Solving the Differential Equation dy/dt y^(1/2) ? 1 and Exploring Its Solutions

Introduction to the Problem:

The differential equation (frac{dy}{dt} y^{1/2} - 1) represents a type of first-order differential equation. This problem aims to explore the methods of solving such an equation, focusing on the conditions given and the solutions derived under specific constraints.

Solving the Differential Equation via Substitution

Let's solve the differential equation step by step using a substitution method. We start by setting (y - 1 u). This implies:

(y u 1)

Substituting (u) into the original differential equation, we get:

(frac{dy}{dt} u^{1/2})

Since (y u 1), we can write:

(frac{du}{dt} frac{1}{2}u^{-1/2})

To separate variables, we rewrite:

(frac{du}{u^{1/2}} - dt)

Integrating both sides:

(int u^{-1/2} du - int dt)

(2u^{1/2} - t C_1)

Substituting back (u y - 1) gives:

(2(y - 1)^{1/2} - t C_1)

Analyzing the Initial Condition

Given (y_0 1), we substitute to find the constant (C_1):

(2(1 - 1)^{1/2} - t C_1)

This implies:

(0 - t C_1)

Hence:

(C_1 t)

Substituting back, we get:

(2(y - 1)^{1/2} - t)

Solving for (t) gives:

(t 2(- (y - 1)^{1/2}))

Thus, the solution is:

(y 1 2left(frac{1 - t}{2}right)^2)

Exploring the Solutions for Different Initial Conditions

Let's break down the solution further into two cases based on initial conditions:

Case 1: y ≥ 1

In this case, the differential equation is:

(frac{dy}{dt} y^{1/2} - 1)

Integrating both sides, we get:

(int y^{-1/2} dy int dt)

(2y^{1/2} t C)

Using the initial condition (y_0 1), we find:

(2(1)^{1/2} 0 C)

Hence:

(C 2)

Thus, the solution is:

(2y^{1/2} t 2)

Solving for (y) gives:

(y 1 left(frac{t 2}{2}right)^2)

Case 2: y ≤ 1

In this case, the differential equation is:

(frac{dy}{dt} 1 - y^{1/2})

Integrating both sides, we get:

(int 1 - y^{-1/2} dy int dt)

(-2y^{1/2} t C)

Using the initial condition (y_0 1), we find:

(-2(1)^{1/2} 0 C)

Hence:

(C -2)

Thus, the solution is:

(-2y^{1/2} t - 2)

Solving for (y) gives:

(y 1 - left(frac{t - 2}{2}right)^2)

Leveraging Trigonometric Substitution

Alternatively, we can solve the differential equation using a trigonometric substitution:

(y sec^2theta)

Then, (dy 2sec^2thetatantheta dtheta). Substituting into the differential equation:

(2sec^2thetatantheta dtheta / (sec^2theta - 1)^{1/2} dtheta)

Simplifying, we get:

(2sec^2thetatantheta dtheta / tantheta dtheta)

This simplifies to:

(2sec^2theta dtheta dtheta)

Thus, the integral becomes:

(2theta theta C)

Therefore, the general solution is:

(2theta t C)

Given (y_0 1), the simple solution (y 1 frac{1}{4} t^2) is derived.

Conclusion

In conclusion, we have explored multiple methods to solve the differential equation (frac{dy}{dt} y^{1/2} - 1), including substitution, trigonometric substitution, and integration. The final solution is:

(y 1 frac{1}{4} t^2)

This solution is derived under the constraint that (y geq 1). For (y leq 1), the equation does not hold.

Understanding these differential equations is crucial in various applications, particularly in physics and engineering, where such equations can describe the behavior of systems in motion or under changing conditions.