Question

(1-x)dx-ydy=0(x+1)dx+e^(y)×dy=0​

Answer 1

Ответ:

Пошаговое объяснение:

1) Диф. уравнение 1-го порядка с разделяющимися переменными  

(1 + x)ydy - (1 + y)xdx = 0  

(1 + x)ydy = (1 + y)xdx  

ydy/(1 + y) = xdx/(1 + x)  

Интегрируем обе части уравнения  

∫ydy/(1 + y) = ∫xdx/(1 + x)  

Решим один из интегралов – второй аналогичный. Проинтегрируем по частям.  

∫xdx/(1 + x)  

Введем замену  

u = x, тогда du = dx  

dv = dx/(1 + x), тогда v = ln|1 + x|  

∫u*dv = u*v - ∫ v*du  

∫xdx/(1 + x) = x*ln|1 + x| - ∫ln|1 + x|*dx = x*ln|1 + x| - (x + 1)*ln|1 + x| + (x + 1) + C = ln|1 + x| + (x + 1) + C  

Получаем решение  

ln|1 + у| + (у + 1) = ln|1 + x| + (x + 1) + C  

 

2) Линейное однородное уравнение второго порядка  

Характеристическое уравнение  

r^2 – 3r = 0  

r1 = 0; r2 = 3 – корни действительные и различные  

Общее решение дифференциального уравнения  

у0 = С1 + С2е^(3x)  

Частное решение найдем из заданных начальных условий  

y(0) = 1 y '(0) = -1  

y(0) = С1 + С2 = 1  

y '(0) = 3С2 = -1  

С2 = -1/3  

С1 = 4/3  

Частное решение  

у = 4/3 - (1/3)*е^(3x)

Answer 2

To solve the system of differential equations (1-x)dx - ydy = 0 and (x+1)dx + e^(y)dy = 0, we can use the method of separation of variables.

Let's start with the first equation:

(1-x)dx - ydy = 0

Rearrange the terms and divide both sides by (1-x):

dx = ydy / (1-x)

Now, we can integrate both sides:

∫dx = ∫(y / (1-x))dy

Integrating, we get:

x + C1 = -ln|1-x| + C2y

where C1 and C2 are constants of integration.

Now, let's move on to the second equation:

(x+1)dx + e^(y)dy = 0

Rearrange the terms and divide both sides by e^y:

dx = -dy / (x+1)

Now, we can integrate both sides:

∫dx = ∫(-dy / (x+1))

Integrating, we get:

x + C3 = -ln|x+1| + C4y

where C3 and C4 are constants of integration.

Now we have two solutions for x in terms of y:

1. x + C1 = -ln|1-x| + C2y
2. x + C3 = -ln|x+1| + C4y

Since these equations represent the same system of differential equations, we can equate them:

-x - ln|1-x| + C2y + C1 = -x - ln|x+1| + C4y + C3

Simplifying the above equation, we get:

ln|1-x| - ln|x+1| + (C4-C2)y = C1 - C3

Using the property of logarithms, we can combine the logarithms:

ln| (1-x) / (x+1) | + (C4-C2)y = C1 - C3

Let's define a new constant C5 = C1 - C3, to simplify further:

ln| (1-x) / (x+1) | + (C4-C2)y = C5

To solve for y, isolate it on one side:

(C4-C2)y = C5 - ln| (1-x) / (x+1) |

Finally, the solution for y is:

y = (C5 - ln| (1-x) / (x+1) |) / (C4-C2)

Please note that C1, C2, C3, C4, and C5 are constants of integration and need to be determined from initial or boundary conditions given in the original problem. Without specific initial conditions, we cannot determine the exact values of these constants.