Question

For each initial value problem, (a) Find the general solution of the differential equation. (b) Impose the initial condition to obtain the solution of the initial value problem. \(2 t y-y^{\prime}=0, \quad y(1)=3\)

Short answer

Question: Solve the initial value problem \(2ty - y^\prime = 0\) with the initial condition \(y(1) = 3\). Answer: The particular solution of the initial value problem is \(y = \frac{3}{e}e^{t^2}\).

Step-by-step solution

Rewrite the given differential equation in the standard form

From the given differential equation, \(2ty- y^\prime = 0\), we can rewrite it in the standard form \(y^\prime = f(t, y)\). To do this, we will isolate the \(y^\prime\) term: \(y^\prime = 2ty\) Now we have the differential equation in the standard form: \(y^\prime = 2ty\).

Solve the differential equation to find the general solution

To solve this first-order linear homogeneous differential equation, we can use the method of separation of variables. Divide both sides by \(y\) and multiply both sides by \(dt\): \(\frac{1}{y}dy = 2tdt\) Now, integrate both sides with respect to the respective variables: \(\int \frac{1}{y} dy = \int 2t dt\) This yields: \(\ln |y| = t^2 + C_1\) To solve for \(y\), we can take the exponential of both sides: \(|y| = e^{t^2 + C_1}\) Since the absolute value of \(y\) is always non-negative, we can use a constant factor \(C = e^{C_1}\) to simplify: \(y = Ce^{t^2}\) Now we have the general solution of the differential equation: \(y = Ce^{t^2}\).

Use the initial condition to find the particular solution

We are given the initial condition \(y(1) = 3\). Let us substitute the values of \(t = 1\) and \(y = 3\) into the general solution to find the value of the constant \(C\): \(3 = Ce^{1^2}\) \(\Rightarrow 3 = Ce\) Now we can solve for \(C\): \(C = \frac{3}{e}\) Substitute the value of \(C\) back into the general solution: \(y = \frac{3}{e}e^{t^2}\) Now we have the particular solution of the initial value problem: \(y = \frac{3}{e}e^{t^2}\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They come into play when a rate of change or a relationship that involves some continuous process is part of a scenario, like how a material cools over time or how populations change. Understanding the core idea behind differential equations is crucial because they are not just abstract mathematical constructs. Instead, they model the behavior of variables in dynamic systems, whether in physics, biology, economics, or engineering.

In the given exercise, we encounter a first-order differential equation. This means that it involves the first derivative of the unknown function, represented by the notation \( y' \). The differential equation \( 2ty - y' = 0 \) captures how the rate of change of the function \( y \) with respect to the variable \( t \) is directly related to the product of \( t \) and \( y \) itself.

General Solution

The general solution to a differential equation represents a family of solutions containing arbitrary constants. In the context of our initial value problem, finding the general solution involves determining the form of the function y that can satisfy the given differential equation for all values of the independent variable t.

Obtaining the General Solution

Given the equation \( y' = 2ty \), the general solution is obtained by separating and integrating both sides. The result, \( y = Ce^{t^2} \), includes an undetermined constant \( C \), signifying an infinite set of possible solutions. This stage is critical as it sets the stage for further refinement of the solution by incorporating initial conditions.

Particular Solution

While the general solution gives us a broad set of solutions, a particular solution is one specific member of this family that satisfies the initial value conditions of the problem. To find it, we use the given initial condition \( y(1) = 3 \).

Applying the Initial Condition

By substituting \( t = 1 \) and \( y = 3 \) into our general solution, we can solve for the constant \( C \). This process tailors the general solution to a unique scenario outlined by the initial conditions, resulting in the particular solution \( y = \frac{3}{e}e^{t^2} \), which uniquely satisfies both the differential equation and the initial condition.

Separation of Variables

Separation of variables is a commonly used method for solving first-order differential equations. It involves rearranging the equation to isolate the derivatives of each variable on opposite sides of the equation.

Implementing Separation of Variables

In our exercise, we have \( y' = 2ty \). By dividing both sides by \( y \) and multiplying both sides by the differential \( dt \), the variables \( y \) and \( t \) become separated. We then proceed to integrate both sides, leading us to the general solution. This technique fundamentally breaks down a complex relationship into simple, integrable parts, allowing us to find the general solution with comparative ease.