Separation of variables is a common method for solving differential equations. Learn how it's done and why it's called this way.
Log in Bryan 4 years agoPosted 4 years ago. Direct link to Bryan's post “How can we say that we're...” How can we say that we're "integrating" both sides, if we're really just adding two long s, then pretending that the dy and dx we're part of the integration operator notation in the first place? How is this valid? Also, you're integrating with respects to two different variables. How does this keep the equality? • (16 votes) Jerry Nilsson 4 years agoPosted 4 years ago. Direct link to Jerry Nilsson's post “It's easier to see if we ...” It's easier to see if we work our way backwards. Let 𝑔(𝑦) = 𝑓(𝑥) + 𝐶 Since these two functions are equal, that implicitly states that 𝑦 is a function of 𝑥, and we can write Also, since the functions are equal, the slopes of their tangent lines at any point must also be equal. 𝑦 = ℎ(𝑥) ⇒ 𝑑𝑦∕𝑑𝑥 = ℎ'(𝑥), and we can write Dividing both sides by 𝑔'(𝑦) we get the separable differential equation To conclude, a separable equation is basically nothing but the result of implicit differentiation, and to solve it we just reverse that process, namely take the antiderivative of both sides. (30 votes) kungfu303 7 years agoPosted 7 years ago. Direct link to kungfu303's post “How do we know what metho...” How do we know what method to choose to solve a differential equations (whether that be by u-substitution, by parts, or by separating the variables)? • (6 votes) Wilson 7 years agoPosted 7 years ago. Direct link to Wilson's post “Separable equations have ...” Separable equations have dy/dx (or dy/dt) equal to some expression. U-substitution is when you see an expression within another (think of the chain rule) and also see the derivative. For example, 2x/(x^2+1), you can see x^2+1 as an expression within another (1/x) and its derivative(2x). Solving by parts is when you see something you can simplify when deriving or integrating by itself. Usually it is deriving something simple like x, or something you can't integrate easily like ln(x). jb3154324 6 months agoPosted 6 months ago. Direct link to jb3154324's post “What if the differential ...” What if the differential equation isn’t separable? How would we find the general/particular solution, y=f(x). Also, is implicit integration possible? • (5 votes) kubleeka 6 months agoPosted 6 months ago. Direct link to kubleeka's post “Differential equations is...” Differential equations is an entire subfield of mathematics in it's own right. If an equation isn't separable, there are dozens of other techniques you might throw at it. Also, yes, if you have an equation, you can integrate both sides, so long as you include a constant of integration on one side. (5 votes) Michael Nich 4 years agoPosted 4 years ago. Direct link to Michael Nich's post “Why when you take the int...” Why when you take the integral of "y" in the left side there is no C? • (4 votes) Iron Programming 4 years agoPosted 4 years ago. Direct link to Iron Programming's post “Whenever we have two cons...” Whenever we have two constants C1 and C2, we can combine them to form one single constant C. So given x + C1 = y + C2 we can simplify by writing y = x + C. Happy learning! :-) (6 votes) lily. 7 months agoPosted 7 months ago. Direct link to lily.'s post “Why do we need to take C ...” Why do we need to take C into account when solving the equation if it can be any arbitrary constant. Like if the answer should be y=10/(e^x+c), why can't the answer be y=10/(e^x)+c? • (3 votes) Venkata 7 months agoPosted 7 months ago. Direct link to Venkata's post “Even though it is arbitra...” Even though it is arbitrary, it needs to be associated with the correct term. c is added to e^(x), irrespective of what it is, not to the whole fraction. So, we add it it e^x and don't pull it out (4 votes) Shehjad Islam 4 years agoPosted 4 years ago. Direct link to Shehjad Islam's post “How do we describe a natu...” How do we describe a natural phenomenon through a differential equation? Is there any video about that? • (2 votes) curiousfermions 4 years agoPosted 4 years ago. Direct link to curiousfermions's post “The introductory ones in ...” The introductory ones in this unit deal with that. Even watering your plants can be written in differential equation! I can not attest it myself but I have heard that university-level physics is all about calculus and physics is basically the part of science that is trying to describe all natural phenomenons.Doing the problems in writing differential equations in this unit will give you a primary idea. (6 votes) Viktor Edlund 5 years agoPosted 5 years ago. Direct link to Viktor Edlund's post “In some cases i have been...” In some cases i have been taught that an derivative dy/dx can't be treated as an ratio of the variables dy and dx. But in other cases like this one they are treated just like a ratio. In this example above you can multiply and divide dy and dx. Why is that? I asked my teachers and they say that its not included in the course but im curious. kind regards, • (2 votes) Marie Bethell 5 years agoPosted 5 years ago. Direct link to Marie Bethell's post “Khan made a video on it. ...” Khan made a video on it. https://www.khanacademy.org/math/ap-calculus-ab/ab-differential-equations-new/ab-7-6/v/addressing-treating-differentials-algebraically (3 votes) paxgole 3 years agoPosted 3 years ago. Direct link to paxgole's post “If I had a step something...” If I had a step something like: (-y = -x + C), why does it become (y = x + C) and not (y = x - C)? • (1 vote) Mazhar Momin 3 years agoPosted 3 years ago. Direct link to Mazhar Momin's post “If you had a step like: _...” If you had a step like: (-y = -x + C), Now, let's multiply both the sides by (-1): Distributing the (-1) on the R.H.S. gives: Here, the constant (C), when multiplied by (-1) will give another constant; which can be denoted by (C) again. In the expression, (-y = -x + C) Thus, (3 votes) john harrington 3 years agoPosted 3 years ago. Direct link to john harrington's post “dy/dx=-xySolve separati...” dy/dx=-xy Solve separation of variables • (1 vote) Jerry Nilsson 3 years agoPosted 3 years ago. Direct link to Jerry Nilsson's post “𝑑𝑦∕𝑑𝑥 = −𝑥𝑦Multip...” 𝑑𝑦∕𝑑𝑥 = −𝑥𝑦 Multiplying both sides by (1∕𝑦)𝑑𝑥, we get Integrating both sides, we get Solving for 𝑦, we get However, because 𝐶 is an arbitrary constant, then ±𝑒^𝐶 is also an arbitrary constant. (3 votes) davidflores 9 months agoPosted 9 months ago. Direct link to davidflores's post “what is the porpise of C?” what is the porpise of C? • (1 vote) Brandon 9 months agoPosted 9 months ago. Direct link to Brandon's post “The derivative of "2x + 3...” The derivative of "2x + 3" is 2. When we try to reverse this by finding the integral of 2, we initially get "2x". Notice how the "+ 3" is missing? So we need to add C, which is a variable that represents a constant value, giving us the result "2x + C". (2 votes)Want to join the conversation?
𝑔(ℎ(𝑥)) = 𝑓(𝑥) + 𝐶
In other words, their derivatives are equal:
𝑔'(ℎ(𝑥)) ∙ ℎ'(𝑥) = 𝑓 '(𝑥)
𝑔'(𝑦) ∙ 𝑑𝑦∕𝑑𝑥 = 𝑓 '(𝑥)
𝑑𝑦∕𝑑𝑥 = 𝑓 '(𝑥)∕𝑔'(𝑦)
At first this may seem like cheating but really all we are saying is C1 and C2 are two mysterious unknown quantities, but we can think of C1 + C2 as 1 mysterious unknown as quantity, and we still haven't changed the value of the equation.
Thank you.
Viktor
He treats the dx's and dy's as super small changes in x and y, which makes them values that can be multiplied and divided. Keep in mind though that he does say it's a rather hand-wavy method, not too rigorous.
let's rewrite this as: (-y) = (-x) + (C)
(-1)*(-y) = (-1)*[(-x) + (C)]
(-1)*(-y) = [(-1)*(-x)] + [(-1)*(C)]
We don't know the value of the constant (C). It can can be either positive or negative. So, multiplying with (-1) can give respective result. It's safe to keep the '+' sign in the original expression intact and the constant (C) in parenthesis.
y = x + C
(1∕𝑦)𝑑𝑦 = −𝑥𝑑𝑥
ln |𝑦| = −𝑥²∕2 + 𝐶, where 𝐶 is an arbitrary constant.
𝑦 = ±𝑒^(−𝑥²∕2 + 𝐶) = ±𝑒^(−𝑥²∕2)⋅𝑒^𝐶
Thus, we can write 𝑦 = 𝐶𝑒^(−𝑥²∕2)
