Question

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$ y^{\prime \prime \prime}+y=t^{3} $$

Short answer

Answer: The general solution to the given third-order differential equation is \(y(t) = C_1e^{-t} + C_2 e^{\frac{1}{2}t} \cos \left(\frac{\sqrt{3}}{2}t\right) + C_3 e^{\frac{1}{2}t} \sin \left(\frac{\sqrt{3}}{2}t\right) + t^3 + 6t\), where \(C_1\), \(C_2\), and \(C_3\) are constants.

Step-by-step solution

Find the complementary function

Solve the homogeneous differential equation, which is given by $$ y^{\prime \prime \prime} + y = 0 $$ The characteristic equation for this differential equation is $$ r^{3} + 1 = 0 $$ To solve for r, we can see that it has a root at r=-1. Then, the characteristic equation becomes $$ (r+1)(r^2-r+1) = 0 $$ The quadratic equation does not have real roots. Solving for the complex roots we get the roots as \(r = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i\). Therefore, the complementary function is given by $$ y_c(t) = C_1e^{-t} + C_2 e^{\frac{1}{2}t} \cos \left(\frac{\sqrt{3}}{2}t\right) + C_3 e^{\frac{1}{2}t} \sin \left(\frac{\sqrt{3}}{2}t\right) $$

Find a particular solution

In order to find a particular solution for the given non-homogeneous differential equation, we can apply the method of undetermined coefficients. Since the non-homogeneous term is a polynomial of degree 3, we will assume a particular solution of the form $$ y_p(t) = At^3 + Bt^2 + Ct + D $$ To calculate the derivatives of \(y_p(t)\): $$ y_p^{(1)}(t) = 3At^2 + 2Bt + C, \quad y_p^{(2)}(t) = 6At + 2B, \quad y_p^{(3)}(t) = 6A $$ Plugging these derivatives into the differential equation, we get: $$ (6A) + (At^3 + Bt^2 + Ct + D) = t^3 $$ Comparing the coefficients, we get: $$ A = 1, \quad B = 0, \quad C = 6A = 6, \quad D = 0 $$ Thus, the particular solution is: $$ y_p(t) = t^3 + 6t $$

Formulate the general solution

Now, we can formulate the general solution by combining the complementary function and the particular solution: $$ y(t) = y_c(t) + y_p(t) = C_1e^{-t} + C_2 e^{\frac{1}{2}t} \cos \left(\frac{\sqrt{3}}{2}t\right) + C_3 e^{\frac{1}{2}t} \sin \left(\frac{\sqrt{3}}{2}t\right) + t^3 + 6t $$ This is the general solution to the given third-order differential equation.

Key concepts

Complementary Function

When we talk about finding a solution to a differential equation, one important component is the complementary function. This refers to the solution of the associated homogeneous differential equation. For our problem, this equation is derived by setting the non-homogeneous part (in this case, "\(t^3\)") of the differential equation to zero. So, it becomes \(y'''+y=0\).

The key to solving this is to determine the **characteristic equation**. It connects the differential equation to algebra, making it easier to find the solution. For our differential equation, this is \(r^3+1=0\). By identifying the roots of this equation, specifically \(r=-1, \frac{1}{2}\pm\frac{\sqrt{3}}{2}i\), we can construct the complementary function. The roots indicate the behavior of the solutions:

• A real root \(r=-1\) leads to an exponential component \(C_1e^{-t}\).
• The complex roots lead to oscillating solutions, bringing in cosine and sine functions.

Thus, the complementary function is \(y_c(t) = C_1e^{-t} + C_2e^{\frac{1}{2}t}\cos\left(\frac{\sqrt{3}}{2}t\right) + C_3e^{\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}t\right)\). It represents the inherent behavior of the system independent of external inputs.

Particular Solution

After determining the complementary function, we aim to find a particular solution of the non-homogeneous differential equation. This solution accounts for the change caused by non-homogeneous terms such as \(t^3\) in the original equation. There is no one-size-fits-all approach to finding this; different methods are used based on the nature of the non-homogeneous term.

In this example, we use the **method of undetermined coefficients**. Since our non-homogeneous term is a polynomial \(t^3\), we assume a solution of the same degree, \(y_p(t) = At^3 + Bt^2 + Ct + D\). This means guessing a polynomial, differentiating it, inserting it into the differential equation, and solving for coefficients \(A, B, C, \) and \(D\) by matching the terms on both sides.

• Once we calculate the derivatives and substitute them, we systematically equate our guessed polynomial to \(t^3\), which results in finding that \(A=1, B=0, C=6, D=0\).

This gives us the particular solution \(y_p(t) = t^3 + 6t\). It's like gaining an additional term that directly balances out the effect of the non-homogeneous element of the differential equation.

General Solution

Combining the complementary function and the particular solution gives us the general solution of the differential equation. This general solution encompasses all possible solutions to the differential equation by considering both intrinsic system properties and the particular effect of an external force or input.

For the given differential equation, the general solution is formulated as follows:

• The complementary function \(y_c(t) = C_1e^{-t} + C_2e^{\frac{1}{2}t}\cos\left(\frac{\sqrt{3}}{2}t\right) + C_3e^{\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}t\right)\) represents the solution of the homogeneous equation.
• The particular solution \(y_p(t) = t^3 + 6t\) accounts for the effect of the non-homogeneous part (\(t^3\) in this equation).

Thus, the general solution assembles these components:\[y(t) = C_1e^{-t} + C_2e^{\frac{1}{2}t}\cos\left(\frac{\sqrt{3}}{2}t\right) + C_3e^{\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}t\right) + t^3 + 6t\]This final solution takes into account the system's standard behavior and adjusts it according to the external influence \(t^3\). By substituting different values for constants \(C_1, C_2,\) and \(C_3\), you can derive specific solutions reflecting various initial conditions or scenarios.