Question

1)yy'=-2/x 2)y'=(2x-1)ctgy, x=2,y=π/2

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

To solve these initial value problems, you can use separation of variables for the first equation and then solve for the constant of integration using the second equation and the given initial condition.

1. yy' = -2/x:

First, let's rewrite this differential equation in the form that can be separated:

yy' = -2/x

Now, separate the variables y and x on opposite sides of the equation:

y dy = (-2/x) dx

Now, integrate both sides:

∫y dy = ∫(-2/x) dx

Integrating both sides gives:

(1/2)y^2 = -2ln|x| + C₁

Where C₁ is the constant of integration.

Now, let's use the second equation and the initial condition to find C₁:

2. y' = (2x - 1)ctg(y)

We know that x = 2 and y = π/2 at the initial condition.

y'(2) = (2(2) - 1)ctg(π/2)

y'(2) = 3 * undefined (ctg(π/2) is undefined)

Since y'(2) is undefined, this initial condition is problematic, and the solution may not exist in this case.

It's possible that there was a mistake in the initial conditions provided or that there is another way to interpret the problem. Please double-check the initial conditions or provide additional information if necessary for a more accurate solution.