Mixing
On the methods used to solve an O.D.E.
Clash Royale CLAN TAG#URR8PPP
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(I'm sorry for the format of the mathematical symbols in advance)
Take the differential equation dy/dx = x * y, for example, the process for solving it goes as follows.
1.) $dy = x * y * dx$
2.) $1/y * dy = x * dx$
3.) $∫ 1/y dy = ∫ x dx$
4.) $ln(y) = (x^2)/2$
5.) e^ln(y) = e^((x^2)/2)
6.) y = e^((x^2)/2)
Now my question is in regards to step 3. When we take the integral of both sides why do we not treat the infinitesimals dy and dx as part of the function that we are integrating? In other words, when taking the integrals why didn't we get $∫ 1/y dy dx = ∫ x dx dx$ instead? What happened to them?
P.S. I am currently taking the regular high school Calculus 2 course so please keep the "advanced" mathematics to a bare minimum when answering the question.
differential-equations self-learning
asked Jul 21 at 13:20
oypus
397
add a comment |Â
up vote
2
down vote
favorite
(I'm sorry for the format of the mathematical symbols in advance)
Take the differential equation dy/dx = x * y, for example, the process for solving it goes as follows.
1.) $dy = x * y * dx$
2.) $1/y * dy = x * dx$
3.) $∫ 1/y dy = ∫ x dx$
4.) $ln(y) = (x^2)/2$
5.) e^ln(y) = e^((x^2)/2)
6.) y = e^((x^2)/2)
Now my question is in regards to step 3. When we take the integral of both sides why do we not treat the infinitesimals dy and dx as part of the function that we are integrating? In other words, when taking the integrals why didn't we get $∫ 1/y dy dx = ∫ x dx dx$ instead? What happened to them?
P.S. I am currently taking the regular high school Calculus 2 course so please keep the "advanced" mathematics to a bare minimum when answering the question.
differential-equations self-learning
asked Jul 21 at 13:20
oypus
397
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
(I'm sorry for the format of the mathematical symbols in advance)
Take the differential equation dy/dx = x * y, for example, the process for solving it goes as follows.
1.) $dy = x * y * dx$
2.) $1/y * dy = x * dx$
3.) $∫ 1/y dy = ∫ x dx$
4.) $ln(y) = (x^2)/2$
5.) e^ln(y) = e^((x^2)/2)
6.) y = e^((x^2)/2)
Now my question is in regards to step 3. When we take the integral of both sides why do we not treat the infinitesimals dy and dx as part of the function that we are integrating? In other words, when taking the integrals why didn't we get $∫ 1/y dy dx = ∫ x dx dx$ instead? What happened to them?
P.S. I am currently taking the regular high school Calculus 2 course so please keep the "advanced" mathematics to a bare minimum when answering the question.
differential-equations self-learning
asked Jul 21 at 13:20
oypus
397
(I'm sorry for the format of the mathematical symbols in advance)
Take the differential equation dy/dx = x * y, for example, the process for solving it goes as follows.
1.) $dy = x * y * dx$
2.) $1/y * dy = x * dx$
3.) $∫ 1/y dy = ∫ x dx$
4.) $ln(y) = (x^2)/2$
5.) e^ln(y) = e^((x^2)/2)
6.) y = e^((x^2)/2)
Now my question is in regards to step 3. When we take the integral of both sides why do we not treat the infinitesimals dy and dx as part of the function that we are integrating? In other words, when taking the integrals why didn't we get $∫ 1/y dy dx = ∫ x dx dx$ instead? What happened to them?
P.S. I am currently taking the regular high school Calculus 2 course so please keep the "advanced" mathematics to a bare minimum when answering the question.
differential-equations self-learning
asked Jul 21 at 13:20
oypus
397
edited Jul 21 at 13:44
asked Jul 21 at 13:20
oypus
397
asked Jul 21 at 13:20
oypus
397
asked Jul 21 at 13:20
oypus
397
397
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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up vote
3
down vote
accepted
$$ y'= xy $$
$$ frac y'y= x $$
Take inetgral now on both sides
$$ colorredint frac y'y colorreddx=colorredint x colorreddx$$
$$ colorredint frac 1y frac dydxcolorreddx=colorredint x colorreddx$$
$$int frac dyy=int x dx$$
I prefer this method
$$frac y'y= x$$
$$(ln(y))'= x$$
Then i take integral on both sides
$$colorredint (ln(y))'colorreddx= colorredint x colorreddx$$
$$int frac d(ln(y))dxdx= int x dx$$
$$int d(ln(y))= int x dx$$
$$ln(y)= frac x^22 +K$$
answered Jul 21 at 13:26
Isham
10.6k3829
1
Thank you. This is so simple that I almost feel stupid for not seeing it. Good job.
– oypus
Jul 21 at 13:47
yw @oypus ...if it helped you thats great...
– Isham
Jul 21 at 13:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$ y'= xy $$
$$ frac y'y= x $$
Take inetgral now on both sides
$$ colorredint frac y'y colorreddx=colorredint x colorreddx$$
$$ colorredint frac 1y frac dydxcolorreddx=colorredint x colorreddx$$
$$int frac dyy=int x dx$$
I prefer this method
$$frac y'y= x$$
$$(ln(y))'= x$$
Then i take integral on both sides
$$colorredint (ln(y))'colorreddx= colorredint x colorreddx$$
$$int frac d(ln(y))dxdx= int x dx$$
$$int d(ln(y))= int x dx$$
$$ln(y)= frac x^22 +K$$
answered Jul 21 at 13:26
Isham
10.6k3829
1
Thank you. This is so simple that I almost feel stupid for not seeing it. Good job.
– oypus
Jul 21 at 13:47
yw @oypus ...if it helped you thats great...
– Isham
Jul 21 at 13:48
add a comment |Â
up vote
3
down vote
accepted
$$ y'= xy $$
$$ frac y'y= x $$
Take inetgral now on both sides
$$ colorredint frac y'y colorreddx=colorredint x colorreddx$$
$$ colorredint frac 1y frac dydxcolorreddx=colorredint x colorreddx$$
$$int frac dyy=int x dx$$
I prefer this method
$$frac y'y= x$$
$$(ln(y))'= x$$
Then i take integral on both sides
$$colorredint (ln(y))'colorreddx= colorredint x colorreddx$$
$$int frac d(ln(y))dxdx= int x dx$$
$$int d(ln(y))= int x dx$$
$$ln(y)= frac x^22 +K$$
answered Jul 21 at 13:26
Isham
10.6k3829
1
Thank you. This is so simple that I almost feel stupid for not seeing it. Good job.
– oypus
Jul 21 at 13:47
yw @oypus ...if it helped you thats great...
– Isham
Jul 21 at 13:48
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$ y'= xy $$
$$ frac y'y= x $$
Take inetgral now on both sides
$$ colorredint frac y'y colorreddx=colorredint x colorreddx$$
$$ colorredint frac 1y frac dydxcolorreddx=colorredint x colorreddx$$
$$int frac dyy=int x dx$$
I prefer this method
$$frac y'y= x$$
$$(ln(y))'= x$$
Then i take integral on both sides
$$colorredint (ln(y))'colorreddx= colorredint x colorreddx$$
$$int frac d(ln(y))dxdx= int x dx$$
$$int d(ln(y))= int x dx$$
$$ln(y)= frac x^22 +K$$
answered Jul 21 at 13:26
Isham
10.6k3829
$$ y'= xy $$
$$ frac y'y= x $$
Take inetgral now on both sides
$$ colorredint frac y'y colorreddx=colorredint x colorreddx$$
$$ colorredint frac 1y frac dydxcolorreddx=colorredint x colorreddx$$
$$int frac dyy=int x dx$$
I prefer this method
$$frac y'y= x$$
$$(ln(y))'= x$$
Then i take integral on both sides
$$colorredint (ln(y))'colorreddx= colorredint x colorreddx$$
$$int frac d(ln(y))dxdx= int x dx$$
$$int d(ln(y))= int x dx$$
$$ln(y)= frac x^22 +K$$
answered Jul 21 at 13:26
Isham
10.6k3829
edited Jul 21 at 13:41
answered Jul 21 at 13:26
Isham
10.6k3829
answered Jul 21 at 13:26
Isham
10.6k3829
answered Jul 21 at 13:26
Isham
10.6k3829
10.6k3829
1
Thank you. This is so simple that I almost feel stupid for not seeing it. Good job.
– oypus
Jul 21 at 13:47
yw @oypus ...if it helped you thats great...
– Isham
Jul 21 at 13:48
add a comment |Â
1
Thank you. This is so simple that I almost feel stupid for not seeing it. Good job.
– oypus
Jul 21 at 13:47
yw @oypus ...if it helped you thats great...
– Isham
Jul 21 at 13:48
1
1
Thank you. This is so simple that I almost feel stupid for not seeing it. Good job.
– oypus
Jul 21 at 13:47
Thank you. This is so simple that I almost feel stupid for not seeing it. Good job.
– oypus
Jul 21 at 13:47
yw @oypus ...if it helped you thats great...
– Isham
Jul 21 at 13:48
yw @oypus ...if it helped you thats great...
– Isham
Jul 21 at 13:48
add a comment |Â
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