Question

Determine the general solution of the given differential equation. \(y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=2 e^{-t}+3\)

Short answer

Question: Determine the general solution for the differential equation \(y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=2e^{-t}+3\). Answer: The general solution for the given differential equation is \(y(t) = C_1e^t + C_2\cos{t} + C_3\sin{t} + e^{-t} + 2\).

Step-by-step solution

Find the Complementary Function (CF)

To find the complementary function, we need to solve the homogeneous differential equation: \(y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=0\) Make the substitution \(y=e^{mt}\) to get the characteristic equation: \(m^3-m^2-m+1=0\) Factor the equation: \((m-1)(m^2+1)=0\) The roots are: \(m_1=1, m_2=i, m_3=-i\). The complementary function (CF) is: \(y_c(t)=C_1e^{t}+C_2\cos{t}+C_3\sin{t}\)

Find the Particular Integral (PI)

Now, we will find the particular integral for the non-homogeneous part, which has the form: \(2e^{-t}+3\). We will guess a function of the form: \(y_p = Ae^{-t}+B\). Compute the first, second, and third derivatives of our guess: \(y_p^{\prime}=-Ae^{-t}\) \(y_p^{\prime\prime}=Ae^{-t}\) \(y_p^{\prime\prime\prime}=-Ae^{-t}\) Substitute this into the original equation: \((-Ae^{-t})-(Ae^{-t})-(-Ae^{-t})+(Ae^{-t}+B)=2e^{-t}+3\) Now, group the terms with \(e^{-t}\) and the constants: \((-A-A+1)Ae^{-t}+(1+B)=2e^{-t}+3\) Equating terms, we get two equations: \((-A-A+1)A=2 \Rightarrow A=1\) \(1+B=3 \Rightarrow B=2\) So, our particular integral (PI) is: \(y_p(t)=e^{-t}+2\)

Write down the General Solution

The general solution of the nonhomogeneous differential equation will be the sum of the complementary function (CF) and the particular integral (PI): \(y(t) = y_c(t) + y_p(t) = C_1e^t + C_2\cos{t} + C_3\sin{t} + e^{-t} + 2\) This is the general solution for the given differential equation.

Key concepts

Complementary Function

In the context of solving differential equations, the complementary function (often abbreviated as CF) plays a crucial role. It represents the solution to the homogeneous version of the differential equation, meaning all parts of the equation without external forces or inputs. This part arises when addressing the equation:

• For our case, the homogeneous differential equation is: \( y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=0 \).

The complementary function, therefore, is a sum of solutions related to the differential equation’s characteristic roots.
By proposing that \( y = e^{mt} \) and substituting it into the homogeneous equation, one obtains the characteristic equation. Solving this will help you determine the characteristic roots, each leading to components of the CF. When you factor the characteristic equation and find its roots like we did, you can construct the CF using exponential, sine, and cosine functions.
For the example provided, the CF involves:

• An exponential term: \( C_1e^t \)
• A combination of sine and cosine terms: \( C_2\cos{t} + C_3\sin{t} \)

These arise from complex roots (\(i\) and \(-i\)), which mathematically simplify into sine and cosine components. This complementary function accounts for the dynamic behavior intrinsic to the system, minus external influences.

Particular Integral

The particular integral, abbreviated as PI, is crucial when dealing with non-homogeneous differential equations. It represents a particular solution that complements the homogeneous solution to satisfy the entire inhomogeneous equation.
Typically, you determine the PI by accounting for the non-homogeneous part of the equation — the term which doesn’t depend on the function itself. For example, given the term \(2e^{-t}+3\), one hypothesizes a potential solution using similar functions, for instance: \(y_p = Ae^{-t}+B\).
To find the values of \(A\) and \(B\), differentiate \(y_p\) accordingly and substitute back into the original differential equation. Simplifying these results will generate equations that allow you to solve for the unknown constants, here \(A = 1\) and \(B = 2\).
Therefore, the PI in our problem is \(y_p(t)=e^{-t} + 2\). The particular integral balances the differential equation by aligning the right side with the non-homogeneous term, resolving the system uniquely.

Characteristic Equation

The characteristic equation is an essential ingredient for solving linear homogeneous differential equations. It is derived from substituting a function of the form \(y = e^{mt}\) into the differential equation.
Here, the starting equation looks like this: \(y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=0\).

• Substitute \(y = e^{mt}\), yields the characteristic equation: \(m^3-m^2-m+1=0\).

The roots of this equation reveal how solutions behave:

• Real roots result in exponential functions. In our case, \(m_1=1\) leads to \(e^t\).
• Complex roots bring about sine and cosine functions. For example, roots \(m_2=i\) and \(m_3=-i\) translate into Cosine and Sine terms.

Factoring the characteristic equation simplifies finding these roots and determines the basis of the Complementary Function (CF). Understanding this equation is key as it reveals the underlying nature and temporal behavior of the differential equation's solutions.

Non-homogeneous Differential Equation

A non-homogeneous differential equation includes external or input terms, setting it apart from its homogeneous counterpart. These equations often appear in real-world applications, illustrating how systems respond to external forces or inputs.
Consider the given equation: \(y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=2e^{-t}+3\). Here, \(2e^{-t}+3\) is what makes it non-homogeneous—these terms are not dependent on the function or its derivatives.
Solving these involves two key parts:

• First, find the Complementary Function (CF), addressing the homogeneous portion without those extra terms.
• Second, derive the Particular Integral (PI), capturing the essence of the non-homogeneous component.

The general solution merges both CF and PI, neatly covering all possible forms of the solutions. This dual approach—handling characteristic behavior and specific influences together—makes tackling these complex situations manageable in the scope of applied mathematics.