1.Solve the differential equation dy/dx= y^2/x^3 for y=f(x) with the condition y(1) = 1.
2.Solve the differential equation y prime equals the product of 2 times x and the square root of the quantity 1 minus y squared.
Explain why the initial value problem y prime equals the product of 2 times x and the square root of the quantity 1 minus y squared with y(0) = 3 does not have a solution.
3.The table below gives selected values for the function f(x). Use a trapezoidal estimation, with 6 trapezoids to approximate the value of the integral from 1 to 2 of f of x, dx. Give 3 decimal places for your answer.
X 1, 1.1, 1.2, 1.5, 1.7, 1.9, 2.0
f(x)1, 2, 4, 6, 7, 9, 10
dy/dx=(y^2)(x^3)
1. Oh, that's a differential equation! Looks like y is getting a bit frisky with x. Let's see what we can do here. We'll separate the variables and integrate both sides. Integrating y^2 gives us (1/3)y^3, and integrating x^-3 gives us (-1/2)x^-2. So we have (1/3)y^3 = (-1/2)x^-2 + C. Now we plug in the initial condition y(1) = 1 and solve for C. Plug in x=1 and y=1 into our equation and solve for C, and we'll have our final answer. But don't worry, this equation won't make you cry! It's just a little different.
2. Now we have a prime problem! This differential equation involves some intense multiplication. We have y' = 2x√(1 - y^2). Unfortunately, when we try to solve this with the initial condition y(0) = 3, we encounter a little trouble. You see, when x=0, y'(0) = 2(0)√(1 - 3^2) = 2(0)√(-8). Uh-oh! The square root of a negative number? That's like trying to find the punchline to an unfinished joke. It just doesn't work. So sadly, this initial value problem doesn't have a solution. But don't worry, there are plenty of other problems out there ready to be solved!
3. Ah, an integral approximation problem! Time to whip out our trusty trapezoids. We have 6 trapezoids, with values of f(x) given at different intervals. We'll start by calculating the width of each trapezoid, which is (2-1)/6 = 1/6. Then, let's calculate the areas of the trapezoids.
For the first trapezoid, (1/6) * (f(1) + f(1.1))/2.
For the second trapezoid, (1/6) * (f(1.1) + f(1.2))/2.
And so on, until the last trapezoid, (1/6) * (f(1.9) + f(2))/2.
Now, add up all these areas, rounding to 3 decimal places, and voila! You have your approximation for the value of the integral from 1 to 2 of f(x), dx. Just remember, approximation is like telling a joke - it may not be exact, but it gets the job done!
1. To solve the differential equation dy/dx = y^2/x^3 with the initial condition y(1) = 1, we can proceed as follows:
Step 1: Separate the variables by multiplying both sides by dx and dividing both sides by y^2:
y^(-2) dy = x^(-3) dx
Step 2: Integrate both sides:
∫y^(-2) dy = ∫x^(-3) dx
This gives us:
-y^(-1) = -x^(-2)/2 + C1
Step 3: Solve for y:
We can multiply both sides by -1 to isolate y:
y^(-1) = x^(-2)/2 - C1
Taking the reciprocal of both sides, we get:
y = 1/(x^(-2)/2 - C1)
Step 4: Apply the initial condition to find the value of C1:
We know that y(1) = 1, so substituting these values into our equation gives us:
1 = 1/(1 - C1)
Solving for C1, we find:
C1 = 1 - 1 = 0
Step 5: Substituting C1 = 0 back into our equation for y, we get the final solution:
y = 1/x^(-2)/2
2. To solve the differential equation y' = 2x√(1 - y^2), we can follow these steps:
Step 1: Separate the variables by dividing both sides by √(1 - y^2):
(1/√(1 - y^2)) dy = 2x dx
Step 2: Integrate both sides:
∫(1/√(1 - y^2)) dy = ∫2x dx
Performing the integration gives us:
arcsin(y) = x^2 + C2
Step 3: Solve for y:
Taking the arcsine of both sides to isolate y, we get:
y = sin(x^2 + C2)
Step 4: Apply the initial condition to find the value of C2:
We know that y(0) = 3, so substituting these values into our equation gives us:
3 = sin(0^2 + C2)
Since sin(0) = 0, this equation cannot be satisfied for y = 3. Therefore, the initial value problem y' = 2x√(1 - y^2) with y(0) = 3 does not have a solution.
3. To approximate the value of the integral from 1 to 2 of f(x) using the trapezoidal rule with 6 trapezoids, we can follow these steps:
Step 1: Calculate the width of each trapezoid, h:
The width of each trapezoid is given by h = (2 - 1)/6 = 1/6.
Step 2: Evaluate the function f(x) at the given x-values:
f(1) = 1, f(1.1) = 2, f(1.2) = 4, f(1.5) = 6, f(1.7) = 7, f(1.9) = 9, and f(2) = 10.
Step 3: Use the trapezoidal rule to approximate the integral:
The trapezoidal rule states that the integral can be approximated by:
∫f(x) dx ≈ h/2 * [f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + 2f(x4) + 2f(x5) + f(x6)]
Substituting the given values, we have:
∫f(x) dx ≈ (1/6)/2 * [1 + 2(2) + 2(4) + 2(6) + 2(7) + 2(9) + 10]
Simplifying and calculating the result, we get:
∫f(x) dx ≈ (1/12) * [1 + 4 + 8 + 12 + 14 + 18 + 10] ≈ 5.917
Therefore, using the trapezoidal rule with 6 trapezoids, the approximate value of the integral from 1 to 2 of f(x) is 5.917 (rounded to 3 decimal places).
dy/dx = y^2/x^3
dy/y^2 = dx/x^3
-1/y = -1/(2x^2)+c
y = 2x^2/(1-cx^2)
what's with all the verbal noise? You want to discuss
y' = 2x√(1-y^2)? Just say so
dy/√(1-y^2) = 2x dx
arcsin(y) = x^2 + c
y = sin(c+x^2)