Separable Differential Equations: general solution to dy/dx = -x/y

Samwise

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Hi,

I am new to differential equations so please go easy on me. I am stuck with a couple of questions for simple 1st order differential equations:

Q1) Find general solution to dy/dx = -x/y
Q2) Find general solution to dy/dx = - y/x and graph at least 2 solutions

My attempt:

Q1) y dy = -x dx (separate variables)
y2 / 2 + C1 = - x2 /2 + C2

y2 / 2 = - x2 /2 + C3 (where C3 = C2 - C1)

y = - x + C


I don't see what I have done wrong here but the solution apparently is circular solution (x2 + y2 = r2 where r2 = 2C) - I can't see anything invalid from what I have done?

Q2) 1/y dy = - 1/x dx

ln (y) = - ln(x) + C
ln (x) + ln (y) = C
ln (xy) = C
y = ec/x

I plotted graphs of e1 and e2 but I'm note sure what do graph when it is e1 or e2 divided by x? What would dividing by x do to a graph of e1 compared to plotting just e1?
Again this solution and graphing is also apparently wrong (but not as much as the last one) - the solution shows taking mod xy (|xy|) so the final solution shown is:

y = + eC / x

The graph for this solution shows hyperbolas symmetric about the y-axis which I understand since they have used the modulus. I really don't understand why they took the modulus? How am I supposed to know to take the modulus rather than what I did?

Any help offered is much appreciated!!!
 
Hi,

I am new to differential equations so please go easy on me. I am stuck with a couple of questions for simple 1st order differential equations:

Q1) Find general solution to dy/dx = -x/y
Q2) Find general solution to dy/dx = - y/x and graph at least 2 solutions

My attempt:

Q1) y dy = -x dx (separate variables)
y2 / 2 + C1 = - x2 /2 + C2

y2 / 2 = - x2 /2 + C (where C = C2 - C1) → Correct upto here

y2 + x2 = 2C = r2 (where 2C = r2 = constant)


y = - x + C ← Incorrect
I don't see what I have done wrong here but the solution apparently is circular solution (x2 + y2 = r2 where r2 = 2C) - I can't see anything invalid from what I have done?

Q2) 1/y dy = - 1/x dx

ln (y) = - ln(x) + C
ln (x) + ln (y) = C
ln (xy) = C
y = ec/x

I plotted graphs of e1 and e2 but I'm note sure what do graph when it is e1 or e2 divided by x? What would dividing by x do to a graph of e1 compared to plotting just e1?
Again this solution and graphing is also apparently wrong (but not as much as the last one) - the solution shows taking mod xy (|xy|) so the final solution shown is:

y = + eC / x

The graph for this solution shows hyperbolas symmetric about the y-axis which I understand since they have used the modulus. I really don't understand why they took the modulus? because

\(\displaystyle \displaystyle{\int \frac{dx}{x} = ln(|x|) + C}\) ←
ln(negative number) is undefined in real domain.


How am I supposed to know to take the modulus rather than what I did?

Any help offered is much appreciated!!!
.
 

Thanks for your help. I understand about the modulus with ln now. However, please could you explain how to go about plotting a graph of e1 /x. I know how to plot a graph of e1 but how does it change when dividing by x? Also in Q1 I can see the solution from my book the solution is a cicular equation but I don't know why? Why can't I multiple both sides by 2 and square root to get y = -x + C?

Thanks again.
 
Thanks for your help. Also in Q1 I can see the solution from my book the solution is a cicular equation but I don't know why? Why can't I multiple both sides by 2 and square root to get y = -x + C?

Thanks again.
because
\(\displaystyle \sqrt{x^2 + y^2} \ne x + y \)...... ...... (x + y)^2 = x^2 + y^2 + 2*x*y
 
...please could you explain how to go about plotting a graph of e1 /x....[?]
It works just like all the other rational functions you've graphed, starting back in algebra. (here) You find the domain, you plot points using x-values in that domain, and you sketch in the graph's curve. This particular function, y = e/x, works exactly like f(x) = 1/x, except that 1/x is multiplied by a constant. ;)
 
It works just like all the other rational functions you've graphed, starting back in algebra. (here) You find the domain, you plot points using x-values in that domain, and you sketch in the graph's curve. This particular function, y = e/x, works exactly like f(x) = 1/x, except that 1/x is multiplied by a constant. ;)

Thanks. Oh yes how silly of me - I was thinking e was changing with x for some reason. So if C is 1 then you get e1 which is 2.72. So for C =1 you would get y = 2.72/x or y = constant/x? So this would just give a graph of 1/x? Does the constant change the relationship? I was thinking not since x is still inversely proportional to y. So what does the constant in this situation versus a graph of y = 1/x with not e1 constant? What is it actually changing in the graph?
 
Thanks. Oh yes how silly of me - I was thinking e was changing with x for some reason. So if C is 1 then you get e1 which is 2.72. So for C =1 you would get y = 2.72/x or y = constant/x? So this would just give a graph of 1/x? Does the constant change the relationship? I was thinking not since x is still inversely proportional to y. So what does the constant in this situation versus a graph of y = 1/x with not e1 constant? What is it actually changing in the graph?

The basic shape of the graph will stay the same, being that of 1/x, but the specific values for any given x will change according to the constant. It seems like you're still just really overthinking it. If the general function is f(C, x) = eC/x, try graphing some simple base cases. For instance, try graphing f(0, x) = 1/x, f(1, x) = e/x and f(2, x) = e2/x. You'll find that these three graphs cannot possibly have the same graph. More specifically, you'll find that, for any given value of x, the values of f(1, x) must be e times greater than f(0, x). Similarly, the values of f(2, x) must be e times greater than f(1, x).

This, of course, makes sense when you think about it because you can pull the eC out front and get: f(C, x) = eC * 1/x
 
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