Mohammad Cannon

2022-06-27

I am taking a beginner linear algebra class and would like a hint on a homework question. I don't need the answer, just a little guidance.

Consider the differential equation:

$\frac{{d}^{2}x}{d{t}^{2}}-\frac{dx}{dt}-x=\mathrm{cos}(t).$

We are told that the differential equation has a solution of the form

$x(t)=a\mathrm{sin}(t)+b\mathrm{cos}(t).$

Find $a$ and $b$, and graph the solution.

I don't know how to set up the system to use elimination to solve.

Consider the differential equation:

$\frac{{d}^{2}x}{d{t}^{2}}-\frac{dx}{dt}-x=\mathrm{cos}(t).$

We are told that the differential equation has a solution of the form

$x(t)=a\mathrm{sin}(t)+b\mathrm{cos}(t).$

Find $a$ and $b$, and graph the solution.

I don't know how to set up the system to use elimination to solve.

Sawyer Day

Beginner2022-06-28Added 30 answers

If you just plug in the given x(t) into the equation, what do we get?

$\begin{array}{rl}x(t)& =a\mathrm{sin}(t)+b\mathrm{cos}(t)\\ {x}^{\prime}(t)& =a\mathrm{cos}(t)-b\mathrm{sin}(t)\\ {x}^{\u2033}(t)& =-a\mathrm{sin}(t)-b\mathrm{cos}(t)\\ \frac{{d}^{2}x}{d{t}^{2}}-\frac{dx}{dt}-x& =(-a\mathrm{sin}t-b\mathrm{cos}t)-(a\mathrm{cos}t-b\mathrm{sin}t)-(a\mathrm{sin}t+b\mathrm{cos}t)\\ & =(-2a+b)\mathrm{sin}t+(-2b-a)\mathrm{cos}t.\end{array}$

Since you know that this is a solution, you should get $\mathrm{cos}t$. That tells you something about the coefficients of $\mathrm{sin}t$ and $\mathrm{cos}t$ that you got by plugging in. Does it look like a system of linear equations?

1: Happens every semester that I teach Calculus II; the book has early mention of Differential Equations as something integrals will be good for, but the students don't have the tools yet to solve even the most simple differential equations. A few problems ask the students to verify that a certain function is a solution, or to find conditions on certain parameters for a given function to be a solution, and invariably 99% of the students have a pavlovian response and try to solve the given equation and if successful compare the solution they find with the given solution, instead of simply plugging in the given answer to check if it works. This happens even though I tell them that to check something is a solution you don't need to solve, even though I do examples on the board before the homework is due, even though I give them quiz problems that specifically say things like "even though you cannot yet find the general solution to this equation"... Sigh.

$\begin{array}{rl}x(t)& =a\mathrm{sin}(t)+b\mathrm{cos}(t)\\ {x}^{\prime}(t)& =a\mathrm{cos}(t)-b\mathrm{sin}(t)\\ {x}^{\u2033}(t)& =-a\mathrm{sin}(t)-b\mathrm{cos}(t)\\ \frac{{d}^{2}x}{d{t}^{2}}-\frac{dx}{dt}-x& =(-a\mathrm{sin}t-b\mathrm{cos}t)-(a\mathrm{cos}t-b\mathrm{sin}t)-(a\mathrm{sin}t+b\mathrm{cos}t)\\ & =(-2a+b)\mathrm{sin}t+(-2b-a)\mathrm{cos}t.\end{array}$

Since you know that this is a solution, you should get $\mathrm{cos}t$. That tells you something about the coefficients of $\mathrm{sin}t$ and $\mathrm{cos}t$ that you got by plugging in. Does it look like a system of linear equations?

1: Happens every semester that I teach Calculus II; the book has early mention of Differential Equations as something integrals will be good for, but the students don't have the tools yet to solve even the most simple differential equations. A few problems ask the students to verify that a certain function is a solution, or to find conditions on certain parameters for a given function to be a solution, and invariably 99% of the students have a pavlovian response and try to solve the given equation and if successful compare the solution they find with the given solution, instead of simply plugging in the given answer to check if it works. This happens even though I tell them that to check something is a solution you don't need to solve, even though I do examples on the board before the homework is due, even though I give them quiz problems that specifically say things like "even though you cannot yet find the general solution to this equation"... Sigh.

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