Question

Use the method of variation of parameters to determine the general solution of the given differential equation. $$y^{\prime \prime \prime }-2y^{\prime \prime }-y^{\prime }+2y=e^{4t}$$

Short answer

Question: Determine the general solution for the given third-order linear differential equation using the method of variation of parameters: $$y^{\prime \prime \prime }-2y^{\prime \prime }-y^{\prime }+2y=e^{4t}$$ Answer: The general solution for the given third-order linear differential equation using the method of variation of parameters is: $$y(t)=C_1{e}^{-t}+C_2{e}^t+C_3{e}^{2t}+(-5t)e^{-t}-t{e}^t$$, where $C_1$, $C_2$, and $C_3$ are arbitrary constants.

Step-by-step solution

Characteristic Equation of the Homogeneous Equation

First, let's find the homogeneous equation related to the given differential equation: $$y^{\prime \prime \prime }-2y^{\prime \prime }-y^{\prime }+2y=0$$ To find the characteristic equation, we will replace each derivative with an exponent. Let's assume a solution of the form $y=e^{rt}$ for the homogeneous equation. For the given equation, the characteristic equation becomes: $$r^3-2r^2-r+2=0$$

Finding the Roots of the Characteristic Equation

The characteristic equation is a cubic equation. We should try to find at least one real root which will help factorize the equation into smaller polynomials to find the remaining roots. By trial and error or synthetic division, one will observe that $r=1$ is a root. Now we can factor the equation as: $$(r-1)(r^2-r-2)=0$$ The quadratic equation has roots: $$r_{1,2}=\frac{1\pm \sqrt{1^2-4(-2)}}{2}=\frac{1\pm \sqrt{9}}{2}=\frac{1\pm 3}{2}$$ Thus, the roots are $r_1=-1$, $r_2=1$, and $r_3=2$.

Finding the Complementary Function

Using the roots $-1$, $1$, and $2$, we can write the complementary function for the given homogeneous equation as: $$y_c=C_1{e}^{-t}+C_2{e}^t+C_3{e}^{2t}$$ where $C_1,C_2$, and $C_3$ are constants.

Applying Variation of Parameters

Next, let's use the method of variation of parameters to find the particular solution. We need to take the first derivative of the complementary solution: $$y_c^{\prime }=-C_1{e}^{-t}+C_2{e}^t+2C_3{e}^{2t}$$ To find the particular solution, we will assume it in the form: $$y_p=u_1{e}^{-t}+u_2{e}^t+u_3{e}^{2t}$$ Now, we need to differentiate $y_p$: $$y_p^{\prime }=-u_1^{\prime }{e}^{-t}+u_1{e}^{-t}+u_2^{\prime }{e}^t+u_2{e}^t+2u_3^{\prime }{e}^{2t}+2u_3{e}^{2t}$$ We will differentiate $y_p^{\prime }$ once more to get: $$y_p^{″}=u_1^{\prime }{e}^{-t}+2u_1{e}^{-t}+u_2^{\prime }{e}^t+2u_2{e}^t+4u_3^{\prime }{e}^{2t}+4u_3{e}^{2t}$$ Now we take the third derivative of $y_p$: $$y_p^{‴}=-u_1^{\prime }{e}^{-t}+3u_1{e}^{-t}+u_2^{\prime }{e}^t+3u_2{e}^t+8u_3^{\prime }{e}^{2t}+8u_3{e}^{2t}$$

Determining the Particular Solution

Now let's substitute $y_p^{\prime }$', $y_p^{″}$, and $y_p^{‴}$ into the given differential equation: $$y_p^{‴}-2y_p^{″}-y_p^{\prime }+2y_p=e^{4t}$$ Substitute: $$(-u_1^{\prime }{e}^{-t}+3u_1{e}^{-t}+u_2^{\prime }{e}^t+3u_2{e}^t+8u_3^{\prime }{e}^{2t}+8u_3{e}^{2t})-2(u_1^{\prime }{e}^{-t}+2u_1{e}^{-t}+u_2^{\prime }{e}^t+2u_2{e}^t+4u_3^{\prime }{e}^{2t}+4u_3{e}^{2t})-(-u_1^{\prime }{e}^{-t}+u_1{e}^{-t}+u_2^{\prime }{e}^t+u_2{e}^t+2u_3^{\prime }{e}^{2t}+2u_3{e}^{2t})+2(u_1{e}^{-t}+u_2{e}^t+u_3{e}^{2t})=e^{4t}$$ Simplify and collect terms: $$(-u_1^{\prime }{e}^{-t}+5u_1{e}^{-t}-u_2^{\prime }{e}^t+u_2{e}^t+6u_3^{\prime }{e}^{2t}+10u_3{e}^{2t})=e^{4t}$$ Now we compare the coefficients of the exponential of like powers: $$-u_1^{\prime }=5,-u_2^{\prime }=1,6u_3^{\prime }=0$$ Integrating each equation with respect to t yields: $$u_1(t)=-5t+C_1,u_2(t)=-t+C_2,u_3(t)=C_3$$ Now, plugging $u_1(t)$, $u_2(t)$, and $u_3(t)$ back into the expression for $y_p$: $$y_p=(-5t+C_1)e^{-t}+(-t+C_2)e^t+C_3{e}^{2t}$$ Since $y_p$ is a particular solution and we want the first function linearly independent of the complementary function, we can take $C_1=C_2=C_3=0$. This gives: $$y_p=(-5t)e^{-t}-t{e}^t$$

General Solution

Now, the general solution is the sum of the complementary and particular solutions: $$y(t)=y_c+y_p=C_1{e}^{-t}+C_2{e}^t+C_3{e}^{2t}+(-5t)e^{-t}-t{e}^t$$ This is the general solution of the given third-order differential equation using the method of variation of parameters.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. They play a crucial role in modeling various physical systems where rates of change are involved, such as in physics, engineering, and biology. A differential equation describes how a quantity changes with respect to another, often time, providing insights into the behavior of dynamic systems.
In mathematical notation, a differential equation typically involves one or more unknown functions and their derivatives. For example, in the exercise, we have a third-order differential equation:

• $y^{‴}-2y^{″}-y^{\prime }+2y=e^{4t}$

This means it's an equation involving the third derivative of $y$ with respect to $t$. The goal is to find the function $y(t)$ that satisfies this equation. There are various methods to solve differential equations, and one such method is the variation of parameters, which we will explore in more detail in subsequent sections.

Complementary Function

The complementary function is a solution to the homogeneous part of a differential equation. The homogeneous equation is derived by setting the non-homogeneous term (like $e^{4t}$ in our exercise) to zero. Thus, our homogeneous equation becomes:

• $y^{‴}-2y^{″}-y^{\prime }+2y=0$

To obtain the complementary function, we use the characteristic equation, which is determined by assuming solutions of the form $y=e^{rt}$. After substituting and simplifying, we find the characteristic equation:

• $r^3-2r^2-r+2=0$

By solving this characteristic equation, we get the roots $r=-1,1,2$. These roots help us to construct the complementary function based on the principle that the solution of a linear differential equation can be expressed as a sum of exponential terms:

• $y_c=C_1{e}^{-t}+C_2{e}^t+C_3{e}^{2t}$

where $C_1,C_2,$ and $C_3$ are arbitrary constants. These constants will later be determined by any initial or boundary conditions if provided.

Particular Solution

The particular solution of a differential equation accounts for the non-homogeneous part, specifically addressing the input function (such as $e^{4t}$) that causes deviation from the homogeneous solution. In the method of variation of parameters, we assume a form like the complementary solution but replace the constants with functions:

• $y_p=u_1{e}^{-t}+u_2{e}^t+u_3{e}^{2t}$

Here, $u_1,u_2,$ and $u_3$ are functions of $t$ that we need to determine. The process involves differentiating $y_p$ and substituting back into the original differential equation to find expressions for $u_1^{\prime },u_2^{\prime },$ and $u_3^{\prime }$.
Through integration, we obtain:

• $u_1(t)=-5t+C_1$
• $u_2(t)=-t+C_2$
• $u_3(t)=C_3$

Finally, substituting these into the assumed form of $y_p$ and simplifying gives a particular solution. This solution ensures that when combined with the complementary function, it satisfies the entire non-homogeneous differential equation.

Characteristic Equation

The characteristic equation is a crucial tool in solving linear homogeneous differential equations. It arises when you seek exponential solutions to a differential equation with constant coefficients, like:

• $y^{‴}-2y^{″}-y^{\prime }+2y=0$

By assuming a solution of the form $y=e^{rt}$, each derivative introduces multiples of $r$, creating a polynomial called the characteristic equation, which, in this case, is:

• $r^3-2r^2-r+2=0$

The roots of this polynomial provide critical information about the nature of the solutions:

• Real roots give rise to exponential terms (like $e^{rt}$).
• Complex roots lead to oscillatory solutions, which we handle using trigonometric functions.

After solving for $r=-1,1,2$, these roots indicate the distinct exponential solutions that form the complementary function $y_c=C_1{e}^{-t}+C_2{e}^t+C_3{e}^{2t}$. Understanding the characteristic equation and its roots is essential in formulating the general solution to the differential equation.