Question

Solve the given initial value problem. What is the interval of existence of the solution? $$ t^{2} y^{\prime \prime}+3 t y^{\prime}+y=8 t+9, \quad y(-1)=1, \quad y^{\prime}(-1)=0 $$

Short answer

A: The method of undetermined coefficients was used to find the particular solution for the ODE. 2. What is the auxiliary equation for the homogeneous ODE? A: The auxiliary equation for the homogeneous ODE is \(r^2 + 2r + 1 = 0\). 3. What are the values for the constants C_1 and C_2 after applying the initial conditions? A: The values for the constants C_1 and C_2 after applying the initial conditions are C_1 = -8e and C_2 = 6e. 4. What is the interval of existence for the solution, considering the continuity of coefficients? A: The interval of existence for the solution is \(t \in (-\infty,\infty)\), except for the point \(t=0\), which is not in the domain of the solution.

Step-by-step solution

Rewrite the ODE in more convenient form

Given the ODE, multiplying both sides by t^2: $$ t^2(t^{2} y^{\prime \prime}+3 t y^{\prime}+y)=t^2(8 t+9). $$ Let \(v = ty\), we will have \(v' = y + ty'\) and \(v'' = ty'' + 2y'\). So, we can rewrite the ODE in terms of \(v\) as: $$ v'' + 2v' + v = 8t^3+9t^2. $$

Find a particular solution

We will now try to find a particular solution using the method of undetermined coefficients. We assume a cubic function for the particular solution in the form of: $$ v_p = At^3 + Bt^2 + Ct + D $$ Differentiating, we obtain: $$ v'_p = 3At^2 + 2Bt + C $$ $$ v''_p = 6At + 2B $$ Substituting the assumed particular solution and its derivatives into the rewritten ODE: $$ (6At + 2B) + 2(3At^2 + 2Bt + C) + (At^3 + Bt^2 + Ct + D) = 8t^3+9t^2 $$ Comparing the coefficients, we get the following system of equations for A, B, C, and D: $$ A = 8 \\ B + 6A = 9 \\ C + 4B = 0 \\ D + 2C = 0 $$ Solving this system, we obtain A=8, B=-3, C=12, and D=-24. Thus, the particular solution is: $$ v_p = 8t^3 - 3t^2 + 12t - 24 $$

Solve the homogeneous ODE

To find the complementary solution, we solve the associated homogeneous ODE: $$ v'' + 2v' + v = 0 $$ Guessing the form of exponential function \(v_c = e^{rt}\), substituting, and dividing by \(e^{rt}\), we get the auxiliary equation: $$ r^2 + 2r + 1 = 0 $$ Solving the auxiliary equation, we find that \(r = -1\) is the only root with multiplicity 2. Therefore, the complementary solution is: $$ v_c = (C_1 + C_2t)e^{-t} $$

Combine the particular and complementary solutions

Adding the particular and complementary solutions yields the general solution for \(v\): $$ v = v_c + v_p = (C_1 + C_2t)e^{-t} + 8t^3 - 3t^2 + 12t - 24 $$ Now, recall that \(v = ty\), so the general solution for \(y\) is given by: $$ y = \frac{1}{t}[(C_1 + C_2t)e^{-t} + 8t^3 - 3t^2 + 12t - 24] $$

Apply the initial conditions

Finally, apply the initial conditions \(y(-1)=1\) and \(y'(-1)=0\) to determine the constants C_1 and C_2: $$ 1 = \frac{1}{-1}[(C_1 - C_2)e + 8 - 3 - 12 - 24] $$ $$ 0 = -C_2e - 8 + 6 - 12 + (C_1 + 2C_2)e $$ Solving this system of equations, we find C_1 = -8e and C_2 = 6e. Therefore, the solution for the initial value problem is: $$ y = \frac{1}{t}((-8e + 6et)e^{-t} + 8t^3 - 3t^2 + 12t - 24) $$ Taking into account the continuity of coefficients, we can safely say that the interval of existence of the solution is \(t \in (-\infty,\infty)\), except for the point \(t=0\) which is not in the domain of the solution.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They express the relationship between the rates of change and the functions themselves. A common example is the equation governing the growth of a population over time, or the motion of a particle under the force of gravity. Solutions to differential equations are functions that satisfy these relationships. With the given exercise, the differential equation presented requires understanding the rate of change, in this case, of a function y, in relation to the variable t. This process often requires that we manipulate the equation to express it in terms more familiar to us and more accessible to solution methods available.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find a particular solution to non-homogeneous linear differential equations with constant coefficients. When the non-homogeneous term is a simple function, like a polynomial, exponential, sine, or cosine, we can make an educated guess on the form of the particular solution. Generally, the guess will take the form of the non-homogeneous term, and possibly its derivatives, based on which you determine the coefficients by substitution and equating coefficients. In our problem, we began by guessing a polynomial function for our particular solution because the non-homogeneous term was a polynomial. After deriving this function and substituting it into the original equation, we compared coefficients to find the values of those undetermined coefficients, effectively customizing our solution to fit the problem.

Interval of Existence

The interval of existence for a solution to a differential equation refers to the range of the independent variable, usually time or space, for which the solution is defined and behaves according to the differential equation. In the context of our exercise, the interval of existence can be influenced by several factors, such as the behavior of the coefficients of the differential equation and points where the solution may become undefined. For example, if the coefficient in front of the highest derivative is zero at some point, it could lead to singularities and affect the interval of existence. Additionally, initial conditions help fix the solution to a specific instance, which could also play a role in determining the interval of existence. When solving the problem, it is crucial to avoid points of discontinuity, like t = 0 in this case, since the original ODE includes a coefficient that is a function of t, and we cannot divide by zero.

Homogeneous ODE

A homogeneous ordinary differential equation (ODE) is one in which every term is a function of the dependent variable and its derivatives. Essentially, a homogeneous ODE is set equal to zero. Solving a homogeneous ODE often involves finding a complementary solution that reflects the behavior of the differential equation without the influence of any external forces or inputs represented by the non-homogeneous part. In our initial value problem, the method to find the general solution combined the complementary solution of the related homogeneous equation and a particular solution to the non-homogeneous equation. This complementary solution often involves exponential functions or trigonometric functions, chosen based on the roots of the characteristic equation derived from the homogeneous ODE.