Question

Найти общее решение дифференциальных уравнений: (y - 1)²dx + (1 - x)³dy = 0 Решить задачу Коши: sin xdx + dy/√y = 0 ; y(0) = 1 Найти общее решение линейного дифференциального уравнения: y' + 2xy = xe ^-x² Прошу с решением,дабы понять принцип,как это решать

Answer 1

$1)\; \; \; (y-1)^2dx+(1-x)^3dy=0\\\\(y-1)^2dx=-(1-x)^3dy\; \; \; \Rightarrow \; \; \; \int \frac{dx}{-(1-x)^3} =\int \frac{dy}{(y-1)^2} \\\\-\int (1-x)^{-3}dx=\int (y-1)^{-2}dy\\\\+\frac{(1-x)^{-2}}{-2}=\frac{(y-1)^{-1}}{-1}+C\\\\-\frac{1}{2(1-x)^2}=-\frac{1}{y-1}+C\\\\2)\; \; sinx\cdot dx+\frac{dy}{\sqrt{y}}=0\; ,\; \; \; y(0)=1\\\\\int sinx\cdot dx=-\int \frac{dy}{\sqrt{y}}\\\\-cosx=-2\sqrt{y}+C\; \; \to \; \; \sqrt{y}=\frac{cosx+C}{2}\; ,\\\\y=\frac{1}{4}(cosx+C)^2\; \; -\; \; obshee\; reshenie$

$y(0)=1\; ,\; \; \sqrt1=\frac{cos0+C}{2}\; ,\; \; 2=1+C\; ,\; \; C=1\\\\y=\frac{1}{4}(cosx+1)\; \; -\; \; chastnoe\; reshenie\\\\3)\; \; y'+2xy=xe^{-x^2}\\\\y=uv\; ,\; y'=u'v+uv'\\\\u'v+uv'+2xuv=xe^{-x^2}\\\\u'v+u(\underbrace {v'+2xv}_{0})=xe^{-x^2}\\\\1)\; \; \frac{dv}{dx}+2xv=0\; ,\; \; \frac{dv}{v}=-2 x\, dx\\\\\int \frac{dv}{v}=-2\int x\, dx\; \; \; \to \; \; lnv=-x^2\; \; \to \; \; v=e^{-x^2}$

$2)\; \; u'\cdot e^{-x^2}=xe^{-x^2}\\\\\frac{du}{dx}=x\; \; \to \; \; \int du=\int x\, dx\; \; \to \; \; u=\frac{x^2}{2}+C$

$3)\; \; y=uv=e^{-x^2}\cdot (\frac{x^2}{2}+C)$

Answer 2

Solving the Differential Equations

To find the general solution for the given differential equations, we will go through each equation one by one.

# Equation 1: (y - 1)²dx + (1 - x)³dy = 0

To solve this equation, we can use the method of exact differential equations. For a differential equation to be exact, the following condition must be satisfied:

∂M/∂y = ∂N/∂x

where M and N are the coefficients of dx and dy, respectively.

In this case, M = (y - 1)² and N = (1 - x)³. Let's check if the condition is satisfied:

∂M/∂y = 2(y - 1) ∂N/∂x = -3(1 - x)²

Since ∂M/∂y is not equal to ∂N/∂x, the equation is not exact. However, we can make it exact by multiplying the entire equation by an integrating factor.

Let's find the integrating factor (IF): IF = e^(∫(∂M/∂y - ∂N/∂x)dx) = e^(∫(2(y - 1) + 3(1 - x)²)dx) = e^(2xy - 2x + y - x³ + x² + C₁)

Multiplying the given equation by the integrating factor, we get: (e^(2xy - 2x + y - x³ + x² + C₁))((y - 1)²dx + (1 - x)³dy) = 0

Expanding and simplifying the equation, we have: (y - 1)²e^(2xy - 2x + y - x³ + x² + C₁)dx + (1 - x)³e^(2xy - 2x + y - x³ + x² + C₁)dy = 0

This equation is now exact. We can find the solution by integrating both sides with respect to x and y, respectively.

Integrating the left side with respect to x, we get: ∫((y - 1)²e^(2xy - 2x + y - x³ + x² + C₁))dx = F(x, y)

Integrating the right side with respect to y, we get: ∫((1 - x)³e^(2xy - 2x + y - x³ + x² + C₁))dy = G(x, y)

Where F(x, y) and G(x, y) are the arbitrary functions of integration.

The general solution for the given differential equation is: F(x, y) = G(x, y)

# Equation 2: sin(x)dx + dy/√y = 0, y(0) = 1

To solve this equation, we can separate the variables and integrate.

Rearranging the equation, we have: dy/√y = -sin(x)dx

Integrating both sides, we get: ∫dy/√y = -∫sin(x)dx

Integrating the left side gives: 2√y = -cos(x) + C₂

Squaring both sides, we have: 4y = cos²(x) - 2cos(x)C₂ + C₂²

Since y(0) = 1, we can substitute x = 0 and y = 1 into the equation to find the value of C₂: 4(1) = cos²(0) - 2cos(0)C₂ + C₂² 4 = 1 - 2C₂ + C₂² 3 = -2C₂ + C₂² C₂² - 2C₂ + 3 = 0 (C₂ - 1)(C₂ - 3) = 0

Therefore, C₂ = 1 or C₂ = 3.

Substituting the values of C₂ back into the equation, we have two possible solutions: 1) 4y = cos²(x) - 2cos(x) + 1 2) 4y = cos²(x) - 2cos(x) + 3

These are the general solutions for the given differential equation.

# Equation 3: y' + 2xy = xe^(-x²)

This is a linear differential equation. To solve it, we can use an integrating factor.

The integrating factor (IF) is given by: IF = e^(∫2xdx) = e^(x²)

Multiplying the given equation by the integrating factor, we get: e^(x²)(y' + 2xy) = xe^(-x²)e^(x²) e^(x²)(y' + 2xy) = x

Simplifying the equation, we have: e^(x²)y' + 2x^2y = x

This equation is now in a form that can be solved using the integrating factor. Let's denote the integrating factor as IF.

Multiplying the equation by the integrating factor, we get: e^(x²)(e^(x²)y' + 2x^2y) = xe^(x²)

Expanding and simplifying the equation, we have: (e^(2x²)y' + 2x^2e^(x²)y) = xe^(x²)

Now, we can integrate both sides of the equation with respect to x.

∫(e^(2x²)y' + 2x^2e^(x²)y)dx = ∫(xe^(x²))dx

Integrating the left side gives: ∫(e^(2x²)y' + 2x^2e^(x²)y)dx = y*e^(2x²) + C₃

Integrating the right side gives: ∫(xe^(x²))dx = ∫(e^(x²))d(x²) = ∫(e^(x²))2xdx = ∫(2xe^(x²))dx = e^(x²) + C₄

Where C₃ and C₄ are the arbitrary constants of integration.

Therefore, the general solution for the given linear differential equation is: y*e^(2x²) + C₃ = e^(x²) + C₄

Simplifying the equation, we have: y*e^(2x²) = e^(x²) + C₄ - C₃

Let's denote C as the difference between C₄ and C₃: C = C₄ - C₃

The general solution can be written as: y = (e^(x²) + C)e^(-2x²)

This is the general solution for the given linear differential equation.

I hope this helps! Let me know if you have any further questions.