# The general solution to the differential equation dy/dx=xy is y=±√(x^2+C). Let y=f(x) be the particular solution to the differential equation with the initial condition f(−5)=−4. What is an expression for f(x) and its domain?

## Well, you see, when it comes to differential equations, I'm more of a jokester than a mathematician. But hang on tight and let me give it a try!

If we have the differential equation dy/dx = xy, and the general solution y = ±√(x^2+C), we need to find a particular solution y = f(x) that satisfies the initial condition f(-5) = -4.

Now, let's start with the general solution and substitute the initial condition to solve for the constant C. Plugging in x = -5 and y = -4, we have:

-4 = ±√((-5)^2 + C)

Now, let's isolate the right side of the equation:

(-4)^2 = ((-5)^2 + C)^2

16 = (25 + C)

Subtracting 25 from both sides:

C = -9

So, the particular solution is y = ±√(x^2 - 9). But hold your laughter, we're not done yet.

Now, to find the expression for f(x) and its domain, we need to choose one of the square root branches. Let's take the positive branch, so we have:

f(x) = √(x^2 - 9)

As for the domain, we need to look out for any negative values inside the square root. Since x^2 - 9 is only defined for x ≤ -3 and x ≥ 3, the domain for f(x) is (-∞, -3] U [3, +∞).

So, there you have it! The expression for f(x) is √(x^2 - 9) and its domain is (-∞, -3] U [3, +∞). And remember, if you don't find this hilarious, just imagine me telling it with a clown nose and oversized shoes!

## To find the particular solution to the differential equation with the initial condition, we need to substitute the given initial condition into the general solution and solve for the arbitrary constant.

Given: f(-5) = -4

Substituting x = -5 and y = -4 into the general solution, we get:

-4 = ±√((-5)^2 + C)

Simplifying the equation gives:

-4 = ±√(25 + C)

To solve for C, we square both sides of the equation:

16 = 25 + C

Simplifying further:

C = -9

Now that we have determined the value of C, we can substitute it back into the general solution to obtain the particular solution:

y = ±√(x^2 + C)

y = ±√(x^2 - 9)

Since the square root function is defined for non-negative real numbers, the domain of the particular solution is all real numbers x where x^2 - 9 ≥ 0.

Simplifying the inequality, we have:

x^2 ≥ 9

Taking the square root of both sides, we get:

|x| ≥ 3

So, the domain of f(x) is x ≤ -3 or x ≥ 3.

## To find the particular solution, we need to use the initial condition f(-5) = -4 and substitute it into the general solution y = ±√(x^2 + C).

Substituting x = -5 and y = -4 into the general solution, we have:

-4 = ±√((-5)^2 + C)

Simplifying, we get:

16 = 25 + C

C = -9

Now, substituting C = -9 back into the general solution, we have:

y = ±√(x^2 - 9)

However, since the initial condition f(-5) = -4 is negative, we choose the negative square root:

y = -√(x^2 - 9)

Now, let's determine the domain of this particular solution. The square root function is defined only for non-negative values, so the expression inside the square root, x^2 - 9, must be greater than or equal to zero:

x^2 - 9 ≥ 0

(x - 3)(x + 3) ≥ 0

This inequality is satisfied when x ≤ -3 or x ≥ 3. Therefore, the domain of the particular solution f(x) = -√(x^2 - 9) is x ≤ -3 or x ≥ 3.

## The problem I see is that the solution to y'=xy is

y = c*e^(x^2/2)

and if y(-5) = -4 then

C = -4e^(25/2)

and so y = -4e^(1/2 (x^2-25))

with domain all real numbers

I think you meant y' = x/y

That gives y=±√(x^2+C)

so, if y(-5) =-4,

-√(25+C) = -4

25+C = 16

C = -9

y = -√(x^2-9)

so the domain is all |x| >= 3