Question

Find the solution of the following initial value problems. $$u^{\prime}(x)=\frac{x e^{2 x}+4 e^{x}}{x e^{x}} ; u(1)=0, x>0$

Short answer

Question: Solve the initial value problem \(u^{\prime}(x)=\frac{x e^{2 x}+4 e^{x}}{x e^{x}}\) with \(u(1) = 0\). Answer: \(u(x) = e^{x}(x-1) + 4x - 4\)

Step-by-step solution

Find the general solution of the differential equation

First, let's simplify the expression given in the differential equation: $$ u^{\prime}(x)=\frac{x e^{2 x}+4 e^{x}}{x e^{x}} $$ Dividing both the numerator and denominator by \(xe^x\), we get: $$ u^{\prime}(x)=\frac{x e^{2 x}+4 e^{x}}{x e^{x}} = \frac{x e^{x} + 4}{x} $$ Now, we will solve this first-order differential equation using the separation of variables method: $$ u^{\prime}(x)=\frac{u'(x)}{dx}=\frac{x e^{x} + 4}{x} $$ Separate the variables: $$ u'(x)dx = (x e^{x} + 4) dx $$ Integrate both sides: $$ \int u'(x) dx = \int (x e^{x} + 4) dx $$ Let's integrate the right-hand side: $$ \int (x e^{x} + 4) dx = e^{x}(x-1) + 4x + C $$ So, we have: $$ u(x) = e^{x}(x-1) + 4x + C $$ This is the general solution of the given differential equation.

Apply the initial condition to find the particular solution

From the given condition, we have \(u(1) = 0\). Now, let's find the value of constant \(C\): $$ u(1) = e^{1}(1-1) + 4(1) + C = 0 $$ Since \(e^{1}(1-1)\) is equal to zero, we get: $$ C = -4 $$ Now, let's plug the value of C back into the general solution: $$ u(x) = e^{x}(x-1) + 4x - 4 $$ This is the particular solution of the given initial value problem.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in differential equations involves finding a function based on its derivative and an initial condition. The goal is to determine the specific solution that passes through a given point on the graph of the solution.

This generally involves:

• Identifying a differential equation that describes a family of functions.
• Applying an initial condition that provides an exact value of the function at a specific point.

In this exercise, the initial condition is given as \(u(1) = 0\). This initial value is crucial because it helps nail down the precise solution from the potentially infinite family of solutions provided by the general solution.

Separation of Variables

Separation of Variables is a common technique used to solve first-order differential equations. The main idea is to rearrange the equation to separate the differentials and place all terms involving one variable on one side of the equation, and all terms involving the other variable on the opposite side.

After rearranging, we can integrate each side separately.

Here, the differential equation \(u^{\prime}(x) = \frac{x e^{x} + 4}{x}\) was restructured to isolate the differential terms. By integrating both sides of the separated equation, we find an expression that includes the arbitrary constant \(C\), which needs further evaluation using an initial value.

First-Order Differential Equation

A First-Order Differential Equation is any differential equation that contains only first derivatives. Here's a basic identifier: if your equation involves only the first derivative of a function, it's first-order.

First-order equations often look like this: \(\frac{dy}{dx} = f(x,y)\). Solutions to these equations describe curves, not just points or values, that represent the changes dictated by the equation across a plane.

In this task, \(u^{\prime}(x) = \frac{x e^{x} + 4}{x}\) is our first-order differential equation. Solving it involves finding a function \(u(x)\) that satisfies this derivative relationship.

Particular Solution

Once a general solution is found, a particular solution can be determined using the initial conditions provided. The general solution includes a constant, often denoted as \(C\), which represents the endless range of possible solutions.

The initial condition is used to calculate this constant. For the problem at hand, using \(u(1) = 0\), we substitute into the general solution to find that \(C = -4\).

We then substitute \(C = -4\) back into the general solution \(u(x) = e^{x}(x-1) + 4x + C\) to derive the particular solution as \(u(x) = e^{x}(x-1) + 4x - 4\).

This solution exactly fits the initial conditions and differential equation, describing a specific curve among all those described by the general form.