Question

Find the solution of the given initial value problem. $$ y^{\prime \prime}-2 y^{\prime}+y=t e^{\prime}+4, \quad y(0)=1, \quad y^{\prime}(0)=1 $$

Short answer

Question: Solve the initial value problem given by the second-order linear non-homogeneous differential equation \(y'' - 2y' + y = te^t + 4\), with the initial conditions \(y(0) = 1\) and \(y'(0) = 1\). Answer: The solution to the initial value problem is \(y(t) = -3e^t + 4te^t + t^2e^t + 4\).

Step-by-step solution

Solve the Homogeneous Equation

First, we need to find the complementary solution to the homogeneous equation: $$ y'' - 2y' + y = 0 $$ To solve this homogeneous equation, we assume a solution of the form \(y=e^{rt}\). We now plug this into the differential equation and obtain the characteristic equation: $$ r^2 - 2r + 1 = 0 $$

Solve the Characteristic Equation

To find \(r\), we need to solve this quadratic equation: $$ r^2 - 2r + 1 = (r - 1)^2 = 0 $$ Therefore, \(r_1 = r_2 = 1\). Since both roots are equal, the complementary solution is given by: $$ y_c(t) = C_1e^t + C_2 te^t $$

Find the Particular Solution using Undetermined Coefficients

The non-homogeneous part is \(te' + 4\). For the particular solution, we consider the form: $$ y_p(t) = At^2e^t + B $$ Now we need to find \(y_p'(t)\) and \(y_p''(t)\) to plug them into the equation. The first derivative: $$ y_p'(t) = 2Ate^t + At^2e^t + Ae^t $$ The second derivative: $$ y_p''(t) = 2Ae^t + 2Ate^t + 2Ate^t + At^2e^t + 2Ae^t + Ae^t $$ Now, group the terms: $$ y_p''(t) = (3A + 2At + At^2)e^t $$

Plug in the Particular Solution

We plug \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the non-homogeneous equation: $$ (3A + 2At + At^2)e^t - 2(2Ate^t + At^2e^t + Ae^t) + (At^2e^t + B) = te^t + 4 $$ Simplify the equation: $$ e^t(At^2 - 4A + B) = te^t + 4 $$ Now we equate the coefficients of like terms: $$ A = 1,\quad B = 4 $$ So, the particular solution is: $$ y_p(t) = t^2e^t + 4 $$

Combine Complementary and Particular Solutions

The general solution is the sum of the complementary and particular solutions: $$ y(t) = y_c(t) + y_p(t) = C_1e^t + C_2 te^t + t^2e^t + 4 $$

Apply Initial Conditions

Now we use the given initial conditions to find the constants \(C_1\) and \(C_2\): $$ y(0) = 1,\quad y'(0) = 1 $$ At \(t = 0\), we have: $$ y(0) = C_1 + 0 + 0 + 4 \Rightarrow C_1 = -3 $$ We'll also need the derivative of \(y(t)\): $$ y'(t) = -3e^t + C_2e^t + te^t + 2t e^t $$ At \(t = 0\), we have: $$ y'(0) = -3 + C_2 + 0 + 0 \Rightarrow C_2 = 4 $$

Find the Complete Solution

Now, we can plug the constants \(C_1\) and \(C_2\) back into the expression for \(y(t)\) to obtain the complete solution: $$ y(t) = -3e^t + 4te^t + t^2e^t + 4 $$

Key concepts

Second-Order Differential Equations

A second-order differential equation is a type of equation that involves the second derivative of a function, along with possibly the first derivative and the function itself. These equations are essential in modeling phenomena where the rate of change is influenced by the current rate and position—think of physics problems like motion under gravity or oscillations of springs.

In our initial exercise, we are asked to solve a second-order differential equation:

• The general form is usually \( ay'' + by' + cy = f(t) \)
• Where \( a, b, \) and \( c \) are constants, and \( f(t) \) is some function of time \( t \).

The equation we're dealing with here is solvable by breaking it into two parts: the homogeneous part and the particular part. Solving them separately allows engineers and mathematicians to build complex models step by step.

Complementary Solution

The complementary solution is the part of the solution that solves the homogeneous equation. This means we set the right-hand side of the differential equation to zero and find solutions that fit this form. These solutions represent the natural response of the system when no external forces are acting on it.

To find the complementary solution in this exercise:

• We consider only the equation \( y'' - 2y' + y = 0 \).
• The solution assumes a format \( y_c(t) = C_1e^{r_1t} + C_2e^{r_2t} \) depending on the roots of the characteristic equation.
• In this case, both roots derive from the characteristic equation being identical, providing a solution: \( y_c(t) = C_1e^t + C_2te^t \).

This complementary solution lines up perfectly with the system's inherent properties without external inputs.

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find the particular solution of a non-homogeneous differential equation. This assumes that the right-hand side of the equation, \( f(t) \), is a function that can be expressed in terms of polynomials, exponentials, or sine and cosine functions.

For our exercise, we approached the method by:

• Guessing a form for the particular solution: \( y_p(t) = At^2e^t + B \), handling the specific non-homogeneous part \( te^t + 4 \).
• Finding derivatives of this guessed form and substituting back into the differential equation to identify coefficients \( A \) and \( B \).
• Matching the coefficients from both sides to determine the values: \( A = 1 \), \( B = 4 \).

This efficient approach allows us to construct the particular solution swiftly by choosing the right format based on the type of non-homogeneous term present.

Characteristic Equation

The characteristic equation is a crucial stepping-stone that arises when seeking the complementary solution of a second-order linear differential equation. To derive it, you assume a trial solution of the form \( y = e^{rt} \), which translates the differential equation into an algebraic form.

The steps involve:

• Substituting \( y = e^{rt} \), which turns differentials into powers of \( r \).
• Reframing the original equation into a quadratic form \( ar^2 + br + c = 0 \)
• Solving the quadratic to find the roots \( r_1 \) and \( r_2 \).

In our equation, the characteristic equation is:\[ r^2 - 2r + 1 = 0 \]This simplifies to \((r-1)^2 = 0\), indicating repeated roots. This resulted in the solution form \( y_c(t) = C_1e^t + C_2te^t \), taking into account the repeated nature of the roots, which modifies the solution form slightly to accommodate for them as per linear differential equation theory.