Question

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$y^{\prime \prime \prime }-y^{\prime }=-4e^t$$

Short answer

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

Step-by-step solution

Find the Complementary Solution

First, we need to solve the homogeneous equation by changing the right side of the given equation to zero: $$y^{\prime \prime \prime }-y^{\prime }=0$$ For this third-order linear homogeneous equation, start by finding the characteristic equation: $$r^3-r=0$$ Factor the characteristic equation: $$r(r^2-1)=r(r-1)(r+1)=0$$ The roots of the cubic equation are: $r_1=0,r_2=1,r_3=-1$. Now, form the complementary (homogeneous) solution as the sum of terms with unknown coefficients: $$y_c(t)=C_1{e}^{0t}+C_2{e}^t+C_3{e}^{-t}$$ Simplify this expression to: $$y_c(t)=C_1+C_2{e}^t+C_3{e}^{-t}$$ This is our complementary solution.

Find the Particular Solution

Now we need to find a particular solution to the non-homogeneous equation: $$y^{\prime \prime \prime }-y^{\prime }=-4e^t$$ Using the method of undetermined coefficients, propose a candidate for the particular solution of the same form as the forcing function: $$y_p(t)=A{e}^t$$ Take the first, second, and third derivatives of the proposed $y_p(t)$ function: $$y_p^{\prime }(t)=A{e}^t$$ $$y_p^{\prime \prime }(t)=A{e}^t$$ $$y_p^{\prime \prime \prime }(t)=A{e}^t$$ Substitute the proposed particular solution and its derivatives into the given non-homogeneous equation: $$A{e}^t-A{e}^t=-4e^t$$ Notice that the terms on the left side cancel out, resulting in an identity. In this case, we modify our assumption for $y_p(t)$ by multiplying it by a factor of t: $$y_p(t)=At{e}^t$$ Now we take the first, second, and third derivatives again: $$y_p^{\prime }(t)=A{e}^t+At{e}^t$$ $$y_p^{\prime \prime }(t)=2A{e}^t+At{e}^t$$ $$y_p^{\prime \prime \prime }(t)=3A{e}^t+At{e}^t$$ Substitute the modified $y_p(t)$ and its derivatives back into the non-homogeneous equation: $$(3A{e}^t+At{e}^t)-(A{e}^t+At{e}^t)=-4e^t$$ This simplifies to: $$2A{e}^t=-4e^t$$ Solving for A: $$A=-2$$ Now, our particular solution is: $$y_p(t)=-2t{e}^t$$

Formulate the General Solution

Combine the complementary solution $y_c(t)$ and the particular solution $y_p(t)$ to form the general solution: $$y(t)=y_c(t)+y_p(t)$$ Plug in our complementary and particular solutions: $$y(t)=C_1+C_2{e}^t+C_3{e}^{-t}-2t{e}^t$$ This is our general solution: $$y(t)=C_1+C_2{e}^t+C_3{e}^{-t}-2t{e}^t$$

Key concepts

Complementary Solution

To crack the case of a differential equation, think of it like a two-part mystery. The first part: the complementary solution. This is the piece of the puzzle that corresponds to the equation without any external influences—that's when we set the equation equal to zero. It represents the system's natural behavior, kind of like a spring bobbing up and down with no one touching it.

To find the complementary solution for a higher-order linear differential equation, you kick off your detective work by crafting a characteristic equation from the differential equation. This is done by replacing each derivative of the function with a power of an unknown variable, usually referred to as 'r'. Solve that, and voilà, you've got your roots! Each root paves way to a term in your complementary solution. For example, if you get roots like 0, 1, and -1, your complementary solution will be a mix of exponential terms mapped to these roots, created with arbitrary coefficients for flexibility. It's like a tailored suit—it has to fit every possible scenario of the homogenous part of your equation.

Particular Solution

Now, dust off your magnifying glass for part two of our mystery: the particular solution. This character enters the stage when external forces come into play—like a constant or a buzzing sine wave—that disturb the peace. The key here is to catch a particular solution that vibes with this specific force.

You play a guessing game called the method of undetermined coefficients. Propose a solution fashioned after the troublemaker (the non-homogeneous part) and differentiate away. If your initial guess gets you nowhere, don't fret! Adjust it with a little twist—maybe throw in a 't' or a 't^2', to escape derivative cancellation traps. Once you've nailed a proposition that fits just right, you've captured the particular solution. It's a unique part that complements your complementary solution—pun intended.

General Solution

With the complementary and particular solutions in your arsenal, you're ready to reveal the general solution. Imagine it as a team-up of both solutions. It's like the full picture of how your differential equation behaves under any given circumstance.

The general solution has the flexibility of the complementary solution—those arbitrary coefficients are still here and ready to take on any value. Plus, it's got the specificity of the particular solution, accounting for that particular disturbance in the force we mentioned earlier. When you add them up, you get a formula capable of describing every twist and turn your function can take. To sum up, the general solution is your master key—it unlocks all possible behaviors of your differential equation.

Method of Undetermined Coefficients

In the realm of differential equations, the method of undetermined coefficients is like having a secret decoder ring. It helps you unearth special solutions when faced with a non-homogeneous equation—those that have an extra term muddling the waters.

To use this method, start by eyeballing the pesky non-homogeneous part. Craft a similar-function-looking guess as your possible solution but leave the coefficients undecided, hence 'undetermined'. Differentiate your guess as needed, plug it into the equation, and solve for these elusive coefficients. It's like solving a riddle: the terms of the equation give you clues, and you fit the pieces together to discover the hidden message (your coefficients). Just remember, this method isn't a one-size-fits-all; it's perfect for when the disturbance is a polynomial, an exponential, or a sine or cosine function. For other mischief-makers, you might need a different detective tool.

Characteristic Equation

The characteristic equation is a pivotal piece of the differential equation puzzle. This algebraic expression holds the secret codes to the complementary solution. By transforming the derivatives in your differential equation into a polynomial, the characteristic equation emerges as a relic with roots waiting to be unearthed.

Once you've set your differential equation to zero, mirroring its homogenous form, switch the derivatives to powers of 'r' and solve that fresh polynomial. The roots you dig up dictate the nature of the complementary solution—real roots give you exponential functions, and complex roots bring sines and cosines to the party in the solution mix. It's a fascinating transformation, seeing how the solution's fate is sealed by the characteristic equation's roots. Solving it is like unlocking a hidden path in an adventure game, revealing the lay of the land for the system's natural dynamics.