dy/dx = sqrt(y/x) Define u = sqrt(y/x). Note that, by our given, this also means that dy/dx = u. Then square both sides of that definition u^2 = y/x Multiply by x u^2 x = y Differentiate both sides (using the product rule) 2ux du/dx + u^2 = dy/dx Note that dy/dx = u, so use that on the RHS to remove the dy/dx term. 2ux du/dx + u^2 = u u is a common factor, so split into two cases. Case 1: u = 0 Then 0 = sqrt(y/x) 0 = y/x y = 0 Case 2: u =/= 0 So we can divide by u 2x du/dx + u = 1 Multiply by 1/(2 sqrt(x)) (I did some work off screen to find this, but it's just an integrating factor) sqrt(x) du/dx + 1/(2 sqrt(x)) u = 1/(2 sqrt(x)) Note that the LHS is the derivative of u sqrt(x) d/dx (u sqrt(x)) = 1/(2 sqrt(x)) Integrate both sides u sqrt(x) = sqrt(x) + c We defined u as u = sqrt(y/x), so u sqrt(x) = sqrt(y) sqrt(y) = sqrt(x) + c y = x + 2 sqrt(x) c + c^2 Solutions are y = 0 or y = x + 2 sqrt(x) c + c^2
Applied math ruins the pleasure of finding the method for finding an annoying answer. It is just belong to the subject were it is applied no more (physics,engineering, chemistry, etc...)
Remember: Newton invented calculus so that he could do physics! Math without application is like dry humping---it may feel good but what does it really accomplish?
Nice problem! However, your solution (to the 1st method) is too simplistic for four reasons. 1. We see in the original equation that x cannot be zero. This splits the solutions into two: one for positive x and another for negative x. Each of these may have multiple solutions (and they do), so the full solution would say that you can choose one equation for y for x < 0 and choose one for x > 0 and these equations may or may not be the same. 2. You wrote that sqrt(y/x) = sqrt(y) / sqrt(x). This is true if x is positive and y is non-negative. But you left out the possibility that both x and y are negative. In that case, sqrt(y/x) = sqrt(-y) / sqrt(-x). If you continue as in your video, you get the additional solution y = -(sqrt(-x) + D)^2 for some real constant D. 3. When you separated the variables, you divided both sides of the equation by sqrt(y). But that is not possible if y is zero. So you left out the additional solution y = 0 4. You squared both sides of the equation sqrt(y) = sqrt(x) + C. But that could and does introduce an extraneous solution. In particular, we need sqrt(x) + C >= 0 Otherwise, the "solution" does not work. Here is a more full solution to your problem. For x > 0 there are two solutions, where both C and D are real constants: y = 0 and y = (sqrt(x) + C)^2 where sqrt(x) >= -C. (If C is non-negative that is a null restriction.) For x < 0 there are two solutions: y = 0 and y = -(sqrt(-x) + D)^2 where sqrt(-x) >= -D. (If D is non-negative that is a null restriction.) A solution that works for all non-zero x is to choose one for each case.
@@TH-cam_username_not_found That would make y(x) exist and continuous for all positive x. However, I am not sure that the two-sided derivative of y would exist at x = C^2. If it doesn't, that would prevent this from being a solution at that point. I'm not able to check that right now.
boss, your method is an overkill. no one does calculus the way you suggest. are you saying that even to differentiate an entity such as 1/x, one has to consider x being 0, x being -ve, x being +ve, and so on? if such hairsplitting is necessary, no progress would ever have been made in engineering or in applied sciences. this notion of 'separation of variables' is something that every calculus student does without worrying whether the denominator is ever likely to be 0. i'm curious as to what's your math background.
Someone isn't posting his detailed analysis anymore so I will do that. Find all functions y: 𝔻→ℝ such that for all x in 𝔻 ; y'(x) = sqrt(y(x)/x) where 𝔻 is any subset of ℝ One 1st notes the restrictions caused by the equation. 1. x is in the denominator so 0 is not in 𝔻 . 2. y(x)/x is inside sqrt so y(x)/x >= 0 . Thus, if x > 0 then y(x) >= 0 and if x < 0 then y(x) = 0 for all a, so by the equation, y'(x) >= 0 . When x > 0 , divide both sides by 2sqrt(y(x)). And, when x < 0 , divide both sides by -2sqrt(-y(x)) but this assumes that y(x) is nonzero so we exclude the values that sets y(x) to zero. we will call the set of these values S. The new equations would be the following: for all x > 0 in 𝔻\S ; y'(x)/(2sqrt(y(x))) = 1/(2sqrt(x)) and for all x < 0 in 𝔻\S ; -y'(x)/(2sqrt(-y(x))) = -1/(2sqrt(-x)) One then recalls the formula for the derivative of sqrt(f(x)) , that the sum of derivatives is the derivative of the sum, and that the derivative of a constant is 0 to conclude that for all x > 0 in an interval of 𝔻\S ; sqrt(y(x)) = sqrt(x) + a and for all x < 0 in an interval of 𝔻\S ; sqrt(-y(x)) = sqrt(-x) + b One should then notice the new restrictions. When x > 0, sqrt(x) + a >= 0. If a >= 0 then no problem. But if a < 0 then x >= (-a)^2. Similarly, when x < 0, sqrt(-x) + b >= 0. If b >= 0 then no problem. But if b < 0 then -x >= (-b)^2 ie. x a^2 and x < -b^2. Back to the equations. rearranging a bit will give us the following: for all x > 0 in an interval of 𝔻\S ; y(x) = (sqrt(x) + a)^2 and for all x < 0 in an interval of 𝔻\S ; y(x) = -(sqrt(-x) + b)^2 It is important to keep in mind that the constants in both cases may not be the same. After all, the derivative of the piecewise function defined by f(x) = a for x > x0 and f(x) = b for x < x0 is the zero function. If the domain is composed of n intervals, there would be n independent constants, one to each interval. The next step is to find the set S. It could be the case that S is the whole domain or in other words, that y(x) = 0 for all x in 𝔻 but this is the trivial case. We will consider now the non-trivial case which is that for s in S, the values arround s do not all set y(x) to 0, that is to say, y(x) is equal to one of the expressions we found somewhere around s. One uses the fact that y is continuous because it is differentiable, ie. lim (x→s) y(x) = y(s) where s is in S, ie lim (x→s) y(x) = y(s). Of course , y(s) = 0 , and we substitute y(x) in the limit by the expressions we found. There would be 2 cases to treat seperately. When x > 0, y(x) = (sqrt(x) + a)^2 and the question would be to find s such that lim (x→s) (sqrt(x) + a)^2 = 0. Since (sqrt(x) + a)^2 is continuous, the question is equivalent to finding s such that (sqrt(s) + a)^2 = 0. If a >= 0 then there is no s (reminder: 0 is not in 𝔻). If however a < 0 then s = (-a)^2. When x < 0, y(x) = -(sqrt(-x) + b)^2 and the question would be to find s such that lim (x→s) -(sqrt(-x) + b)^2 = 0. Since -(sqrt(-x) + b)^2 is continuous, the question is equivalent to finding s such that -(sqrt(-s) + b)^2 = 0. If b >= 0 then there is no s. If however a < 0 then s = -(-b)^2. Now here is where the magic happens: For the case where a and b are negative, y(x) is equal to the expressions (sqrt(x) + a)^2 and -(sqrt(-x) + b)^2 only for x > a^2 and x < -b^2 respectively but we can extend y beyond those values by setting y(x) = 0 for x = -b^2 (reminder: x can't be 0) which is to say there exist piecewise solutions! Nothing is left but to verify that all restrictions are fulfilled. For x > 0; y(x) = (sqrt(x) + a)^2 or y(x) = 0 which are both and y'(x) = 2(sqrt(x) + a)/(2sqrt(x)) >= 0 For x < 0; y(x) = -(sqrt(-x) + b)^2 = -a1^2 or if a1 >= 0, y(x) = -(sqrt(-x) + a1)^2 for all x. for x in [-1,0[ , if a2 < 0, y(x) = -(sqrt(-x) + a2)^2 if x < -a2^2 and y(x) = 0 if x >= -a2^2 or if a2 >= 0, y(x) = -(sqrt(-x) + a2)^2 for all x. for x in ]1,3[ , if a3 < 0, y(x) = (sqrt(x) + a3)^2 if x > a3^2 and y(x) = 0 if x = 0, y(x) = -(sqrt(-x) + a1)^2 for all x. That was a lot. If you read all of this, Thank you 😄!!
Why not squaring both sides.then it will be y'/y=1/x when integrated gives ln(y)=ln(x)+c. We write it as ln(y)= ln(x)+ln(k) where ln(k)=c. Or ln(y)=ln(nk) or y= nk. Is it correct?
dy/dx = sqrt(y/x)
Define u = sqrt(y/x). Note that, by our given, this also means that dy/dx = u.
Then square both sides of that definition
u^2 = y/x
Multiply by x
u^2 x = y
Differentiate both sides (using the product rule)
2ux du/dx + u^2 = dy/dx
Note that dy/dx = u, so use that on the RHS to remove the dy/dx term.
2ux du/dx + u^2 = u
u is a common factor, so split into two cases.
Case 1: u = 0
Then 0 = sqrt(y/x)
0 = y/x
y = 0
Case 2: u =/= 0
So we can divide by u
2x du/dx + u = 1
Multiply by 1/(2 sqrt(x)) (I did some work off screen to find this, but it's just an integrating factor)
sqrt(x) du/dx + 1/(2 sqrt(x)) u = 1/(2 sqrt(x))
Note that the LHS is the derivative of u sqrt(x)
d/dx (u sqrt(x)) = 1/(2 sqrt(x))
Integrate both sides
u sqrt(x) = sqrt(x) + c
We defined u as u = sqrt(y/x), so u sqrt(x) = sqrt(y)
sqrt(y) = sqrt(x) + c
y = x + 2 sqrt(x) c + c^2
Solutions are y = 0 or y = x + 2 sqrt(x) c + c^2
Take C to be negative and y(x) = 0 when sqrt(x) = -C.
Applied math ruins the pleasure of finding the method for finding an annoying answer. It is just belong to the subject were it is applied no more (physics,engineering, chemistry, etc...)
Remember: Newton invented calculus so that he could do physics! Math without application is like dry humping---it may feel good but what does it really accomplish?
Nice problem! However, your solution (to the 1st method) is too simplistic for four reasons.
1. We see in the original equation that x cannot be zero. This splits the solutions into two: one for positive x and another for negative x. Each of these may have multiple solutions (and they do), so the full solution would say that you can choose one equation for y for x < 0 and choose one for x > 0 and these equations may or may not be the same.
2. You wrote that sqrt(y/x) = sqrt(y) / sqrt(x). This is true if x is positive and y is non-negative. But you left out the possibility that both x and y are negative. In that case, sqrt(y/x) = sqrt(-y) / sqrt(-x). If you continue as in your video, you get the additional solution
y = -(sqrt(-x) + D)^2 for some real constant D.
3. When you separated the variables, you divided both sides of the equation by sqrt(y). But that is not possible if y is zero. So you left out the additional solution
y = 0
4. You squared both sides of the equation sqrt(y) = sqrt(x) + C. But that could and does introduce an extraneous solution. In particular, we need
sqrt(x) + C >= 0
Otherwise, the "solution" does not work.
Here is a more full solution to your problem.
For x > 0 there are two solutions, where both C and D are real constants:
y = 0 and
y = (sqrt(x) + C)^2 where sqrt(x) >= -C. (If C is non-negative that is a null restriction.)
For x < 0 there are two solutions:
y = 0 and
y = -(sqrt(-x) + D)^2 where sqrt(-x) >= -D. (If D is non-negative that is a null restriction.)
A solution that works for all non-zero x is to choose one for each case.
Try this solution:
Take C to be negative and y(x) = 0 when sqrt(x) = -C. Similar thing for x < 0.
@@TH-cam_username_not_found That would make y(x) exist and continuous for all positive x. However, I am not sure that the two-sided derivative of y would exist at x = C^2. If it doesn't, that would prevent this from being a solution at that point. I'm not able to check that right now.
@@rorydaulton6858 It does.
boss, your method is an overkill. no one does calculus the way you suggest. are you saying that even to differentiate an entity such as 1/x, one has to consider x being 0, x being -ve, x being +ve, and so on? if such hairsplitting is necessary, no progress would ever have been made in engineering or in applied sciences. this notion of 'separation of variables' is something that every calculus student does without worrying whether the denominator is ever likely to be 0. i'm curious as to what's your math background.
@@rorydaulton6858 I didn't know he is your boss! 😂
Someone isn't posting his detailed analysis anymore so I will do that.
Find all functions y: 𝔻→ℝ such that for all x in 𝔻 ; y'(x) = sqrt(y(x)/x) where 𝔻 is any subset of ℝ
One 1st notes the restrictions caused by the equation.
1. x is in the denominator so 0 is not in 𝔻 .
2. y(x)/x is inside sqrt so y(x)/x >= 0 . Thus, if x > 0 then y(x) >= 0 and if x < 0 then y(x) = 0 for all a, so by the equation, y'(x) >= 0 .
When x > 0 , divide both sides by 2sqrt(y(x)). And, when x < 0 , divide both sides by -2sqrt(-y(x)) but this assumes that y(x) is nonzero so we exclude the values that sets y(x) to zero. we will call the set of these values S.
The new equations would be the following: for all x > 0 in 𝔻\S ; y'(x)/(2sqrt(y(x))) = 1/(2sqrt(x)) and for all x < 0 in 𝔻\S ; -y'(x)/(2sqrt(-y(x))) = -1/(2sqrt(-x))
One then recalls the formula for the derivative of sqrt(f(x)) , that the sum of derivatives is the derivative of the sum, and that the derivative of a constant is 0 to conclude that
for all x > 0 in an interval of 𝔻\S ; sqrt(y(x)) = sqrt(x) + a
and for all x < 0 in an interval of 𝔻\S ; sqrt(-y(x)) = sqrt(-x) + b
One should then notice the new restrictions. When x > 0, sqrt(x) + a >= 0. If a >= 0 then no problem. But if a < 0 then x >= (-a)^2. Similarly, when x < 0, sqrt(-x) + b >= 0. If b >= 0 then no problem. But if b < 0 then -x >= (-b)^2 ie. x a^2 and x < -b^2.
Back to the equations. rearranging a bit will give us the following:
for all x > 0 in an interval of 𝔻\S ; y(x) = (sqrt(x) + a)^2 and
for all x < 0 in an interval of 𝔻\S ; y(x) = -(sqrt(-x) + b)^2
It is important to keep in mind that the constants in both cases may not be the same. After all, the derivative of the piecewise function defined by f(x) = a for x > x0 and f(x) = b for x < x0 is the zero function. If the domain is composed of n intervals, there would be n independent constants, one to each interval.
The next step is to find the set S. It could be the case that S is the whole domain or in other words, that y(x) = 0 for all x in 𝔻 but this is the trivial case. We will consider now the non-trivial case which is that for s in S, the values arround s do not all set y(x) to 0, that is to say, y(x) is equal to one of the expressions we found somewhere around s.
One uses the fact that y is continuous because it is differentiable, ie. lim (x→s) y(x) = y(s) where s is in S, ie lim (x→s) y(x) = y(s). Of course , y(s) = 0 , and we substitute y(x) in the limit by the expressions we found. There would be 2 cases to treat seperately.
When x > 0, y(x) = (sqrt(x) + a)^2 and the question would be to find s such that
lim (x→s) (sqrt(x) + a)^2 = 0. Since (sqrt(x) + a)^2 is continuous, the question is equivalent to finding s such that (sqrt(s) + a)^2 = 0. If a >= 0 then there is no s (reminder: 0 is not in 𝔻). If however a < 0 then s = (-a)^2.
When x < 0, y(x) = -(sqrt(-x) + b)^2 and the question would be to find s such that
lim (x→s) -(sqrt(-x) + b)^2 = 0. Since -(sqrt(-x) + b)^2 is continuous, the question is equivalent to finding s such that -(sqrt(-s) + b)^2 = 0. If b >= 0 then there is no s. If however a < 0
then s = -(-b)^2.
Now here is where the magic happens: For the case where a and b are negative, y(x) is equal to the expressions (sqrt(x) + a)^2 and -(sqrt(-x) + b)^2 only for x > a^2 and x < -b^2 respectively but we can extend y beyond those values by setting y(x) = 0 for x = -b^2 (reminder: x can't be 0) which is to say there exist piecewise solutions!
Nothing is left but to verify that all restrictions are fulfilled.
For x > 0; y(x) = (sqrt(x) + a)^2 or y(x) = 0 which are both and y'(x) = 2(sqrt(x) + a)/(2sqrt(x)) >= 0
For x < 0; y(x) = -(sqrt(-x) + b)^2 = -a1^2 or if a1 >= 0, y(x) = -(sqrt(-x) + a1)^2 for all x.
for x in [-1,0[ , if a2 < 0, y(x) = -(sqrt(-x) + a2)^2 if x < -a2^2 and y(x) = 0 if x >= -a2^2 or if a2 >= 0, y(x) = -(sqrt(-x) + a2)^2 for all x.
for x in ]1,3[ , if a3 < 0, y(x) = (sqrt(x) + a3)^2 if x > a3^2 and y(x) = 0 if x = 0, y(x) = -(sqrt(-x) + a1)^2 for all x.
That was a lot. If you read all of this, Thank you 😄!!
I forgot that one should verify the differentiability of the piecewise function, by computing the right-sided and left-sided limits at the points s.
Why don't you put ABS under ln?
For ex., integral(dx/x) = ln|x|, NOT ln(x).
y=x(1+c/√x)^2
😊😊😊 👏🎉👏🎉👏
I used the first method.
Why not squaring both sides.then it will be y'/y=1/x when integrated gives ln(y)=ln(x)+c. We write it as ln(y)= ln(x)+ln(k) where ln(k)=c. Or
ln(y)=ln(nk) or y= nk. Is it correct?
There is no root on the LHS