittella
524
edited 28 mins ago
answered 48 mins ago
Marco Pittella
524
answered 48 mins ago
Marco Pittella
524
answered 48 mins ago
Marco Pittella
524
524
add a comment |Â
add a comment |Â
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You may set $z=y^-3$.
– dmtri
1 hour ago
Do you mean you cannot solve for $C$ ?
– dmtri
1 hour ago
it's exponent $-1/3$ not $-3$ since we have that$$y(x)=frac 1 (Ke^-3x^2/2+1)^1/3$$
– Isham
1 hour ago