Solve $(frac{dy}{dx}-1)e^{frac{dy}{dx}}=y$
$begingroup$
Solve $$left(frac{dy}{dx}-1right)e^{frac{dy}{dx}}=y$$
I just found a particular solution $y'=ln x$ hence $y=x(ln x-1)$ from 'observing'. And check it by differentiating it. But I don't know how to make a strict proof. Any hints would be helpful.
ordinary-differential-equations
edited Dec 24 '18 at 14:59
Larry
2,43331130
asked Dec 24 '18 at 14:23
LOISLOIS
4338
$endgroup$
add a comment |
$begingroup$
Solve $$left(frac{dy}{dx}-1right)e^{frac{dy}{dx}}=y$$
I just found a particular solution $y'=ln x$ hence $y=x(ln x-1)$ from 'observing'. And check it by differentiating it. But I don't know how to make a strict proof. Any hints would be helpful.
ordinary-differential-equations
edited Dec 24 '18 at 14:59
Larry
2,43331130
asked Dec 24 '18 at 14:23
LOISLOIS
4338
$endgroup$
add a comment |
$begingroup$
Solve $$left(frac{dy}{dx}-1right)e^{frac{dy}{dx}}=y$$
I just found a particular solution $y'=ln x$ hence $y=x(ln x-1)$ from 'observing'. And check it by differentiating it. But I don't know how to make a strict proof. Any hints would be helpful.
ordinary-differential-equations
edited Dec 24 '18 at 14:59
Larry
2,43331130
asked Dec 24 '18 at 14:23
LOISLOIS
4338
$endgroup$
Solve $$left(frac{dy}{dx}-1right)e^{frac{dy}{dx}}=y$$
I just found a particular solution $y'=ln x$ hence $y=x(ln x-1)$ from 'observing'. And check it by differentiating it. But I don't know how to make a strict proof. Any hints would be helpful.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 24 '18 at 14:59
Larry
2,43331130
asked Dec 24 '18 at 14:23
LOISLOIS
4338
edited Dec 24 '18 at 14:59
Larry
2,43331130
asked Dec 24 '18 at 14:23
LOISLOIS
4338
edited Dec 24 '18 at 14:59
Larry
2,43331130
edited Dec 24 '18 at 14:59
Larry
2,43331130
edited Dec 24 '18 at 14:59
Larry
2,43331130
2,43331130
asked Dec 24 '18 at 14:23
LOISLOIS
4338
asked Dec 24 '18 at 14:23
LOISLOIS
4338
asked Dec 24 '18 at 14:23
LOISLOIS
4338
4338
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take the derivative of the equation to get, outside the constant solution $y=-1$,
$$
y''y'e^{y'}=y'implies y''e^{y'}=1implies e^{y'}=x+C, ~~ y'=ln(x+C)
$$
Now insert into the original equation to get
$$
y(x)=(ln(x+C)-1)(x+C)
$$
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
Hint: This equation is a separable one,we get $$frac{dy(x)}{dx}=Wleft(frac{y(x)}{e}right)+1$$ and $$frac{frac{dy(x)}{dx}}{Wleft(frac{y(x)}{e}right)+1}=1$$
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Take the derivative of the equation to get, outside the constant solution $y=-1$,
$$
y''y'e^{y'}=y'implies y''e^{y'}=1implies e^{y'}=x+C, ~~ y'=ln(x+C)
$$
Now insert into the original equation to get
$$
y(x)=(ln(x+C)-1)(x+C)
$$
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
Take the derivative of the equation to get, outside the constant solution $y=-1$,
$$
y''y'e^{y'}=y'implies y''e^{y'}=1implies e^{y'}=x+C, ~~ y'=ln(x+C)
$$
Now insert into the original equation to get
$$
y(x)=(ln(x+C)-1)(x+C)
$$
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
Take the derivative of the equation to get, outside the constant solution $y=-1$,
$$
y''y'e^{y'}=y'implies y''e^{y'}=1implies e^{y'}=x+C, ~~ y'=ln(x+C)
$$
Now insert into the original equation to get
$$
y(x)=(ln(x+C)-1)(x+C)
$$
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
$endgroup$
Take the derivative of the equation to get, outside the constant solution $y=-1$,
$$
y''y'e^{y'}=y'implies y''e^{y'}=1implies e^{y'}=x+C, ~~ y'=ln(x+C)
$$
Now insert into the original equation to get
$$
y(x)=(ln(x+C)-1)(x+C)
$$
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
answered Dec 24 '18 at 15:14
LutzLLutzL
59.3k42057
59.3k42057
add a comment |
add a comment |
$begingroup$
Hint: This equation is a separable one,we get $$frac{dy(x)}{dx}=Wleft(frac{y(x)}{e}right)+1$$ and $$frac{frac{dy(x)}{dx}}{Wleft(frac{y(x)}{e}right)+1}=1$$
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
add a comment |
$begingroup$
Hint: This equation is a separable one,we get $$frac{dy(x)}{dx}=Wleft(frac{y(x)}{e}right)+1$$ and $$frac{frac{dy(x)}{dx}}{Wleft(frac{y(x)}{e}right)+1}=1$$
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
add a comment |
$begingroup$
Hint: This equation is a separable one,we get $$frac{dy(x)}{dx}=Wleft(frac{y(x)}{e}right)+1$$ and $$frac{frac{dy(x)}{dx}}{Wleft(frac{y(x)}{e}right)+1}=1$$
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
$endgroup$
Hint: This equation is a separable one,we get $$frac{dy(x)}{dx}=Wleft(frac{y(x)}{e}right)+1$$ and $$frac{frac{dy(x)}{dx}}{Wleft(frac{y(x)}{e}right)+1}=1$$
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
answered Dec 24 '18 at 14:31
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77k42866
77k42866
add a comment |
add a comment |
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