- Separable equations introduction
- Addressing treating differentials algebraically
- Separable differential equations
- Separable differential equations: find the error
- Worked example: separable differential equations
- Separable differential equations
- Worked example: identifying separable equations
- Identifying separable equations
- Identify separable equations
Separable equations can be written in the form dy/dx=f(x)g(y). See how we analyze various differential equations to see if they are separable.
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- At3:10Sal says #2 in the video cannot be written as a product of x and function of y, therefore its not separable. If the criteria for an equation to be separable is to isolate x and y as stated in previous videos, why does it have to be a product? In other words, why couldnt we manipulate #2 to be dy + y = dx - x + (1/2) and that be separable?(22 votes)
- I didn't understand this at first until my professor asked us to try it. In the case of the second problem, he simplifies it to dy/dx=-x-y+1/2 and you are correct that we can add the y term to both sides getting us y(dx)+dy=(-x+1/2)dx (I just reordered the terms in your equation and multiplied the dx last). Now with this form, y(dx)+dy=(-x+1/2)dx, try to integrate both sides and you will see that you can't. In fact, you would have to come up with a new mathematical proof for a way to solve this problem.(21 votes)
- In the second and fourth equations, Sal said that they didn't "feel separable" so they weren't separable. While this seems like a reasonable way to prove that an equation isn't separable, are there any more rigorous ways of proving that a differential equation is not separable?(4 votes)
- The idea is that when you have y' or dy/dx by itself on one side of the equation, you can divide both sides by all y termsand maybe a constant and not effect any of the xs.
Another way of putting it is that after getting the derivative on one side, you can put all xs and maybe a constant in one set of parenthesis and all ys with maube a constant in another set, even if those parenthesis might hold just an x or y.
I'm not sure if there is a simple method of determining if this is possible by just looking at the initial equation, I think you just have to try getting y' by itself and see if it fits the description.(4 votes)
- can't dy/dx(x+y)=x be (x+y)dy=xdx and hence be separable?(1 vote)
- The term 'separable' means that both variables will be isolated to their respective sides. However, in the second equation that you provide, even though the x is isolated on the right side, the left side has both variables x and y. Therefore, the differential equation is not separable.(7 votes)
- dy/dx = yx^−1
1/y dy= 1/x dx
please help i cannot continue(2 votes)
- After integrating both sides, you get:
ln(y) + constant = ln(x) + constant
A constant minus a constant is just another constant, so:
ln (y) = ln(x) + constant
Now you can put e in the base:
e^(ln (y)) = e^(ln(x) + c)
y = e^(ln(x) + c)
There's your solution,
Hope this helps!
(Note: 'c' represents the constant, I started putting c because writing constant every time is pretty annoying).(4 votes)
- So can someone summarize for me that when do the differential equations not separable?I think when there are y'' in certain equations and if y' times or plus any form of (x+y),that will not be separable.(3 votes)
- How can be solved an equation that is not separable? What method do I have to use?
Example: dy/dx = y+2(1 vote)
- I believe that that equation is separable. Although you can't see any function of x, I believe that 1 will count as a function of x. So, get dy and y+2 on the same side and get dx on the other side. Integrate both sides, and you should get your answer.(5 votes)
- "Solve the differential equation by separation of variables."
dy/dx = y/x
I'm having trouble solving this, when I separate I end up with the differentials on the bottom. Not sure how to integrate that.(1 vote)
- Can anyone help me figure out how to solve the differential equation (2+2y^2)y'=e^x(y) in terms of x? I'm super stuck!!(1 vote)
- The equation is separable, so you can move the y from the RHS to the LHS. Note that y' = dy/dx.
If you're still stuck, could you show your working until where you get stuck?(3 votes)
- I'm quite bothered by the last one. can we gather up the xy' with the x in the LHS and then move the yy' to the RHS, leaving us with:
(x-1)dy/dx = -y dy/dx
hence y= 1-x ?
can I call that separable?(1 vote)
- 𝑥 ∙ 𝑑𝑦∕𝑑𝑥 + 𝑦 ∙ 𝑑𝑦∕𝑑𝑥 = 𝑥
is not a separable equation.
𝑦 = 1 − 𝑥 is not a solution, because then 𝑑𝑦∕𝑑𝑥 = −1, and as we plug that into the equation we get
𝑥 ∙ (−1) + (1 − 𝑥) ∙ (−1) = 𝑥 ⇒ 𝑥 = −1,
which is obviously not true for all 𝑥.(2 votes)
- show that dy\dx=2xy-2x=y-1 is a separable equation(0 votes)
- [Sal] Which of the differential equations are separable? And I encourage you to pause this video and see which of these are actually separable. Now, the way that I approach this is I try to solve for the derivative, and if when I solve for the derivative, if I get dy, dx is equal to some function of y times some other function of x then I say okay, this is separable. 'Cause I could rewrite this as, I could divide both sides by g of y, and I get one over g of y, which is itself a function of y, times dy is equal to h of x dx. You would go from this first equation to the second equation just by dividing both sides by g of y and multiplying both sides by dx. And then it's clear you have a separable equation you can integrate both sides. But the key is let's solve for the derivative and see if we can put this in a form where we have the product of a function of y times a function of x. So let's do it with this first one here. So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x. I'm gonna get x times, I'll write y prime is the derivative of y with respect to x, is equal to 3 minus y. So I subtracted y from both sides. Let's see if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually I'm gonna write it this way. I'm gonna write it 3 minus y times 1 over x. And so it's clear I'm able to write the derivative as a product of a function of y, and a function of x. So this indeed is separable. And I could show you, I could multiply both sides by dx and I could divide both sides by 3 minus y now, and I would get 1 over 3 minus y dy is equal to 1 over x dx. So clearly, this one right over here is separable. Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a slightly different colour so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the 2x, the 2y from both sides, so actually, lemme just do, whoops, lemme do a couple things at once. I'm gonna subtract 2x from both sides. I am going to subtract 2y from both sides. So I'm gonna subtract 2y from both sides. I'm gonna add one to both sides, so I'm gonna add one to both sides. And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have 2 times the derivative of y with respect to x is equal to negative 2x minus 2y plus 1. And now see I can divide everything by two; I would get the derivative of y with respect to x is equal to, and actually, yeah I would get, I'm just gonna divide by two, so I'm gonna get negative x minus y and then I'm going to get plus 1/2. So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y, so this one I'm not gonna, I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating I can rewrite this as, well this is dy dx. If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y dy is equal to x squared plus x dx. So clearly separable. Alright now this last choice, this is interesting, they've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y, is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x over x plus y. And here my algebraic tool kit of how do I separate x and y so I can write this as a function of x times a function of y, not obvious to me here. So this one is not separable. So only the first one and the third one.