Mixing
How do you solve the following differential equation (Proof of Lemma 3 of Hermalin, 1998)?
$begingroup$
$(e(theta) - s theta)e'(theta) = s(1-s)theta$
The solution to the differential equation is given by:
$e(theta)=frac{1}{2}(s+sqrt{4s-3s^2})theta$
Here, the dependent variable is $e(theta)$, while $s$ can be treated like a constant.
ordinary-differential-equations game-theory economics
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
$endgroup$
add a comment |
$begingroup$
$(e(theta) - s theta)e'(theta) = s(1-s)theta$
The solution to the differential equation is given by:
$e(theta)=frac{1}{2}(s+sqrt{4s-3s^2})theta$
Here, the dependent variable is $e(theta)$, while $s$ can be treated like a constant.
ordinary-differential-equations game-theory economics
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
$endgroup$
add a comment |
$begingroup$
$(e(theta) - s theta)e'(theta) = s(1-s)theta$
The solution to the differential equation is given by:
$e(theta)=frac{1}{2}(s+sqrt{4s-3s^2})theta$
Here, the dependent variable is $e(theta)$, while $s$ can be treated like a constant.
ordinary-differential-equations game-theory economics
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
$endgroup$
$(e(theta) - s theta)e'(theta) = s(1-s)theta$
The solution to the differential equation is given by:
$e(theta)=frac{1}{2}(s+sqrt{4s-3s^2})theta$
Here, the dependent variable is $e(theta)$, while $s$ can be treated like a constant.
ordinary-differential-equations game-theory economics
ordinary-differential-equations game-theory economics
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
edited Jan 12 at 17:05
Daniel Challam
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
asked Jan 12 at 14:38
Daniel ChallamDaniel Challam
113
113
add a comment |
add a comment |
1 Answer
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$begingroup$
The equation is homogeneous in the equal degrees sense, thus set $e(θ)=θv(θ)$ to get
$$
θ(v-s)(θv'+v)=s(1−s)θimplies (v-s)θv'=s(1−s)-v(v-s)\~\
frac{(v-s)dv}{(v-frac s2)^2+frac34s^2-s}=-frac{dθ}{θ}.
$$
Now apply a separation in logarithmic and arcus tangent derivative on the left side and integrate, then try to solve for $v$.
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The equation is homogeneous in the equal degrees sense, thus set $e(θ)=θv(θ)$ to get
$$
θ(v-s)(θv'+v)=s(1−s)θimplies (v-s)θv'=s(1−s)-v(v-s)\~\
frac{(v-s)dv}{(v-frac s2)^2+frac34s^2-s}=-frac{dθ}{θ}.
$$
Now apply a separation in logarithmic and arcus tangent derivative on the left side and integrate, then try to solve for $v$.
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
$endgroup$
add a comment |
$begingroup$
The equation is homogeneous in the equal degrees sense, thus set $e(θ)=θv(θ)$ to get
$$
θ(v-s)(θv'+v)=s(1−s)θimplies (v-s)θv'=s(1−s)-v(v-s)\~\
frac{(v-s)dv}{(v-frac s2)^2+frac34s^2-s}=-frac{dθ}{θ}.
$$
Now apply a separation in logarithmic and arcus tangent derivative on the left side and integrate, then try to solve for $v$.
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
$endgroup$
add a comment |
$begingroup$
The equation is homogeneous in the equal degrees sense, thus set $e(θ)=θv(θ)$ to get
$$
θ(v-s)(θv'+v)=s(1−s)θimplies (v-s)θv'=s(1−s)-v(v-s)\~\
frac{(v-s)dv}{(v-frac s2)^2+frac34s^2-s}=-frac{dθ}{θ}.
$$
Now apply a separation in logarithmic and arcus tangent derivative on the left side and integrate, then try to solve for $v$.
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
$endgroup$
The equation is homogeneous in the equal degrees sense, thus set $e(θ)=θv(θ)$ to get
$$
θ(v-s)(θv'+v)=s(1−s)θimplies (v-s)θv'=s(1−s)-v(v-s)\~\
frac{(v-s)dv}{(v-frac s2)^2+frac34s^2-s}=-frac{dθ}{θ}.
$$
Now apply a separation in logarithmic and arcus tangent derivative on the left side and integrate, then try to solve for $v$.
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
answered Jan 12 at 19:58
LutzLLutzL
58.3k42054
58.3k42054
add a comment |
add a comment |
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