dy/dx=(y-1)(y-2) Prove that if y = f(x), x ∈ R, is a solution of (1), then y = f(x + c) is also a solution, where c ∈ R is constant.

To prove that y = f(x + c) is also a solution of the differential equation dy/dx=(y-1)(y-2), we need to show that:

(d/dx)f(x+c) = (f(x+c)-1)(f(x+c)-2)

To do this, we can use the chain rule of differentiation:

(d/dx)f(x+c) = df/d(x+c) * d(x+c)/dx

Since d(x+c)/dx = 1, we can simplify this to:

(d/dx)f(x+c) = df/d(x+c)

Now, let's substitute y = f(x) into the differential equation:

dy/dx = (y-1)(y-2)

Substituting x + c for x and f(x + c) for y, we get:

df/d(x+c) = (f(x+c)-1)(f(x+c)-2)

This is the same as what we need to prove, so we have shown that y = f(x + c) is also a solution of the differential equation.

相关问题

Let M be a set and let X, Y, Z, W ⊂ M. We define the symmetric difference: X △ Y := (X − Y ) ∪ (Y − X) (i) (1 pt) Show that X △ Y = (X ∪ Y ) − (X ∩ Y ). (ii) (1 pt) Show that (M − X) △ (M − Y ) = X △ Y . (iii) (1 pt) Show that the symmetric difference is associative, i.e., (X △ Y ) △ Z = X △ (Y △ Z). (iv) (1 pt) Show that X ∩ (Y △ Z) = (X ∩ Y ) △ (X ∩ Z). (v) (1 pt) Show that X △ Y = Z △ W iff X △ Z = Y △ W. (vi) (1 pt) Indicate the region of X △ Y △ Z in a Venn diagram. (vii) (1 pt) Sketch a Venn diagram for 4 distinct sets.

(i) To show that X △ Y = (X ∪ Y ) − (X ∩ Y ), we need to prove two inclusions:

First, suppose that x ∈ X △ Y. Then either x ∈ X but x ∉ Y or x ∈ Y but x ∉ X. In either case, we have x ∈ X ∪ Y but x ∉ X ∩ Y, so x ∈ (X ∪ Y ) − (X ∩ Y ). Therefore, X △ Y ⊆ (X ∪ Y ) − (X ∩ Y ).

Conversely, suppose that x ∈ (X ∪ Y ) − (X ∩ Y ). Then x ∈ X ∪ Y and x ∉ X ∩ Y. If x ∈ X, then x ∉ Y, so x ∈ X △ Y. Otherwise, x ∈ Y but x ∉ X, so x ∈ Y − X, and therefore x ∈ X △ Y. Therefore, (X ∪ Y ) − (X ∩ Y ) ⊆ X △ Y.

Combining both inclusions, we get X △ Y = (X ∪ Y ) − (X ∩ Y ).

(ii) To show that (M − X) △ (M − Y ) = X △ Y, we can use the result from part (i) twice:

(M − X) △ (M − Y ) = ((M − X) ∪ (M − Y )) − ((M − X) ∩ (M − Y )) = ((M ∩ Y ) − X) ∪ ((M ∩ X) − Y )

= (X ∪ Y ) − (X ∩ Y ) = X △ Y.

(iii) To show that the symmetric difference is associative, we need to prove that (X △ Y ) △ Z = X △ (Y △ Z).

Using the definition of symmetric difference, we have

(X △ Y ) △ Z = ((X − Y ) ∪ (Y − X)) △ Z = (((X − Y ) ∪ (Y − X)) − Z) ∪ (Z − ((X − Y ) ∪ (Y − X)))

= ((X − Y ) ∩ (Z − Y )) ∪ ((Y − X) ∩ (Z − X)),

and

X △ (Y △ Z) = X △ ((Y − Z) ∪ (Z − Y)) = ((X − (Y − Z)) ∪ ((Y − Z) − X)) ∪ ((Y ∩ Z) − (X ∪ (Y − Z)))

= ((X − Y ) ∪ ((Y − Z) ∩ (X − Z))) ∪ (((Y ∩ Z) ∩ X) − (Y − Z)).

To show that these sets are equal, it suffices to show that each is a subset of the other.

Suppose that x ∈ (X △ Y ) △ Z. Then either x ∈ (X − Y ) ∩ (Z − Y ) or x ∈ (Y − X) ∩ (Z − X). In the first case, we have x ∈ X − Y and x ∈ Z − Y, so x ∈ X − Y ⊆ X. Since x ∈ Z − Y, we have x ∉ Y, so x ∈ X △ Y. Similarly, in the second case, we have x ∈ X − Y, x ∉ Z, and x ∈ Y △ Z, so again x ∈ X △ (Y △ Z).

Conversely, suppose

Let x, y ∈ R, show that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0)

To prove that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0), we need to prove both directions of the equivalence:

First, let's prove that if ∀x∃y(xy = 0), then ∃y∀x(xy = 0):

Assume ∀x∃y(xy = 0). We want to show that there exists a y such that for all x, xy = 0.

Let's choose an arbitrary value of y such that for any x, either xy = 0 (if x ≠ 0) or y = 0 (if x = 0). Such a value of y exists since for any nonzero x, there exists a y = 0 such that xy = 0, and for x = 0, any y would satisfy xy = 0.

Now, we need to show that this y satisfies ∀x(xy = 0):

Take an arbitrary x. If x = 0, then xy = 0 (since y can be any value). If x ≠ 0, then y was chosen so that xy = 0. Therefore, in either case, xy = 0, and we have shown that ∃y∀x(xy = 0).

Now let's prove the other direction, that if ∃y∀x(xy = 0), then ∀x∃y(xy = 0):

Assume ∃y∀x(xy = 0). We want to show that for any x, there exists a y such that xy = 0.

Take an arbitrary x. Since ∃y∀x(xy = 0), there exists a y such that for all x, xy = 0. In particular, this y satisfies xy = 0 for our chosen value of x. Therefore, we have shown that for any x, there exists a y such that xy = 0, and thus we have proved ∀x∃y(xy = 0).

Since we have proved both directions of the equivalence, we can conclude that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0).