Question

Use the method of Problem 32 to solve the given differential $$ 2 y^{\prime \prime}+3 y^{\prime}+y=t^{2}+3 \sin t \quad(\text { see Problem } 7) $$

Short answer

Question: Determine the general solution for the given differential equation: $$ 2 y^{\prime \prime}+3 y^{\prime}+y = t^{2}+3 \sin t. $$ Answer: The general solution for the given differential equation is: $$ y(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t} + \frac{1}{2}t^2 - 3t + \frac{5}{2} + \sin t - \frac{2}{3}\cos t. $$

Step-by-step solution

Find the complementary function

First, find the complementary function by solving the homogeneous equation, i.e. the equation $$ 2 y^{\prime \prime} +3 y^{\prime} + y = 0. $$ To do this, let \(y = e^{rt}\) and substitute into the homogeneous equation to get the auxiliary equation: $$ 2r^2 + 3r + 1 = 0. $$ Apply the quadratic formula to solve for \(r\): $$ r = \frac{-3 \pm \sqrt{3^2 - 4(2)(1)}}{4} = \frac{-3 \pm \sqrt{1}}{4}. $$ The solutions are \(r_1 = -\frac{1}{2}\) and \(r_2 = -1\). Hence, the complementary function can be represented as: $$ y_c(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t}. $$

Find the particular solution

To find the particular solution, let's guess a solution and adjust for the appropriate order. Given that the right side of the equation shows a quadratic function in \(t\) and a trigonometric function, we can make the following guesses: $$ y_p(t) = At^2 + Bt + C + D\sin t + E\cos t. $$ Now differentiate \(y_p(t)\) twice: $$ y_p'(t) = 2At + B + D\cos t - E\sin t, $$ and $$ y_p''(t) = 2A - D\sin t - E\cos t. $$ Substitute \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the given differential equation and simplify: $$ 2(2A - D\sin t - E\cos t) + 3(2At + B + D\cos t - E\sin t) + (At^2 + Bt + C + D\sin t + E\cos t) = t^2 + 3\sin t. $$ This leads to the following system of equations: $$ \begin{cases} A = \frac12, \\ 6A + B = 0, \\ 2C + 3D + E = 3, \\ 2A + 3E = 0, \\ 3B - 2D + D = 0 \end{cases}. $$ Solving this system, we find the coefficients: $$ A = \frac12, \quad B = -3, \quad C = \frac{5}{2}, \quad D = 1, \quad E = -\frac23. $$ Thus, the particular solution has the form: $$ y_p(t) = \frac{1}{2}t^2 - 3t + \frac{5}{2} + \sin t - \frac{2}{3}\cos t. $$

Find the general solution

Finally, add both the complementary function and the particular solution to obtain the general solution: $$ y(t) = y_c(t) + y_p(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t} + \frac{1}{2}t^2 - 3t + \frac{5}{2} + \sin t - \frac{2}{3}\cos t. $$

Key concepts

Complementary Function

In solving differential equations, finding the complementary function is a crucial step. The complementary function, denoted as \( y_c(t) \), represents the general solution of the associated homogeneous differential equation. To derive it, we begin with the differential equation:

• Write the homogeneous part of the equation. For example: \( 2 y^{\prime \prime} + 3 y^{\prime} + y = 0 \). Here, we exclude any terms that are not dependent on \( y \) or its derivatives.
• Assume a trial solution in the form \( y = e^{rt} \) (a typical approach for linear equations with constant coefficients).
• Substitute this trial solution into the homogeneous equation to form the auxiliary equation.

By solving this auxiliary equation, we find values of \( r \), which can then be used to construct the complementary function. For example, if the solution to the auxiliary equation yields \( r_1 = -\frac{1}{2} \) and \( r_2 = -1 \), the complementary function becomes:\[y_c(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t}\]Here, \( C_1 \) and \( C_2 \) are constants determined by initial conditions if they are provided.

Particular Solution

Finding a particular solution for a differential equation involves constructing a specific solution that satisfies the non-homogeneous part of the differential equation. This is typically represented by \( y_p(t) \). Given an equation that includes terms such as polynomials or trigonometric functions (like \( t^2 + 3 \sin t \)), we must tailor our guess for the particular solution. Here's how:

• Identify the form of the non-homogeneous terms: in this case, a quadratic in \( t \) and a trigonometric term.
• Guess a trial function that includes similar terms. For instance, \( At^2 + Bt + C + D\sin t + E\cos t \), where \( A, B, C, D, \) and \( E \) are coefficients we need to determine.

Substitute this guessed function into the differential equation and solve for these coefficients by equating coefficients on both sides of the equation.Once determined, the particular solution is constructed. For example:\[y_p(t) = \frac{1}{2}t^2 - 3t + \frac{5}{2} + \sin t - \frac{2}{3}\cos t\]This particular solution, when added to the complementary function, will yield the general solution of the differential equation.

Auxiliary Equation

The auxiliary equation is a polynomial equation used to determine the complementary function for a homogeneous differential equation with constant coefficients. It arises from substituting a guessed form \( y = e^{rt} \) into the homogeneous equation.To find the auxiliary equation:

• Start with the homogeneous equation. For instance, \( 2 y^{\prime \prime} + 3 y^{\prime} + y = 0 \).
• Assume the solution form \( y = e^{rt} \) and plug it into the equation, leading to terms like \( 2r^2, 3r, \) and \( 1 \).
• The resulting expression is the auxiliary equation: \( 2r^2 + 3r + 1 = 0 \).

By solving this auxiliary equation using algebraic methods or the quadratic formula, we can find the roots \( r_1 \) and \( r_2 \). These roots are essential for forming the complementary function of the differential equation.Solving the auxiliary equation becomes straightforward when using the quadratic formula, especially when it involves higher-degree terms.

Quadratic Formula

The quadratic formula is a method to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). It is particularly useful when determining roots that contribute to the complementary function of a differential equation. The general formula for solving a quadratic expression is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here’s how it applies in the context of the differential equation's auxiliary equation:

• Identify the coefficients \( a, b, \) and \( c \) from the auxiliary equation. For example, in \( 2r^2 + 3r + 1 = 0 \), \( a = 2, b = 3, \text{ and } c = 1 \).
• Substitute these values into the quadratic formula to find the roots.

For the sample equation, the roots become:\[r = \frac{-3 \pm \sqrt{3^2 - 4(2)(1)}}{4} = \frac{-3 \pm \sqrt{1}}{4}\]Which simplifies to \( r_1 = -\frac{1}{2} \) and \( r_2 = -1 \). These solutions of \( r \) guide the construction of the complementary function for the associated differential equation.