Problem 36 - Advanced Engineering Mathematics

Question

For \(y^{\prime}+e^{x} y=1\) an integrating factor is \(e^{e^{x}}\). Thus \\[ \frac{d}{d x}\left[e^{e^{x}} y\right]=e^{e^{x}} \quad \text { and } \quad e^{e^{x}} y=\int_{0}^{x} e^{e^{t}} d t+c \\] From \(y(0)=1\) we get \(c=e,\) so \(y=e^{-e^{x}} \int_{0}^{x} e^{e^{t}} d t+e^{1-e^{x}}\) When \(y^{\prime}+e^{x} y=0\) we can separate variables and integrate: \\[ \frac{d y}{y}=-e^{x} d x \text { and } \ln |y|=-e^{x}+c \\] Thus \(y=c_{1} e^{-e^{x}} .\) From \(y(0)=1\) we get \(c_{1}=e,\) so \(y=e^{1-e^{x}}\) When \(y^{\prime}+e^{x} y=e^{x}\) we can see by inspection that \(y=1\) is a solution.

Step-by-Step Solution

Answer

The general solution is \( y = e^{1-e^x} + e^{-e^x} \int_0^x e^{e^t} dt \). The particular solution for given conditions is \( y = e^{1-e^x} \).

Step 1: Rewrite the original differential equation

The given differential equation is: \[ y' + e^x y = 1 \] Identify this as a first-order linear differential equation and seek an integrating factor.

Step 2: Identify the integrating factor

The integrating factor given is \( e^{e^x} \). Multiplying the whole differential equation by this integrating factor yields: \[ e^{e^x} y' + e^{e^x} e^x y = e^{e^x} \] This transforms the equation into an exact derivative form.

Step 3: Integrate the transformed equation

Recognize that the left side of the equation is the derivative of \( e^{e^x} y \). Thus, the equation becomes \[ \frac{d}{dx}(e^{e^x} y) = e^{e^x} \]. Integrate both sides with respect to \( x \): \[ e^{e^x} y = \int e^{e^x} \, dx + c \]

Step 4: Substitute initial condition and solve for integration constant

Substitute \( x = 0 \) and \( y = 1 \) into the equation to find \( c \). Since \( y = e^{-e^x} \int_0^x e^{e^t} dt + e^{1-e^x} \), substituting the given values you get: \( c = e \).

Step 5: Separate variables for the homogeneous equation

For the homogeneous part \( y' + e^x y = 0 \), separate variables: \[ \frac{dy}{y} = -e^x \, dx \] Integrate both sides: \[ \ln |y| = -e^x + c \] This solves to \( y = c_1 e^{-e^x} \) after exponentiating both sides.

Step 6: Solve for constant using initial condition for homogeneous equation

Utilize \( y(0)=1 \) to find \( c_1 \): hence \( c_1 = e \). Therefore, \( y = e^{1-e^x} \) is the solution to the homogeneous differential equation.

Step 7: Inspect for particular solution for the non-homogeneous equation

For the non-homogeneous equation \( y' + e^x y = e^x \), assume \( y = 1 \) and substitute into the equation: \( 0 + e^x \times 1 = e^x \), this confirms \( y=1 \) is a valid solution.

Key Concepts

Integrating Factors

To solve first-order linear differential equations, we often employ a method called integrating factors. This technique is particularly effective because it transforms a non-exact equation into an exact one, which is easier to integrate.

Consider the differential equation, \( y' + P(x)y = Q(x) \). Here, the function \( \mu(x) = e^{\int P(x)\, dx} \) is called the integrating factor. By multiplying the entire differential equation by \( \mu(x) \), the left part of the equation becomes the derivative of \( \mu(x)y \).

This simplifies the equation to \( \frac{d}{dx}[\mu(x)y] = \mu(x)Q(x) \), and you can integrate both sides to find \( y \).

• Multiply the original equation with the integrating factor.
• Transform the left side into a single derivative term.
• Integrate both sides with respect to \( x \) to solve for \( y \).
• Apply any initial conditions to find constants.

Homogeneous Differential Equations

A homogeneous differential equation is one where all terms depend only on the dependent variable and its derivatives, usually set to zero. These equations often have the form \( y' + P(x)y = 0 \).

You can solve these equations using separation of variables, which involves isolating \( y \) terms on one side and \( x \) terms on the other. For example, dividing both sides by \( y \) gives \( \frac{dy}{y} = -P(x)dx \).

By integrating both sides, you can solve for \( y \). The solution will typically incorporate a constant \( C \), found using initial conditions.

• Isolate \( y \) terms and \( x \) terms on opposite sides.
• Integrate both sides separately.
• Use initial conditions to find the integration constant.

Non-Homogeneous Differential Equations

Non-homogeneous differential equations include a non-zero term on the right side, making them \( y' + P(x)y = Q(x) \) where \( Q(x) eq 0 \). These can be more complex than homogeneous equations because the presence of \( Q(x) \) demands a different approach.

One effective method is to find a particular solution by inspection or other techniques. For instance, if you suspect a simple function, like a constant, might be a solution, substitute it back into the equation to check its validity.

After finding a particular solution, the general solution combines it with the solution of the associated homogeneous equation, such as \( y = y_h + y_p \), where \( y_h \) is the solution to the homogeneous part and \( y_p \) is the particular solution.

• Identify the non-homogeneous term \( Q(x) \).
• Find a particular solution by trying simple forms or other methods.
• Combine the particular solution with the homogeneous solution.

Separation of Variables

Separation of Variables is a strategy often applied to solve differential equations where the variables can be rearranged so that each belongs to a different side of the equation. This technique is especially useful for equations like \( \frac{dy}{dx} = g(x)h(y) \).

To separate the variables, rearrange it as \( \frac{1}{h(y)}dy = g(x)dx \). Then, integrate both sides with respect to their respective variables. This process leads to solutions involving logarithms or exponential functions,
showing that you have separated the \( y \) from the \( x \) terms effectively.

• Express the equation such that all \( y \) and its derivatives are on one side.
• Integrate each side independently.
• Solve for \( y \) after integration.
• Use initial conditions to find specific solutions.