Linear system of equations for a differential equation

In summary, the equation m\frac{d^{2}x}{dt^{2}} + c\frac{dx}{dt} + kx = (ax + b)^{2} + c^{2} can be solved for the constants m, c, k by creating a linear system of equations using MapleSoft. However, the equation cannot be put into linear system form as it is non-linear. The standard method of solving would be to first solve the corresponding homogeneous equation and then try to find a solution to the original equation. Converting the equation into two systems of first order differentials could also lead to a solution.
  • #1
Abyssnight
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Solve the equation [itex]m\frac{d^{2}x}{dt^{2}} + c\frac{dx}{dt} + kx = (ax + b)^{2} + c^{2}[/itex] for the constants [itex]m, c, k[/itex]

The right hand side a, b, and c are arbitrary digits. For me they are a = 2, b = 3, and c = 8.
The problem recommends creating a linear system of equations for me to solve. This is to be done using MapleSoft.

I tried making a set of equations using (m*r^2 + c*r + k) = (2+3)^2 + 64 and then using various values of r to get a system of equations. But the result got me [itex]m = 0, c = 0[/itex] and k equaling some number. However this is incorrect because the left part of the main equations with the constants will be used to make a vibration model that I would solve. With m and c equaling 0, I would not have any derivatives and therefore not being able to solve the equation.

So now I'm at a loss. I also tried converting the equation into 2 systems of first order differentials but that lead to me really nowhere.
 
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  • #2
Abyssnight said:
Solve the equation [itex]m\frac{d^{2}x}{dt^{2}} + c\frac{dx}{dt} + kx = (ax + b)^{2} + c^{2}[/itex] for the constants [itex]m, c, k[/itex]

The right hand side a, b, and c are arbitrary digits. For me they are a = 2, b = 3, and c = 8.
The problem recommends creating a linear system of equations for me to solve. This is to be done using MapleSoft.

I tried making a set of equations using (m*r^2 + c*r + k) = (2+3)^2 + 64
I'm afraid you are completely confused as to what you are doing. You titled this "linear system of equations" but this cannot be put into "linear system" form because your equation is NOT linear to begin with: x(t) is the dependent variable and you have a non-linear function of the form [itex](ax+ b)^2[/itex].

A linear equation would have to have a function of the independent variable, t, on the right, not a function of the dependent variable x.

That is [itex]m d^2x/dt^2+ c dx/dt+ kx= (at+ )^2+ c[/itex] would be a linear equation, not the equation you have.

The standard method of solution would be to solve the corresponding homogeneous equation first: solve the characteristic equation [itex]mr^2+ cr+ k= 0[/itex] to get a general solution to the associated homogeneous equation then try to find a solution of the form "[itex]x= At^2+ Bt+ C[/itex]".

and then using various values of r to get a system of equations. But the result got me [itex]m = 0, c = 0[/itex] and k equaling some number. However this is incorrect because the left part of the main equations with the constants will be used to make a vibration model that I would solve. With m and c equaling 0, I would not have any derivatives and therefore not being able to solve the equation.

So now I'm at a loss. I also tried converting the equation into 2 systems of first order differentials but that lead to me really nowhere.
I'm not sure what you mean by "using various values of r to get a system of equations." You need to get a system of first order differential equations from the orginal equation. To do that defined a new dependent variable, y(t), as the derivative of x: y(t)= dx/dt. Then the second derivative of x is just the first derivative of y and "[itex]md^2x/dt+ cdx/dt+ kx[/itex]" becomes "[itex]m dy/dt+ c dx/dt+ kx[/itex]" and the entire differential equation becomes
[itex]mdy/dt+ c dx/dt+ kx= (at+ b)^2+ c[/itex]
and, of course, [itex]dx/dt- y= 0[/itex]

That could be written as a matrix equation:
[tex]m\frac{\begin{pmatrix}x \\ y\end{pmatrix}}{dt}+ \begin{pmatrix}k & c \\ 0 & -1 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}(at+ b)^2+ c \\ 0 \end{pmatrix}[/tex]
 
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FAQ: Linear system of equations for a differential equation

What is a linear system of equations for a differential equation?

A linear system of equations for a differential equation is a set of equations that describe the relationship between the variables in a differential equation. These equations are linear, meaning that the variables are raised to the first power and there are no products of variables. Solving this system of equations will give the solution to the differential equation.

How is a linear system of equations different from a nonlinear system of equations?

A linear system of equations only contains terms that are raised to the first power, while a nonlinear system of equations may contain terms raised to higher powers or products of variables. This makes solving a linear system of equations for a differential equation much easier, as it can be solved using techniques such as substitution or elimination.

What is the purpose of using a linear system of equations to solve a differential equation?

The purpose of using a linear system of equations to solve a differential equation is to find the general solution to the differential equation. This general solution will contain an arbitrary constant, which can then be determined by using initial conditions or boundary conditions.

Can a linear system of equations be used to solve any differential equation?

No, not all differential equations can be solved using a linear system of equations. Only linear differential equations can be solved in this way. Nonlinear differential equations may require different methods, such as separation of variables or using a numerical approach.

What are some examples of real-world applications of linear systems of equations for differential equations?

Linear systems of equations for differential equations are used in many fields of science and engineering, including physics, chemistry, and economics. Some examples include modeling the motion of objects under the influence of forces, predicting chemical reactions, and analyzing economic systems.

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