Solve the following initial value problem
$begingroup$
I want to solve the following initial initial value problem
$x dy + (y-y^2ln(x))dx = 0, y(1) = 1/4$
So far i've approached it by trying to separate it and get
$dx/x = dy/(y-y^2ln(x))$
When I go to the the integral with respect to y for the left side i'm unsure what to do for the $ln(x)$ term.
So I tried to approach it as an exact type of problem instead and get that its not exact, but when I put it into the form $(diM/diY - diN/diX)/N$ I get $2yln(x)/x$.
Im not sure how to do symbol for partial derivative so I said $di$ instead.
Is that my integrating factor? It isn't in terms of only x so i'm confused.
Any help would be appreciated.
ordinary-differential-equations
edited Dec 20 '18 at 0:11
Botond
6,54531034
asked Dec 19 '18 at 23:12
dragophdragoph
111
$endgroup$
add a comment |
$begingroup$
I want to solve the following initial initial value problem
$x dy + (y-y^2ln(x))dx = 0, y(1) = 1/4$
So far i've approached it by trying to separate it and get
$dx/x = dy/(y-y^2ln(x))$
When I go to the the integral with respect to y for the left side i'm unsure what to do for the $ln(x)$ term.
So I tried to approach it as an exact type of problem instead and get that its not exact, but when I put it into the form $(diM/diY - diN/diX)/N$ I get $2yln(x)/x$.
Im not sure how to do symbol for partial derivative so I said $di$ instead.
Is that my integrating factor? It isn't in terms of only x so i'm confused.
Any help would be appreciated.
ordinary-differential-equations
edited Dec 20 '18 at 0:11
Botond
6,54531034
asked Dec 19 '18 at 23:12
dragophdragoph
111
$endgroup$
add a comment |
$begingroup$
I want to solve the following initial initial value problem
$x dy + (y-y^2ln(x))dx = 0, y(1) = 1/4$
So far i've approached it by trying to separate it and get
$dx/x = dy/(y-y^2ln(x))$
When I go to the the integral with respect to y for the left side i'm unsure what to do for the $ln(x)$ term.
So I tried to approach it as an exact type of problem instead and get that its not exact, but when I put it into the form $(diM/diY - diN/diX)/N$ I get $2yln(x)/x$.
Im not sure how to do symbol for partial derivative so I said $di$ instead.
Is that my integrating factor? It isn't in terms of only x so i'm confused.
Any help would be appreciated.
ordinary-differential-equations
edited Dec 20 '18 at 0:11
Botond
6,54531034
asked Dec 19 '18 at 23:12
dragophdragoph
111
$endgroup$
I want to solve the following initial initial value problem
$x dy + (y-y^2ln(x))dx = 0, y(1) = 1/4$
So far i've approached it by trying to separate it and get
$dx/x = dy/(y-y^2ln(x))$
When I go to the the integral with respect to y for the left side i'm unsure what to do for the $ln(x)$ term.
So I tried to approach it as an exact type of problem instead and get that its not exact, but when I put it into the form $(diM/diY - diN/diX)/N$ I get $2yln(x)/x$.
Im not sure how to do symbol for partial derivative so I said $di$ instead.
Is that my integrating factor? It isn't in terms of only x so i'm confused.
Any help would be appreciated.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 20 '18 at 0:11
Botond
6,54531034
asked Dec 19 '18 at 23:12
dragophdragoph
111
edited Dec 20 '18 at 0:11
Botond
6,54531034
asked Dec 19 '18 at 23:12
dragophdragoph
111
edited Dec 20 '18 at 0:11
Botond
6,54531034
edited Dec 20 '18 at 0:11
Botond
6,54531034
edited Dec 20 '18 at 0:11
Botond
6,54531034
6,54531034
asked Dec 19 '18 at 23:12
dragophdragoph
111
asked Dec 19 '18 at 23:12
dragophdragoph
111
asked Dec 19 '18 at 23:12
dragophdragoph
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As equation for $y$ it is also Bernoulli. Which means that replacing $y=z^{-1}$, $dy=-z^{-2}dz$ gives
$$
-xdz+(z-ln x)dx=0.
$$
Now that the equation is linear in $z$ the integrating factor is obvious as $x^{-2}$ where then
$$
-d((xy)^{-1})-x^{-2}ln(x)dx=0
$$
is exact.
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
$endgroup$
add a comment |
$begingroup$
As LutzL answered, using $y=frac 1 z$ the equation becomes
$$-frac{x z'-z+log (x)}{z^2}=0$$ So, consider
$$x z'-z=-log (x)$$ The homogeneous is simple
$$xz'-z=0implies z=C x$$ Now, if you prefer variation of parameters instead of integrating factor, then
$$x^2 C'=-log (x) implies C'=-frac {log(x)} {x^2}$$ and $C$ is easily obtained using integration by parts.
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
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
$begingroup$
As equation for $y$ it is also Bernoulli. Which means that replacing $y=z^{-1}$, $dy=-z^{-2}dz$ gives
$$
-xdz+(z-ln x)dx=0.
$$
Now that the equation is linear in $z$ the integrating factor is obvious as $x^{-2}$ where then
$$
-d((xy)^{-1})-x^{-2}ln(x)dx=0
$$
is exact.
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
$endgroup$
add a comment |
$begingroup$
As equation for $y$ it is also Bernoulli. Which means that replacing $y=z^{-1}$, $dy=-z^{-2}dz$ gives
$$
-xdz+(z-ln x)dx=0.
$$
Now that the equation is linear in $z$ the integrating factor is obvious as $x^{-2}$ where then
$$
-d((xy)^{-1})-x^{-2}ln(x)dx=0
$$
is exact.
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
$endgroup$
add a comment |
$begingroup$
As equation for $y$ it is also Bernoulli. Which means that replacing $y=z^{-1}$, $dy=-z^{-2}dz$ gives
$$
-xdz+(z-ln x)dx=0.
$$
Now that the equation is linear in $z$ the integrating factor is obvious as $x^{-2}$ where then
$$
-d((xy)^{-1})-x^{-2}ln(x)dx=0
$$
is exact.
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
$endgroup$
As equation for $y$ it is also Bernoulli. Which means that replacing $y=z^{-1}$, $dy=-z^{-2}dz$ gives
$$
-xdz+(z-ln x)dx=0.
$$
Now that the equation is linear in $z$ the integrating factor is obvious as $x^{-2}$ where then
$$
-d((xy)^{-1})-x^{-2}ln(x)dx=0
$$
is exact.
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
answered Dec 19 '18 at 23:27
LutzLLutzL
60.2k42057
60.2k42057
add a comment |
add a comment |
$begingroup$
As LutzL answered, using $y=frac 1 z$ the equation becomes
$$-frac{x z'-z+log (x)}{z^2}=0$$ So, consider
$$x z'-z=-log (x)$$ The homogeneous is simple
$$xz'-z=0implies z=C x$$ Now, if you prefer variation of parameters instead of integrating factor, then
$$x^2 C'=-log (x) implies C'=-frac {log(x)} {x^2}$$ and $C$ is easily obtained using integration by parts.
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
As LutzL answered, using $y=frac 1 z$ the equation becomes
$$-frac{x z'-z+log (x)}{z^2}=0$$ So, consider
$$x z'-z=-log (x)$$ The homogeneous is simple
$$xz'-z=0implies z=C x$$ Now, if you prefer variation of parameters instead of integrating factor, then
$$x^2 C'=-log (x) implies C'=-frac {log(x)} {x^2}$$ and $C$ is easily obtained using integration by parts.
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
As LutzL answered, using $y=frac 1 z$ the equation becomes
$$-frac{x z'-z+log (x)}{z^2}=0$$ So, consider
$$x z'-z=-log (x)$$ The homogeneous is simple
$$xz'-z=0implies z=C x$$ Now, if you prefer variation of parameters instead of integrating factor, then
$$x^2 C'=-log (x) implies C'=-frac {log(x)} {x^2}$$ and $C$ is easily obtained using integration by parts.
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
As LutzL answered, using $y=frac 1 z$ the equation becomes
$$-frac{x z'-z+log (x)}{z^2}=0$$ So, consider
$$x z'-z=-log (x)$$ The homogeneous is simple
$$xz'-z=0implies z=C x$$ Now, if you prefer variation of parameters instead of integrating factor, then
$$x^2 C'=-log (x) implies C'=-frac {log(x)} {x^2}$$ and $C$ is easily obtained using integration by parts.
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
answered Dec 20 '18 at 6:28
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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