Question

Consider the nonhomogeneous differential equation $$ t^{3} y^{\prime \prime \prime}+a t^{2} y^{\prime \prime}+b t y^{\prime}+c y=g(t), \quad t>0 . $$ In each exercise, the general solution of the differential equation is given, where \(c_{1}, c_{2}\), and \(c_{3}\) represent arbitrary constants. Use this information to determine the constants \(a, b, c\) and the function \(g(t)\) $$ y=c_{1} t+c_{2} t^{2}+c_{3} t^{4}+2 \ln t $$

Short answer

#tag_title# Question: Find the values of constants a, b, c, and the function g(t) for the following nonhomogeneous differential equation: $$ t^3y'''+aty''+bty'+cy=g(t) $$ Given the general solution of the equation is: $$ y=c_{1} t+c_{2} t^{2}+c_{3} t^{4}+2 \ln t $$ #tag_title# Answer: The values of the constants are a=0, b=0, c=0 and the function g(t)=2 ln t.

Step-by-step solution

Differentiate the general solution

Differentiate the given general solution y three times with respect to t to find its first, second, and third derivatives: $$ y'=c_{1}+2 c_{2} t+4 c_{3} t^{3}+\frac{2}{t} $$ $$ y''=2 c_{2}+12 c_{3} t^{2}-\frac{2}{t^{2}} $$ $$ y'''=24 c_{3} t+\frac{4}{t^{3}} $$

Substitute into the differential equation

Now plug these derivatives into the given differential equation and simplify the equation: $$ t^{3}(24 c_{3} t+\frac{4}{t^{3}})+a t^{2}(2 c_{2}+12 c_{3} t^{2}-\frac{2}{t^{2}})+b t(c_{1}+2 c_{2} t+4 c_{3} t^{3}+\frac{2}{t})+c(c_{1} t+c_{2} t^{2}+c_{3} t^{4}+2 \ln t)=g(t) $$

Equate coefficients and find constants

Equate the coefficients of corresponding powers of t and compare constant terms to find the values of a, b, and c: Coefficient of \(t^5\): \(24c_3=0\); therefore, \(c_3=0\). Coefficient of \(t^4\): \(12ac_3 + 4bc_3 = 0\); since \(c_3=0\), this equation is automatically satisfied. Coefficient of \(t^3\): \(a(2c_2) + b(2c_1) + c(1)=0\). Without loss of generality, let \(c_1=c_2=1\). Then, \(2a+2b+c=0\). Coefficient of \(t^2\): \(-2a+\frac{2b}{t^3}+2ct=0\). By comparing powers of t, we get: \(-2a=0,~2b=0,~2c=0\). Thus, \(a=0,~b=0,\) and \(c=0\).

Determine the function g(t)

Finally, determine the function g(t) by finding the remaining term after equating coefficients and simplifying: $$ g(t)=0(t^{5})+0(t^{4})+0(t^{3})+0(t^{2})+2\ln t $$ Therefore, the values of the constants are \(a=0,~b=0,~c=0\) and the function \(g(t)=2\ln t\).

Key concepts

Differential Equation Solution

Understanding how to solve a nonhomogeneous differential equation can be quite daunting, but with the right approach, it can become more accessible. A differential equation essentially relates a function with its derivatives, and finding a solution means determining a function that satisfies the equation. For nonhomogeneous equations, the solution comprises two parts: the complementary (or homogeneous) solution, which solves the equation without the nonhomogeneous part, and a particular solution that specifically addresses the nonhomogeneity of the equation.

To compound understanding, imagine you are given a complex puzzle. The complementary solution is like sorting out the edge pieces to form the frame; it sets the basic structure. The particular solution, on the other hand, is like finding the exact pieces that fill a uniquely-shaped space within that frame - it deals with the aspects that make the puzzle unique. The general solution, like the one given in the exercise, combines both, ensuring that all aspects of the equation are addressed.

Coefficients Equating

The technique of coefficients equating is akin to making sure every ingredient in a recipe is accounted for and in perfect proportion. When we substitute the general solution and its derivatives back into the original differential equation, we get an equation consisting of various terms involving the independent variable (in this exercise, 't'). Each term is associated with a coefficient. Equating coefficients is a method where we compare each term on both sides of the differential equation, ensuring that the terms of the same power have identical coefficients.

Now, why is this important? It allows us to isolate and solve for the unknown constants in the differential equation, much like determining how much salt or sugar is needed for our recipe. Without this step, we couldn't balance the equation or find a precise solution—our mathematical dish would be incomplete!

Arbitrary Constants

In the context of differential equations, arbitrary constants play a critical role. These constants emerge naturally when integrating differential equations and represent the infinite number of solutions that a particular differential equation may have. Think of these constants as dials on a machine; by adjusting them, you can navigate through the myriad of potential solutions to target the specific one that meets the initial conditions or the constraints of the problem at hand.

In the solved exercise, the constants are represented by \(c_1\), \(c_2\), and \(c_3\). The beauty of these constants is that they provide us with the flexibility to mold the general solution to fit any additional information or boundary conditions that we are given. In other words, they allow the solution to be tailored to the specific scenario we're dealing with—and that's a powerful tool in applying differential equations to real-world problems.