Solving Differential equation on variable separation can't solve please help
$begingroup$
I can't it don't know where am I being wrong. I wasn't able to separate variables or make to a linear form and it isn't even homogeneous and I can't think of any substitution that would help me solve this can someone please help me with this question
$$frac{dy}{dx}=frac{(x+y)^2}{(x+2)(y-2)}$$
My approach= $$frac{dy}{dx}=frac{(x+y+2-2)^2}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)^2}{(x+2)(y-2)} + frac{(y-2)^2}{(x+2)(y-2)} + frac{2(x+2)(y-2)}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)}{(y-2)}+frac{(y-2)}{(x+2)}+2$$
Now I can't figure out how to solve it please help
ordinary-differential-equations
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
$endgroup$
|
show 5 more comments
$begingroup$
I can't it don't know where am I being wrong. I wasn't able to separate variables or make to a linear form and it isn't even homogeneous and I can't think of any substitution that would help me solve this can someone please help me with this question
$$frac{dy}{dx}=frac{(x+y)^2}{(x+2)(y-2)}$$
My approach= $$frac{dy}{dx}=frac{(x+y+2-2)^2}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)^2}{(x+2)(y-2)} + frac{(y-2)^2}{(x+2)(y-2)} + frac{2(x+2)(y-2)}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)}{(y-2)}+frac{(y-2)}{(x+2)}+2$$
Now I can't figure out how to solve it please help
ordinary-differential-equations
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
$endgroup$
$begingroup$
Try $u=y-2$, $v=x+2$.
$endgroup$
– Hans Lundmark
Dec 17 '18 at 7:56
$begingroup$
If I face a problem on a certain question and want to ask it so it matches the standard of this forum? Can u please help me with it kinda new here
$endgroup$
– aditya prakash
Dec 17 '18 at 8:02
$begingroup$
Two advices: the first one is to have a look at the basic Math Jax syntax. This will help you in producing nicely formatted mathematical texts, improving the understandability (and the beauty) of your posts. The second advice is to provide a little context for your question: is it an exercise or something else? What did you tried in order to solve the it? And you may add any othe observation which may help other members help you.
$endgroup$
– Daniele Tampieri
Dec 17 '18 at 8:19
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I did the edits you suggested is anything else needed?
$endgroup$
– aditya prakash
Dec 17 '18 at 9:00
$begingroup$
At this stage of insight, you can just set $u=frac{y-2}{x+2}$ to get $y'=(x+2)u'+u=u+u^{-1}+2$ which is separable.
$endgroup$
– LutzL
Dec 17 '18 at 11:25
|
show 5 more comments
$begingroup$
I can't it don't know where am I being wrong. I wasn't able to separate variables or make to a linear form and it isn't even homogeneous and I can't think of any substitution that would help me solve this can someone please help me with this question
$$frac{dy}{dx}=frac{(x+y)^2}{(x+2)(y-2)}$$
My approach= $$frac{dy}{dx}=frac{(x+y+2-2)^2}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)^2}{(x+2)(y-2)} + frac{(y-2)^2}{(x+2)(y-2)} + frac{2(x+2)(y-2)}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)}{(y-2)}+frac{(y-2)}{(x+2)}+2$$
Now I can't figure out how to solve it please help
ordinary-differential-equations
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
$endgroup$
I can't it don't know where am I being wrong. I wasn't able to separate variables or make to a linear form and it isn't even homogeneous and I can't think of any substitution that would help me solve this can someone please help me with this question
$$frac{dy}{dx}=frac{(x+y)^2}{(x+2)(y-2)}$$
My approach= $$frac{dy}{dx}=frac{(x+y+2-2)^2}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)^2}{(x+2)(y-2)} + frac{(y-2)^2}{(x+2)(y-2)} + frac{2(x+2)(y-2)}{(x+2)(y-2)}$$
$$frac{dy}{dx}=frac{(x+2)}{(y-2)}+frac{(y-2)}{(x+2)}+2$$
Now I can't figure out how to solve it please help
ordinary-differential-equations
ordinary-differential-equations
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
edited Dec 17 '18 at 8:56
aditya prakash
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
asked Dec 17 '18 at 7:38
aditya prakashaditya prakash
423
423
$begingroup$
Try $u=y-2$, $v=x+2$.
$endgroup$
– Hans Lundmark
Dec 17 '18 at 7:56
$begingroup$
If I face a problem on a certain question and want to ask it so it matches the standard of this forum? Can u please help me with it kinda new here
$endgroup$
– aditya prakash
Dec 17 '18 at 8:02
$begingroup$
Two advices: the first one is to have a look at the basic Math Jax syntax. This will help you in producing nicely formatted mathematical texts, improving the understandability (and the beauty) of your posts. The second advice is to provide a little context for your question: is it an exercise or something else? What did you tried in order to solve the it? And you may add any othe observation which may help other members help you.
$endgroup$
– Daniele Tampieri
Dec 17 '18 at 8:19
$begingroup$
I did the edits you suggested is anything else needed?
$endgroup$
– aditya prakash
Dec 17 '18 at 9:00
$begingroup$
At this stage of insight, you can just set $u=frac{y-2}{x+2}$ to get $y'=(x+2)u'+u=u+u^{-1}+2$ which is separable.
$endgroup$
– LutzL
Dec 17 '18 at 11:25
|
show 5 more comments
$begingroup$
Try $u=y-2$, $v=x+2$.
$endgroup$
– Hans Lundmark
Dec 17 '18 at 7:56
$begingroup$
If I face a problem on a certain question and want to ask it so it matches the standard of this forum? Can u please help me with it kinda new here
$endgroup$
– aditya prakash
Dec 17 '18 at 8:02
$begingroup$
Two advices: the first one is to have a look at the basic Math Jax syntax. This will help you in producing nicely formatted mathematical texts, improving the understandability (and the beauty) of your posts. The second advice is to provide a little context for your question: is it an exercise or something else? What did you tried in order to solve the it? And you may add any othe observation which may help other members help you.
$endgroup$
– Daniele Tampieri
Dec 17 '18 at 8:19
$begingroup$
I did the edits you suggested is anything else needed?
$endgroup$
– aditya prakash
Dec 17 '18 at 9:00
$begingroup$
At this stage of insight, you can just set $u=frac{y-2}{x+2}$ to get $y'=(x+2)u'+u=u+u^{-1}+2$ which is separable.
$endgroup$
– LutzL
Dec 17 '18 at 11:25
$begingroup$
Try $u=y-2$, $v=x+2$.
$endgroup$
– Hans Lundmark
Dec 17 '18 at 7:56
$begingroup$
Try $u=y-2$, $v=x+2$.
$endgroup$
– Hans Lundmark
Dec 17 '18 at 7:56
$begingroup$
If I face a problem on a certain question and want to ask it so it matches the standard of this forum? Can u please help me with it kinda new here
$endgroup$
– aditya prakash
Dec 17 '18 at 8:02
$begingroup$
If I face a problem on a certain question and want to ask it so it matches the standard of this forum? Can u please help me with it kinda new here
$endgroup$
– aditya prakash
Dec 17 '18 at 8:02
$begingroup$
Two advices: the first one is to have a look at the basic Math Jax syntax. This will help you in producing nicely formatted mathematical texts, improving the understandability (and the beauty) of your posts. The second advice is to provide a little context for your question: is it an exercise or something else? What did you tried in order to solve the it? And you may add any othe observation which may help other members help you.
$endgroup$
– Daniele Tampieri
Dec 17 '18 at 8:19
$begingroup$
Two advices: the first one is to have a look at the basic Math Jax syntax. This will help you in producing nicely formatted mathematical texts, improving the understandability (and the beauty) of your posts. The second advice is to provide a little context for your question: is it an exercise or something else? What did you tried in order to solve the it? And you may add any othe observation which may help other members help you.
$endgroup$
– Daniele Tampieri
Dec 17 '18 at 8:19
$begingroup$
I did the edits you suggested is anything else needed?
$endgroup$
– aditya prakash
Dec 17 '18 at 9:00
$begingroup$
I did the edits you suggested is anything else needed?
$endgroup$
– aditya prakash
Dec 17 '18 at 9:00
$begingroup$
At this stage of insight, you can just set $u=frac{y-2}{x+2}$ to get $y'=(x+2)u'+u=u+u^{-1}+2$ which is separable.
$endgroup$
– LutzL
Dec 17 '18 at 11:25
$begingroup$
At this stage of insight, you can just set $u=frac{y-2}{x+2}$ to get $y'=(x+2)u'+u=u+u^{-1}+2$ which is separable.
$endgroup$
– LutzL
Dec 17 '18 at 11:25
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint.
Make the change of variables
$$
u = x+2\
v = y-2
$$
giving
$$
frac{dv}{du}=frac{(u+v)^2}{u v}
$$
and then $v = lambda u$ arriving into a separable DE.
$$
frac{dlambda}{frac{1}{lambda} + 2} = frac{du}{u}
$$
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
$endgroup$
$begingroup$
Yup it's solved didn't thought it that way thanks you were a great help
$endgroup$
– aditya prakash
Dec 17 '18 at 9:44
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint.
Make the change of variables
$$
u = x+2\
v = y-2
$$
giving
$$
frac{dv}{du}=frac{(u+v)^2}{u v}
$$
and then $v = lambda u$ arriving into a separable DE.
$$
frac{dlambda}{frac{1}{lambda} + 2} = frac{du}{u}
$$
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
$endgroup$
$begingroup$
Yup it's solved didn't thought it that way thanks you were a great help
$endgroup$
– aditya prakash
Dec 17 '18 at 9:44
add a comment |
$begingroup$
Hint.
Make the change of variables
$$
u = x+2\
v = y-2
$$
giving
$$
frac{dv}{du}=frac{(u+v)^2}{u v}
$$
and then $v = lambda u$ arriving into a separable DE.
$$
frac{dlambda}{frac{1}{lambda} + 2} = frac{du}{u}
$$
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
$endgroup$
$begingroup$
Yup it's solved didn't thought it that way thanks you were a great help
$endgroup$
– aditya prakash
Dec 17 '18 at 9:44
add a comment |
$begingroup$
Hint.
Make the change of variables
$$
u = x+2\
v = y-2
$$
giving
$$
frac{dv}{du}=frac{(u+v)^2}{u v}
$$
and then $v = lambda u$ arriving into a separable DE.
$$
frac{dlambda}{frac{1}{lambda} + 2} = frac{du}{u}
$$
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
$endgroup$
Hint.
Make the change of variables
$$
u = x+2\
v = y-2
$$
giving
$$
frac{dv}{du}=frac{(u+v)^2}{u v}
$$
and then $v = lambda u$ arriving into a separable DE.
$$
frac{dlambda}{frac{1}{lambda} + 2} = frac{du}{u}
$$
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
edited Dec 17 '18 at 10:03
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
answered Dec 17 '18 at 9:40
CesareoCesareo
9,3963517
9,3963517
$begingroup$
Yup it's solved didn't thought it that way thanks you were a great help
$endgroup$
– aditya prakash
Dec 17 '18 at 9:44
add a comment |
$begingroup$
Yup it's solved didn't thought it that way thanks you were a great help
$endgroup$
– aditya prakash
Dec 17 '18 at 9:44
$begingroup$
Yup it's solved didn't thought it that way thanks you were a great help
$endgroup$
– aditya prakash
Dec 17 '18 at 9:44
$begingroup$
Yup it's solved didn't thought it that way thanks you were a great help
$endgroup$
– aditya prakash
Dec 17 '18 at 9:44
add a comment |
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$begingroup$
Try $u=y-2$, $v=x+2$.
$endgroup$
– Hans Lundmark
Dec 17 '18 at 7:56
$begingroup$
If I face a problem on a certain question and want to ask it so it matches the standard of this forum? Can u please help me with it kinda new here
$endgroup$
– aditya prakash
Dec 17 '18 at 8:02
$begingroup$
Two advices: the first one is to have a look at the basic Math Jax syntax. This will help you in producing nicely formatted mathematical texts, improving the understandability (and the beauty) of your posts. The second advice is to provide a little context for your question: is it an exercise or something else? What did you tried in order to solve the it? And you may add any othe observation which may help other members help you.
$endgroup$
– Daniele Tampieri
Dec 17 '18 at 8:19
$begingroup$
I did the edits you suggested is anything else needed?
$endgroup$
– aditya prakash
Dec 17 '18 at 9:00
$begingroup$
At this stage of insight, you can just set $u=frac{y-2}{x+2}$ to get $y'=(x+2)u'+u=u+u^{-1}+2$ which is separable.
$endgroup$
– LutzL
Dec 17 '18 at 11:25