Help with differential equations please.

I am stuck on solving separable differential equations, specifically the step where you go from ∫(1/𝑔(𝑦) * 𝑑𝑦/𝑑𝑡) 𝑑𝑡 = ∫ℎ(𝑡) 𝑑𝑡 to ∫(1/𝑔(𝑦) 𝑑𝑦 = ∫ℎ(𝑡) 𝑑𝑡. I understand that ∫(𝑑𝑦/𝑑𝑡) 𝑑𝑡 equals 𝑑𝑦, but with the additional factor of 1/𝑔(𝑦) I cannot for the life of me grasp why we are able to just 'cancel out' the 𝑑𝑡's and ignore the 1/𝑔(𝑦). Can someone provide an explanation for this? I have looked for an explanation almost everywhere but everyone seems to gloss over this particular step. Any help would be appreciated.

Reply 1

mqb2766

21

Original post by babushka22

I am stuck on solving separable differential equations, specifically the step where you go from ∫(1/𝑔(𝑦) * 𝑑𝑦/𝑑𝑡) 𝑑𝑡 = ∫ℎ(𝑡) 𝑑𝑡 to ∫(1/𝑔(𝑦) 𝑑𝑦 = ∫ℎ(𝑡) 𝑑𝑡. I understand that ∫(𝑑𝑦/𝑑𝑡) 𝑑𝑡 equals 𝑑𝑦, but with the additional factor of 1/𝑔(𝑦) I cannot for the life of me grasp why we are able to just 'cancel out' the 𝑑𝑡's and ignore the 1/𝑔(𝑦). Can someone provide an explanation for this? I have looked for an explanation almost everywhere but everyone seems to gloss over this particular step. Any help would be appreciated.

Just denote 1/g(y) as f(y) say and suppose the integral is F(y), then you have
∫f(y) 𝑑𝑦/𝑑𝑡 𝑑𝑡 = F(y)
You could recognise the integrand
f(y) 𝑑𝑦/𝑑𝑡
is the chain rule on F so
dF/dt = dF/dy dy/dt = f(y) dy/dt
Then integrating both sides with respect to t gives the result youre after (first equation above).

Rather than just cancelling the dts, really youre doing the reverse chain rule but the effect is the same. Slightly neater
https://www.math-cs.gordon.edu/courses/mat225/handouts/sepvar.pdf

(edited 10 months ago)

Reply 2

RobertMills

The step you're referring to involves understanding the concept of separation of variables in differential equations. Let's break it down:
When you have the equation:
∫1/g(y)(dy/dt)=∫h(t)dt
This equation represents a separable differential equation, where the variables y and t can be separated on opposite sides of the equation. Here's how it works:

1.

Separation of Variables: The goal is to isolate y and t on separate sides of the equation. To do this, you're essentially integrating with respect to t on both sides. The left side has dy/dt and dt, so you can think of it as integrating dy with respect to y, and the right side is integrating dt with respect to t.

2.

Integrating with Respect to t: The integral of dy/dt with respect to t is simply y because the derivative of y with respect to t is dy/dt. So, ∫1/g(y)(dy/dt)becomes ∫1/g(y)dy. This step involves a change of variables from t to y, and the dt terms effectively cancel out.

3.

Integrating the Right Side: Similarly, integrating h(t) with respect to t on the right side just yields ∫h(t)dt.

So, by integrating both sides with respect to t, you're essentially "undoing" the derivative operation and isolating the variables y and t on separate sides of the equation. The dt terms cancel out because you're integrating with respect to t, and you're left with the integral of 1/g(y) with respect to y on the left side.
In summary, the cancellation of dt occurs because the integral undoes the derivative operation, and you're left with the integral of 1/g(y) with respect to y on the left side of the equation.

It is not the easy task to complete the questions related to differential equation. You need to practice it more and solve a lot of assignments to get complete command on it.

Reply 3

chavvo

7

Original post by RobertMills

The step you're referring to involves understanding the concept of separation of variables in differential equations. Let's break it down:
When you have the equation:
∫1/g(y)(dy/dt)=∫h(t)dt
This equation represents a separable differential equation, where the variables y and t can be separated on opposite sides of the equation. Here's how it works:

1.

Separation of Variables: The goal is to isolate y and t on separate sides of the equation. To do this, you're essentially integrating with respect to t on both sides. The left side has dy/dt and dt, so you can think of it as integrating dy with respect to y, and the right side is integrating dt with respect to t.

2.

Integrating with Respect to t: The integral of dy/dt with respect to t is simply y because the derivative of y with respect to t is dy/dt. So, ∫1/g(y)(dy/dt)becomes ∫1/g(y)dy. This step involves a change of variables from t to y, and the dt terms effectively cancel out.

3.

Integrating the Right Side: Similarly, integrating h(t) with respect to t on the right side just yields ∫h(t)dt.

So, by integrating both sides with respect to t, you're essentially "undoing" the derivative operation and isolating the variables y and t on separate sides of the equation. The dt terms cancel out because you're integrating with respect to t, and you're left with the integral of 1/g(y) with respect to y on the left side.
In summary, the cancellation of dt occurs because the integral undoes the derivative operation, and you're left with the integral of 1/g(y) with respect to y on the left side of the equation.

It is not the easy task to complete the questions related to differential equation. You need to practice it more and solve a lot of assignments to get complete command on it.

Is all this posted from an AI response or something?

The issue here isn't really expanding the process in laborious detail, it's actually in understanding why you can think of one side as "integrating with respect to y" and why the dt terms "effectively cancel out". Both these points were dealt with by mqb's response and the link he provided :smile:

Reply 4

babushka22

OP

7

Original post by mqb2766

Just denote 1/g(y) as f(y) say and suppose the integral is F(y), then you have
∫f(y) 𝑑𝑦/𝑑𝑡 𝑑𝑡 = F(y)
You could recognise the integrand
f(y) 𝑑𝑦/𝑑𝑡
is the chain rule on F so
dF/dt = dF/dy dy/dt = f(y) dy/dt
Then integrating both sides with respect to t gives the result youre after (first equation above).

Rather than just cancelling the dts, really youre doing the reverse chain rule but the effect is the same. Slightly neater
https://www.math-cs.gordon.edu/courses/mat225/handouts/sepvar.pdf

Thank you, I finally understand now. This was very helpful!

Help with differential equations please.

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