Question

Y'² −1=2yy"; y(2) = 1, y'(2) = 2

Answer 1

$(y')^2-1=2yy''$

Пусть $y' = f(y)$  тогда

$y'' = f'(y)y' = f'f$

$\displaystyle f^2-1 = 2yf'f\\ff' = \frac{f^2-1}{2y}\\f\frac{df}{dy} = \frac{f^2-1}{2y}\\\frac{2fdf}{f^2-1} = \frac{dy}{y}\\\frac{d(f^2)}{f^2-1} = d(\ln y)\\\ln(f^2-1) = \ln y+\tilde{C}\\f^2-1 = Cy\\\frac{dy}{dx} = \pm\sqrt{1+Cy}$

Подставим начальные условия: при y=1 y'=2 или

$2 = \pm\sqrt{1+C} \Rightarrow C = 3$ и знак выбирается +

$\displaystyle \frac{dy}{dx} = \sqrt{1+3y}\\\frac{dy}{\sqrt{1+3y}} = dx\\\frac{2}{3}\sqrt{1+3y} = x + C_1$

Подставим начальные условия при x=2 y=1 или

$\frac{2}{3}\sqrt{4} = 2+C_1 \Rightarrow C_1 = -2/3$

$\displaystyle 2\sqrt{1+3y} = 3x-2\\1+3y = (1.5x-1)^2\\y = \frac{(1.5x-1)^2-1}{3}$

Answer 2

The given equation is a second-order nonlinear differential equation in terms of y. To solve this equation, you can use the method of power series or expand the solution as a Taylor series. Here's how you can approach it:

Let's represent the solution y(x) as a Taylor series:

y(x) = ∑[n=0 to ∞] a_n (x - 2)^n

To find the solution, we need to calculate the derivatives of y with respect to x:

y'(x) = ∑[n=0 to ∞] a_n * n * (x - 2)^(n-1) y''(x) = ∑[n=0 to ∞] a_n * n * (n - 1) * (x - 2)^(n-2)

Now, we can substitute these derivatives into the given differential equation:

(y'(x))^2 - 1 = 2 * y(x) * y''(x)

(∑[n=0 to ∞] a_n * n * (x - 2)^(n-1))^2 - 1 = 2 * y(x) * (∑[n=0 to ∞] a_n * n * (n - 1) * (x - 2)^(n-2))

Now, let's calculate these series terms one by one. First, calculate the square of y'(x):

(∑[n=0 to ∞] a_n * n * (x - 2)^(n-1))^2 = ∑[n=0 to ∞] ∑[m=0 to ∞] (a_n * n * (x - 2)^(n-1)) * (a_m * m * (x - 2)^(m-1))

Now, let's focus on the right side of the equation. You'll need to calculate the product of y(x) and y''(x) as a series:

2 * y(x) * (∑[n=0 to ∞] a_n * n * (n - 1) * (x - 2)^(n-2))

Now, you can equate the coefficients of corresponding powers of (x - 2) on both sides of the equation. This will give you a set of equations for the coefficients a_n. Start with the lowest power and work your way up.

Here's what the equation for the lowest power (n = 0) looks like:

a_0^2 - 1 = 0

Solve for a_0:

a_0^2 = 1 a_0 = ±1

Now, let's find the equation for the next term (n = 1):

2 * a_0 * a_1 - 2 * a_0 * 0 * 1 = 0

a_1 = 0

Proceed with the rest of the terms, calculating the coefficients a_n one by one, and use the initial conditions y(2) = 1 and y'(2) = 2 to determine their values.

The final result will be a power series that represents the solution y(x) to the given differential equation.