Find a general solution to the first – degree ordinary differential equation. If an initial condition is given, find the corresponding particular solution. Write solution explicitly if possible:
dx/dt=sqrt(x+t+1)
To find the general solution to the first-degree ordinary differential equation and the corresponding particular solution if an initial condition is given, we can use the method of separation of variables.
1. Rewrite the given differential equation:
dx/dt = √(x + t + 1)
2. Move the variables to separate sides of the equation:
1/√(x + t + 1) dx = dt
3. Now, integrate both sides of the equation with respect to their respective variables:
∫(1/√(x + t + 1)) dx = ∫dt
4. Evaluate the integrals:
2√(x + t + 1) = t + C₁
Where C₁ is the constant of integration.
5. To isolate x, divide both sides of the equation by 2:
√(x + t + 1) = (t + C₁)/2
6. Square both sides of the equation to remove the square root:
x + t + 1 = ((t + C₁)/2)²
7. Simplify the equation:
x + t + 1 = (t² + 2C₁t + C₁²)/4
8. Rearrange the equation to isolate x:
x = (t² + 2C₁t + C₁²)/4 - t - 1
9. Further simplify the equation:
x = (t² + 2C₁t + C₁² - 4t - 4)/4
10. Combine like terms and rewrite the equation:
x = (t² + (2C₁ - 4)t + (C₁² - 4))/4
This is the general solution to the given differential equation.
If an initial condition, such as x(t₀) = x₀, is given, we can use it to find the corresponding particular solution.
1. Substitute the initial condition values into the general solution:
x₀ = (t₀² + (2C₁ - 4)t₀ + (C₁² - 4))/4
2. Rearrange the equation to solve for C₁:
C₁ = (4x₀ - t₀² + 4t₀ + 4)/2
3. Plug the value of C₁ back into the general solution:
x = (t² + (2((4x₀ - t₀² + 4t₀ + 4)/2) - 4)t + (((4x₀ - t₀² + 4t₀ + 4)/2)² - 4))/4
4. Simplify the equation further to obtain the explicit form of the particular solution.
Note: Depending on the specific values of t₀ and x₀, the particular solution may vary.