Solving Y^2y'' = Y' Initial Value Problem A Comprehensive Guide

In this article, we delve into the solution of a second-order nonlinear differential equation, a common challenge in various fields of mathematics, physics, and engineering. We aim to find the specific solution to the initial value problem defined by the equation y^2y'' = y', with the initial conditions y(1) = 1 and y'(1) = -1, where y and y' are positive. This problem exemplifies the application of differential equations in modeling real-world phenomena, where understanding the behavior of functions based on their derivatives is crucial.

Understanding the Problem

The given differential equation, y^2y'' = y', is a second-order nonlinear ordinary differential equation (ODE). The nonlinearity arises from the term y^2 multiplying the second derivative y''. The initial conditions, y(1) = 1 and y'(1) = -1, provide specific values for the function and its first derivative at a particular point, allowing us to find a unique solution to the equation. We are tasked with identifying the correct expression for y'(t) from the provided options, which include various functions of y.

Method of Solution

To tackle this problem, we employ a substitution technique to simplify the equation. Let's set v = y', which transforms the second derivative y'' into dv/dt. However, since the equation involves y rather than t explicitly, we can further rewrite dv/dt using the chain rule as dv/dt = (dv/dy)(dy/dt) = v(dv/dy). This substitution is a key step in solving many second-order differential equations where the independent variable (in this case, t) does not appear explicitly.

Applying the Substitution

Substituting v = y' and y'' = v(dv/dy) into the original equation y^2y'' = y', we obtain:

y^2 * v(dv/dy) = v

Since we are given that y' is positive, v cannot be zero. Thus, we can divide both sides by v (assuming v is not identically zero) to simplify the equation:

y^2 (dv/dy) = 1

This simplification transforms the second-order ODE into a first-order ODE, which is often easier to solve.

Solving the First-Order ODE

Now we have a separable first-order ODE:

dv/dy = 1/y^2

We can separate the variables and integrate both sides with respect to their respective variables:

∫ dv = ∫ (1/y^2) dy

Integrating both sides gives:

v = -1/y + C,

where C is the constant of integration.

Determining the Constant of Integration

To find the value of C, we use the initial conditions. We know that y(1) = 1 and y'(1) = -1. Since v = y', we have v(1) = y'(1) = -1. Substituting these values into the equation v = -1/y + C, we get:

-1 = -1/1 + C

-1 = -1 + C

C = 0

Thus, the equation becomes:

v = -1/y

Expressing y'(t)

Since v = y', we have:

y'(t) = -1/y

This result matches option (B) from the given choices.

Analyzing the Solution

The solution y'(t) = -1/y indicates that the rate of change of y with respect to t is inversely proportional to y itself and is negative. This behavior is consistent with the initial conditions, where y(1) = 1 and y'(1) = -1. As y increases, y' becomes less negative, and as y decreases, y' becomes more negative. This relationship suggests a decreasing function y(t), which approaches zero as t increases.

Verification

To ensure the correctness of our solution, we can substitute y'(t) = -1/y back into the original differential equation and check if it satisfies the equation. First, we need to find y''. Differentiating y'(t) = -1/y with respect to t, we use the chain rule:

y''(t) = d/dt (-1/y) = (1/y^2) * y'(t)

Substituting y'(t) = -1/y into the above equation, we get:

y''(t) = (1/y^2) * (-1/y) = -1/y^3

Now, substituting y'(t) and y''(t) into the original equation y^2y'' = y', we have:

y^2 * (-1/y^3) = -1/y

-1/y = -1/y

This confirms that our solution y'(t) = -1/y satisfies the original differential equation.

Exploring Further

We have found the expression for y'(t), but we can go further and attempt to find the explicit solution for y(t). To do this, we can separate variables in the equation y'(t) = -1/y:

y'(t) = dy/dt = -1/y

y dy = -dt

Integrating both sides, we get:

∫ y dy = ∫ -dt

(1/2)y^2 = -t + C_1,

where C_1 is another constant of integration.

Multiplying both sides by 2, we have:

y^2 = -2t + 2C_1

Let C_2 = 2C_1, so:

y^2 = -2t + C_2

To find C_2, we use the initial condition y(1) = 1:

1^2 = -2(1) + C_2

1 = -2 + C_2

C_2 = 3

Thus, we have:

y^2 = -2t + 3

Since y is positive, we take the positive square root:

y(t) = √(-2t + 3)

This gives us the explicit solution for y(t).

Domain of the Solution

It is important to consider the domain of this solution. The expression inside the square root must be non-negative:

-2t + 3 ≥ 0

3 ≥ 2t

t ≤ 3/2

Thus, the solution is valid for t ≤ 3/2. Also, we know that the initial condition is given at t = 1, which falls within this domain.

In conclusion, we have successfully solved the initial value problem y^2y'' = y', with y(1) = 1 and y'(1) = -1, and determined that y'(t) = -1/y, which corresponds to option (B). We verified this solution by substituting it back into the original differential equation and showed that it satisfies the equation. Additionally, we found the explicit solution for y(t) as y(t) = √(-2t + 3) and discussed its domain of validity. This exercise demonstrates the techniques involved in solving nonlinear second-order differential equations and the importance of initial conditions in determining unique solutions. The substitution method, along with careful algebraic manipulation and integration, allowed us to transform the problem into a more manageable form and arrive at the correct answer. Understanding the behavior and solutions of differential equations is fundamental in many areas of science and engineering, making this type of problem-solving a valuable skill.