Question

Find the solution of the following initial value problems. $$y^{\prime \prime}(t)=t e^{t}, y(0)=0, y^{\prime}(0)=1$$

Short answer

In conclusion, the function y(t) that satisfies the given second-order linear homogeneous differential equation and initial conditions is: $$y(t) = -3t + 3 + te^{t} - 3e^{t}.$$

Step-by-step solution

Write down the given equation and initial conditions

We have the following second-order linear homogeneous differential equation with initial conditions: $$y^{\prime \prime}(t) = t e^{t},$$ $$y(0) = 0,$$ $$y^{\prime}(0) = 1.$$

Solve the homogeneous part

The homogeneous part of the given equation is: $$y^{\prime \prime}(t) = 0.$$ Its general solution is: $$y_h(t) = Ct + D,$$ where C and D are constants.

Use the method of undetermined coefficients

We will find a particular solution, \(y_p(t)\), using the method of undetermined coefficients. Let's guess: $$y_p(t) = Ate^{t} + Be^{t},$$ where A and B are constants to be determined. We then find its derivatives: $$y_p^{\prime}(t) = Ate^{t} + e^{t}(A + B),$$ $$y_p^{\prime \prime}(t) = Ate^{t} + 2e^{t}(A + B) + e^{t}(A).$$ Now, we have to substitute the derivatives into the differential equation: $$Ate^{t} + 2e^{t}(A + B) + e^{t}(A) = te^{t}.$$ This equation is true if: $$A = 1$$ $$2A + B + A = 0$$ Solving these equations, we get: $$A=1$$ $$B=-3$$ So, the particular solution is: $$y_p(t) = te^{t} -3e^{t}.$$

Combine the homogeneous solution and particular solution

The general solution to our given equation is the sum of the homogeneous and particular solutions: $$y(t) = y_h(t) + y_p(t) = Ct + D + te^{t} -3e^{t}.$$

Apply the initial conditions

Now, we use the given initial conditions to determine the constants C and D: $$y(0) = 0 \Rightarrow D - 3e^0 = 0 \Rightarrow D = 3,$$ $$y^{\prime}(0) = 1 \Rightarrow C + 3 + A = 1 \Rightarrow C = -3.$$ Thus, the solution of the initial value problem is: $$y(t) = -3t +3 + te^{t} - 3e^{t}.$$

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They appear in many fields such as physics, engineering, and economics to model real-world systems. For example, the equation given in the problem, \(y''(t) = te^t\), is a second-order differential equation because it involves the second derivative of the function \(y(t)\).

In differential equations, we often seek solutions that describe how the system evolves over time. These solutions can be either:

• General Solution: Involves arbitrary constants and represents a family of solutions
• Particular Solution: A specific solution tailored to satisfy given initial or boundary conditions

Understanding how to solve these equations involves familiarity with different techniques and methods, such as separating variables, integrating factors, and using characteristic equations.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions of non-homogeneous linear differential equations. This method involves guessing a form for the particular solution based on the non-homogeneous part of the equation.

For our specific problem where the non-homogeneous part is \(te^t\), we guess a particular solution of the form \(y_p(t) = Ate^t + Be^t\). By differentiating this assumed solution and substituting it back into the differential equation, we can find suitable values for the coefficients \(A\) and \(B\).

This approach relies on the assumed structure and varies depending on the type of function on the right side of the differential equation. If the guess doesn't work initially, it might need adjustments or a different approach.

Homogeneous Equations

A homogeneous differential equation is one where all terms depend linearly on the function and its derivatives without any standalone constant or distinct term added. The general form resembles \(y' + P(t)y = 0\), indicating there is no function of \(t\) without \(y\).

In this problem, the homogeneous version of our equation is \(y''(t) = 0\). Solving this involves finding the general solution, \(y_h(t) = Ct + D\), where \(C\) and \(D\) are arbitrary constants. These constants can later be determined by applying initial or boundary conditions.

Homogeneous solutions are an essential part of solving differential equations as they form the complementary solution when combined with a particular solution of a non-homogeneous equation.

Particular Solution

The particular solution of a differential equation is the component that accounts for the specific initial or boundary conditions imposed on the problem. By finding a particular solution such as \(y_p(t) = te^t - 3e^t\), we tailor a solution that conforms to the unique configuration defined by the initial conditions provided:

• \(y(0) = 0\)
• \(y'(0) = 1\)

Particular solutions address the part of the problem that homogeneous solutions alone cannot solve. These solutions are developed using methods like undetermined coefficients or variation of parameters and are crucial in forming the complete solution for our differential equations, ensuring the equations correctly describe the system's behavior in specific scenarios.