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# Worked example: identifying 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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• At Sal 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?
• 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.
• 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?
• 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.
• 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.
• 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.
• dy/dx = yx^−1
1/y dy= 1/x dx
In(y)= In(x)
• 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)

Hope this helps!

(Note: 'c' represents the constant, I started putting c because writing constant every time is pretty annoying).
• 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.
• "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 you please show your working until where you get stuck? It'll help me explain to you how to solve it.
• 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.
• 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?