Find a general solution to each first ¨C degree ordinary differential equation. If an initial condition is given, find the corresponding particular solution. Write solution explicitly if possible.
dx/dt=¡Ì(x+t+1)
To find the general solution to the first-degree ordinary differential equation dx/dt = √(x + t + 1), we can use the method of separation of variables.
First, let's rewrite the equation as:
1/√(x + t + 1) dx = dt
Now, we separate the variables by multiplying both sides of the equation by √(x + t + 1):
dx = √(x + t + 1) dt
Next, we integrate both sides of the equation separately. The integral on the left side will be ∫ dx and the integral on the right side will be ∫ √(x + t + 1) dt.
Integrating both sides, we get:
∫ dx = ∫ √(x + t + 1) dt
The integral of dx is simply x, and for the integral of √(x + t + 1) dt, we need to apply a substitution. Let's substitute u = x + t + 1. Then du = dx, and the integral becomes:
∫ du = ∫ √u dt
Simplifying, we have:
u = t + C₁, where C₁ is the constant of integration.
Substituting back u = x + t + 1, we get:
x + t + 1 = t + C₁
Rearranging the equation, we find:
x = C₁ - 1
This is the general solution to the given first-degree ordinary differential equation.
Now, if an initial condition is given, let's say x(0) = 2, we can find the particular solution by plugging in the initial condition into the general solution:
x = C₁ - 1
When t = 0, x = 2:
2 = C₁ - 1
Solving for C₁, we find:
C₁ = 3
So, the corresponding particular solution is:
x = 3 - 1
Simplifying further, we get:
x = 2
Hence, the particular solution for the given initial condition is x = 2.