Question

Y*(1-lny)*y''+(1+lny)*y'2=0

Answer 1

Ответ:

у=5

Пошаговое объяснение:

не кто не видел кроме меня

Answer 2

The given differential equation is:

Y*(1-ln(Y))Y'' + (1+ln(Y))(Y')^2 = 0

To solve this differential equation, we can use a substitution method. Let's make the substitution:

u = ln(Y)

Differentiating u with respect to Y gives:

du/dY = 1/Y [Differentiation of ln(Y) with respect to Y]

Now let's find the second derivative of Y with respect to the variable u:

d²Y/du² = d(du/dY)/du = d(1/Y)/du

Using the chain rule, we can express this derivative in terms of Y:

d(1/Y)/du = d(1/Y)/dY * dY/du = -1/Y² * dY/du

Substituting these values into the given differential equation, we get:

Y*(1-u)(-1/Y² * dY/du) + (1+u)(dY/du)² = 0

Simplifying this equation gives:

-(1-u)dY/du + (1+u)(dY/du)² = 0

Now we can solve this equation for dY/du:

-(1-u) + (1+u)(dY/du) = 0 (1+u)(dY/du) = 1-u dY/du = (1-u)/(1+u)

Separating variables, we get:

dY/(1-u) = (1-u)/(1+u) * du

Integrating both sides with respect to their respective variables:

∫ dY/(1-u) = ∫ (1-u)/(1+u) * du

This integration leads to:

ln|1-u| = ∫ (1-u)/(1+u) * du

The integral on the right side can be solved using partial fractions or other integration techniques. Once you solve it, you can substitute back u = ln(Y) to find Y as a function of the original variable.