Question

v\(z^{\prime}=3 z+t^{2} e^{-t}\) with \(z(0)=1\) over \([0,2]\)

Short answer

Solve using integrating factor and initial condition; follow steps for detailed calculation.

Step-by-step solution

Recognize the Type of Equation

The given equation is a first-order linear differential equation of the form \(z' = g(t) + p(t)z\), where \(p(t) = 3\) and \(g(t) = t^2 e^{-t}\).

Solve Homogeneous Equation

Solve the associated homogeneous equation \(z' = 3z\), which gives \(z_h = Ce^{3t}\).

Find the Integrating Factor

Compute the integrating factor \(I(t) = e^{\int p(t)dt} = e^{3t}\).

Apply Integrating Factor

Multiply the entire differential equation by the integrating factor: \(e^{3t}z' = e^{3t}(3z + t^2 e^{-t})\).

Simplify Equation Using Product Rule

Recognize that the left side \(e^{3t}z' = \frac{d}{dt}(e^{3t}z)\). Hence, the equation simplifies to \(\frac{d}{dt}(e^{3t}z) = e^{3t}t^2e^{-t}\).

Integrate Both Sides

Integrate both sides over the interval to find \(z(t)\): \[ e^{3t}z = \int e^{3t}t^2e^{-t} \, dt + C \].

Evaluate Particular Integral

Evaluate the integral on the right. Let \( u = t^2 \) and \( dv = e^{(3-1)t} dt \); using integration by parts, evaluate \(\int t^2 e^{2t} \, dt\). This integral must be done by parts twice.

Combine Homogeneous and Particular Solutions

The complete solution is \(z(t) = e^{-3t}(\text{solution to } \int e^{3t}t^2e^{-t} \, dt + C)\).

Apply Initial Condition

Use the initial condition \(z(0) = 1\) to find \(C\). Substitute \(t=0\) into the solution: \(1 = e^{-3(0)} \left( \text{evaluation of integral at } t=0 + C \right)\) and solve for \(C\).

Verify the Solution

Verify the solution satisfies the original differential equation and the initial condition over the interval \([0, 2]\).

Key concepts

First-order linear differential equations

First-order linear differential equations are equations that can be written in the form \( rac{dy}{dx} + p(x)y = g(x) \). These equations are called 'linear' because they feature the unknown function \(y\) and its derivative \(y'\) in linear terms. Solving these equations often requires special techniques, as the solution is not always straightforward.

Understanding the structure of these equations is crucial. In our exercise, the equation can be rewritten in the form \( z' - 3z = t^2 e^{-t} \), identifying \( p(t) = -3 \) and \( g(t) = t^2 e^{-t} \). Recognizing this layout helps us to correctly apply the necessary solving techniques for these equations.

Integrating factor method

The integrating factor method is a powerful technique used to solve first-order linear differential equations. The basic idea is to multiply the entire equation by an 'integrating factor' which is designed to turn the left-hand side into a derivative of a product of functions.

The integrating factor \( I(t) \) is found by calculating \( e^{\int p(t) dt} \). In our exercise, since \( p(t) = 3 \), we calculate \( I(t) = e^{3t} \).

Multiplying the original differential equation by this integrating factor simplifies the left side into a product rule form: \( \frac{d}{dt}(e^{3t}z) \). This clever trick allows easier integration of the equation, making it feasible to solve for the solution \( z(t) \).

Initial value problems

An initial value problem involves solving a differential equation along with a condition that specifies the value of the function at a particular point. This extra piece of information is vital for determining the "constant of integration" that appears in the general solution of the differential equation.

In our case, the initial value problem is given by \( z(0) = 1 \). This means that after solving the differential equation, we apply this condition to find the particular constant \( C \) that fits the solution to the specific context. This step ensures that the final solution isn't just a general formula, but one that satisfies the stated initial condition, effectively pinning down one particular solution among potentially infinite others.

Integration by parts

Integration by parts is a technique from calculus used to integrate products of functions. It's based on the rule \( \int u \, dv = uv - \int v \, du \). This method is particularly useful when faced with an integral that is not easily solvable by standard methods.

In our problem, we've encountered the integral \( \int t^2 e^{2t} \, dt \). This requires applying integration by parts more than once, as each step simplifies a part of the integral. Although it might seem complex initially, breaking it down step-by-step allows for the integration to eventually become manageable.

By continually applying integration by parts, we eventually solve the integral. This process enables us to find the particular solution of the differential equation when substituted back with the integrating factor method.