Problem 11 - Calculus: Early Transcendentals

Question

Solve the initial value problem. $$t y^{\prime}-2 y=0, y(1)=4$$

Step-by-Step Solution

Answer

The solution to the initial value problem is \( y = 4t^2 \).

Step 1: Identify the Type of Differential Equation

The equation given is a first-order linear homogeneous differential equation of the form \(t y' - 2y = 0\). Such equations can often be solved using techniques like separation of variables or integrating factor.

Step 2: Rearrange to Standard Form

Let's rewrite the differential equation as \( y' - \frac{2}{t}y = 0\). This helps in matching it to the standard linear form \( y' + P(x)y = 0 \).

Step 3: Recognize and Apply Separation of Variables

Since it appears in the separable form after divide through by \( t \): \( y' = \frac{2}{t}y \), we can separate variables as \( \frac{dy}{y} = \frac{2}{t}dt \).

Step 4: Integrate Both Sides

Integrate both sides to find: \( \int \frac{dy}{y} = \int \frac{2}{t} dt \), which gives \( \ln|y| = 2 \ln|t| + C \), where \( C \) is the constant of integration.

Step 5: Simplify the Equation

Apply exponentiation to both sides to remove the logarithm, yielding \( |y| = |t|^2 e^C \). To simplify, represent \( e^C \) by a new constant, \( C_1 \), leading to \( y = C_1 t^2 \).

Step 6: Apply Initial Conditions

Use the initial condition \( y(1) = 4 \) to find \( C_1 \). Substituting \( t = 1 \) and \( y = 4 \) into \( y = C_1 t^2 \), we obtain \( 4 = C_1 (1)^2 \), thus \( C_1 = 4 \).

Step 7: Write the Particular Solution

Substitute \( C_1 = 4 \) back into the general solution, yielding \( y = 4t^2 \). This is the particular solution to the initial value problem.

Key Concepts

Separation of Variables

Separation of variables is a powerful method used to solve first-order differential equations. It is especially effective when the equation can be rearranged so that each variable and its derivative are on opposite sides of the equation.

In our given problem, the equation initially looked like this: \( ty' - 2y = 0 \). By rearranging, we obtained \( y' = \frac{2}{t}y \). This shows that the equation is now in a form where variables can be separated.

The next step is to separate the variables: move all terms involving \( y \) to one side and terms involving \( t \) to the other. This leads to:

• \( \frac{dy}{y} = \frac{2}{t} dt \)

By integrating both sides, we solve for \( y \) in terms of \( t \). This gives a solution that can be further worked upon using initial or boundary conditions. The integration results in:

• \( \ln|y| = 2 \ln|t| + C \)

Exponential functions are then used to solve the logarithm, giving us the solution form.

Initial Value Problem

When we talk about an initial value problem, we refer to a type of differential equation that comes with specific initial conditions provided.

In our exercise, the initial value condition is given as \( y(1) = 4 \). This initial condition is crucial, as it allows us to determine the specific constant in our general solution derived from the differential equation.

After finding a general solution, which was \( y = C_1 t^2 \), we apply the initial condition. By substituting \( t = 1 \) and \( y = 4 \) into the equation:

• We solve for the constant \( C_1 \).

Finding \( C_1 = 4 \), we then use this to write our particular solution as \( y = 4t^2 \). This initial value provided is what tailors our general solution to solve the specific case described by the problem.

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a type of differential equation where the function and its derivatives fulfill a linear relationship, and there is no constant term involved apart from zero.

In the problem \( ty' - 2y = 0 \), the equation can be rewritten as \( y' - \frac{2}{t}y = 0 \), showcasing its linear homogeneous nature. Here:

• No terms are isolated from \( y \) or its derivative \( y' \).
• The equation maintains linearity.

Such equations have solutions that are either exponential or polynomial in form. By rewriting it, recognizing its form, and using either separation of variables or integrating factor, one finds the solution.

In this problem's solution, separating variables proved effective. This is because the absence of a non-zero term on the right-hand side allows us to capture relationships that rely entirely on functional forms like \( y = C_1t^2 \). This comes from recognizing the structure intrinsic to linear homogeneous equations, which are simpler and elegant to handle under conditions like constant coefficients.