Question

Найти частное решение дифференциального уравнения y’=2x+4,если y=5, при x=-1.

Answer 1

Ответ:

см рис

Пошаговое объяснение:

Answer 2

Ответ:

Пошаговое объяснение:

Answer 3

Finding the Particular Solution of the Differential Equation

To find the particular solution of the differential equation y' = 2x + 4 with the initial condition y = 5 when x = -1, we can use the method of separation of variables and integrate both sides of the equation.

Let's solve the differential equation step by step:

1. Start with the given differential equation: y' = 2x + 4.

2. Separate the variables by moving all terms involving y to one side and all terms involving x to the other side: dy = (2x + 4)dx.

3. Integrate both sides of the equation with respect to their respective variables: - Integrating dy gives us y = ∫(2x + 4)dx. - Integrating (2x + 4)dx gives us y = x^2 + 4x + C, where C is the constant of integration.

4. Now, we need to find the value of the constant C using the initial condition y = 5 when x = -1. - Substituting x = -1 and y = 5 into the equation y = x^2 + 4x + C, we get 5 = (-1)^2 + 4(-1) + C. - Simplifying the equation, we have 5 = 1 - 4 + C. - Solving for C, we find C = 8.

5. Finally, substitute the value of C back into the equation y = x^2 + 4x + C to get the particular solution of the differential equation: - The particular solution is y = x^2 + 4x + 8.

Therefore, the particular solution of the differential equation y' = 2x + 4 with the initial condition y = 5 when x = -1 is y = x^2 + 4x + 8.