Question

find the solution of the given initial value problem. $$ y^{\prime}-y=2 t e^{2 t}, \quad y(0)=1 $$

Short answer

Question: Given the differential equation \(y' - y = 2te^{2t}\) with initial condition \(y(0) = 1\), find the solution y(t). Answer: The solution to the initial value problem is \(y(t) = (2t-2)e^{2t} + 3e^{t}\).

Step-by-step solution

Rewrite as a first-order linear equation in standard form

We can notice that the given differential equation is already in standard form: $$ y^{\prime}-y=2 t e^{2 t} $$ Here, the function p(t) is -1, and the function g(t) is 2te^{2t}.

Determine the integrating factor

We'll now find the integrating factor \(μ(t)\) by calculating the exponential of the integral of p(t) with respect to t: $$ μ(t) = e^{\int -1 dt} = e^{-t} $$

Multiply by the integrating factor

Next, multiply both sides of the equation by the integrating factor, \(e^{-t}\): $$ e^{-t} (y^{\prime}-y)=2 t e^{t} $$

Integrate both sides with respect to t

Now, we'll find that the left side of the equation is the derivative of the product \(y(t)e^{-t}\). Let's integrate both sides with respect to t: $$ \int y(t) e^{-t} dt = \int 2 t e^{t} dt $$

Solve the integral on the right side

To solve the integral on the right side, we will use integration by parts: $$ \int 2 t e^{t} dt = (2t-2)e^{t} + C $$

Integrate left side and solve for y(t)

Now the equation looks like: $$ y(t) e^{-t} = (2t-2)e^{t} + C $$ Let's multiply both sides by \(e^{t}\) to solve for y(t): $$ y(t) = (2t-2)e^{2t} + Ce^{t} $$

Apply the initial condition and find C

Lastly, we'll apply the initial condition y(0) = 1 to find C: $$ 1 = (2(0)-2)e^{2(0)} + Ce^{(0)} $$ This simplifies to: $$ 1 = -2 + C $$ So, C = 3.

Write the final solution

Now, we have found the solution to the initial value problem: $$ y(t) = (2t-2)e^{2t} + 3e^{t} $$

Key concepts

First-order Linear Equation

In mathematics, a first-order linear differential equation takes the form \( y' + p(t)y = g(t) \). In our initial value problem, the given equation is \( y' - y = 2te^{2t} \). We can see that it fits the standard form by rewriting it as \( y' + (-1)y = 2te^{2t} \). Here:

• \( y' \) represents the derivative of \( y \) with respect to \( t \).
• \( p(t) = -1 \) and \( g(t) = 2te^{2t} \).

The first-order linear equation provides a systematic way to find an accurate solution for such problems, simplifying the process of determining the function \( y(t) \) that satisfies the equation.

Integrating Factor

The integrating factor is a powerful tool used to solve first-order linear differential equations. It transforms the equation into a form that is easier to integrate.To find the integrating factor \( \mu(t) \), we compute the exponential of the integral of \( p(t) \), which is simply \( e^{-t} \) in our example. The integrating factor formula is:

• \( \mu(t) = e^{\int p(t) \, dt} = e^{-t} \)

Multiplying the entire differential equation by this integrating factor, we obtain:\( e^{-t} (y' - y) = 2t e^t \).This step ensures the left side of the equation becomes an exact derivative, making integration straightforward.

Integration by Parts

Integration by parts is a technique used to integrate products of functions, such as \( \int 2t e^t \, dt \), which we encounter on the right side of our equation.The integration by parts formula is:

• \( \int u \, dv = uv - \int v \, du \)

For our integral:

• Let \( u = 2t \) and \( dv = e^t \, dt \)
• Then \( du = 2 \, dt \) and \( v = e^t \)

Using the formula, we find:\( \int 2t e^t \, dt = (2t - 2)e^t + C \).Integration by parts breaks down complicated integrals into simpler parts, leading us to the solution.

Differential Equation Solution

The solution to a differential equation like our initial value problem combines all the processes described.After finding and applying the integrating factor, the equation simplifies so that the left is the derivative of \( y(t)e^{-t} \).Upon integrating both sides, including using integration by parts on the right, we solve for \( y(t) \):\( y(t) e^{-t} = (2t - 2)e^t + C \).Solving for \( y(t) \) by multiplying through by \( e^t \), results in:\( y(t) = (2t - 2)e^{2t} + Ce^t \).Applying the initial condition \( y(0) = 1 \), we find \( C = 3 \). Finally, the specific solution to the initial value problem is:\( y(t) = (2t - 2)e^{2t} + 3e^t \).This embodies the complete understanding and application of mathematical techniques to solve a differential equation.