Yongjian Xu
Ms Faber
Mathematics IA HL
Solving Ordinary Differential Equations and its Applications
Introduction
One of the goals of mathematics is to model the behavior of systems, whether these systems are theoretical or real physical systems. For example, functions are used to model static systems, such as how the velocity of an object changes over time, or how the population of rabbits change over time. However, these systems assume that all other variables are kept constant, but in reality, there are other factors which affect how these functions behave. For example, the velocity with respect to time of an object depends on the acceleration of the object, so these systems vary depending on different underlying variables. To model such dynamic systems, differential equations can be used, which are equations that relates a function to their derivatives. These derivative relations are prominent throughout nature, and the invention of differential by Isaac Newton allowed scientists to create accurate models of natural phenomenon.
Similar how solutions to algebraic equations are come number, solutions to differential equations are functions, but unlike algebraic equations, most differential equations cannot be solved by rearranging the equation and getting the function to one side of the equation. Methods to solving differential equations vary for different classes of differential equations, and some differential equations are not just extremely tedious to solve, but impossible to solve. There are also multiple solutions to a differential equation, so usually the solutions are defined as classes of functions that takes on a general form.
This IA aims to explore some simple types of first and second order linear ordinary differential equations and their general solutions. Also, applications of the differential equations discussed in this IA will also be explored, so that the physical significance of differential equations could be better understood.
Classifications of Differential Equations
Since the topics of discussion are first and second order linear homogeneous ordinary differential equations, it is important to understand what they actually mean and how they are mathematically represented. The first step to categorizing a differential equation is the number of variables that its solution takes into account. For differential equations with solutions of a single variable x, they are called ordinary differential equations, and the variable differentials in the equation dx, which are small changes in x, only relates to the single variable. The equation below is an example of an ordinary differential equation:
2 (d^2 y)/〖dx〗^2 +3x dy/dx-xy=0
Notice that in the equation above, the highest order of derivative is 2, and so the differential equation is also second order. The definition of the linearity and the homogeneity of a differential equation is more mathematical, as they very for different cases. In general, a differential equation is said to be linear if it can be expressed as follow for some nth order differential equation:
a_0 (x)y+a_1 (x) dy/dx+a_2 (x) (d^2 y)/〖dx〗^2 +⋯+a_n (x) (d^n y)/〖dx〗^n +b(x)=0
Note that a_n (x) in the equation is not a function of y in order for the differential equation to remain linear. This becomes especially important for linear differential equations of the first order, as it allows for a solution in a general form to be found. The homogeneity of a linear differential equation is determined by the value of the b(x) term. If b(x)=0, then the equation is homogeneous:
a_0 (x)y+a_1 (x) dy/dx+a_2 (x) (d^2 y)/〖dx〗^2 +⋯+a_n (x) (d^n y)/〖dx〗^n =0
Thus, for the differential equation given as an example, it is a second order linear homogeneous ordinary differential equation. The solutions to first and second order linear homogeneous ordinary differential equations are relatively simplistic, as they have certain properties that allows for the formulation of a general solution. However, these two types of differential equations have are very important, as its practical uses in the natural science is prominent throughout.
First Order Linear Homogeneous ODE
First order linear ODEs take on the general differential form of:
h(x) dy/dx+p(x)y+b(x)=0
The functions h(x) p(x) and b(x) must be functions of the variable x, as the equation is linear, and for the simplest case of the first order linear ODE, let p(x)=0, b(x)=0 and h(x)=1, which makes the differential equation homogeneous as b(x)=0:
dy/dx=0
What the equation above is stating is that the first derivative of some function y=f(x) is equal to 0. While the solution to this equation may seem obvious, as for any function y=c, where c is a constant, its first derivative will always be zero, as the function does not change with respect to x. The same result can be obtained when separating the differentials in the derivative operator and integrating both sides:
dy=0dx
∫▒dy=∫▒0dx
y=c
For the ODE above, the equation y=c is the general form of the solution, where c∈C. This method of solving the equation is called the separation of variables, which breaks down the derivative operator into its components and puts them on two different sides of the equation. If the equation above is non-homogeneous but p(x)=0 and h(x)=1, the differential equation becomes:
dy/dx+b(x)=0
Similarly, rearranging the equation and separating the variables allows us to solve the differential equation:
dy=-b(x)dx
∫▒dy=∫▒〖-b(x)dx〗
y=-∫▒〖b(x)dx〗
which is a general solution to the ODE above. The solution of the ODE in case of in which h(x) is non-zero can also be obtained quite easily:
h(x)dy/dx+b(x)=0
To separate the variables, simply divide h(x) to the RHS of the equation after moving b(x) to the RHS, which gives the general solution:
∫▒〖dy=∫▒〖-(b(x))/(h(x)) dx〗〗
y=-∫▒〖b(x)/(h(x)) dx〗
By plugging the functions b(x) and h(x) and integrating them with respect to x, the solution of the differential equation in its general form can be found. For example, let h(x)=e^2x and b(x)=-sin(x), which lets the differential equation become:
e^2x dy/dx-sin(x)=0
and plugging them into the general solution gives:
y=∫▒〖sin(x)/e^2x dx〗=∫▒〖sin(x)〗 e^(-2x) dx
Solving the integral would give the specific general solution of the differential equation with h(x)=e^2x and b(x)=-sin(x), which can be solved by using integration by parts:
Sign (±)
Differentiate
Integrate
+ u=e^(-2x) v=sin(x)
– du/dx=-2e^(-2x) ∫▒〖v dx〗=-cos(x)
+ (d^2 u)/〖dx〗^2 =〖4e〗^(-2x) ∫▒∫▒〖v 〖dx〗^2 〗=-sin(x)
By taking the sum of the product of the sign of each row and the terms indicated by the diagonal arrows plus the integral of the product of the sign of the last and the horizontal arrow term, indicated in the table above, the following equation could be obtained:
∫▒〖e^(-2x) sin(x)〗 dx=〖-e〗^(-2x) cos(x)-2e^(-2x) sin(x)-4∫▒〖e^(-2x) sin(x)〗 dx
By moving the integrals to the LHS of the equation and rearranging, the solution to the integral can be found:
∫▒〖sin(x)〗 e^(-2x)=-1/5 [e^(-2x) cos(x)+2e^(-2x) sin(x) ]+c
where c∈R, as for any value of c, the solution would always remain valid. The solution above can be shown to be true by differentiating it with respect to x, but it would be painstakingly long. Thus, a graphical proof is shown where the function above is differentiated, and can be seen to match that of the original function sin(x)e^(-2x), which by the fundamental theorem of calculus proves it to be true:
Since the two curves are identical, the function found is indeed a solution to the differential equation that was used as an example. Although the integration by parts was grueling, solving the general solution to the equation itself was straight forward, as the variables in the equation could be separated, and integrating both sides would give the solution.
Now, let’s consider a more complex form of the first order linear ODE, the case for which the ODE is linear and homogeneous, with h(x) and p(x) being non-zero:
h(x) dy/dx+p(x)y=0
Based on the variable separation method, the same process could be attempted:
dy/dx=-p(x)/h(x) y
1/y dy=-p(x)/h(x) dx
One again, the variables could be separated, and by integrating both sides would again yield a general solution:
∫▒〖1/y dy〗=-∫▒〖p(x)/h(x) dx〗
ln|y|=-∫▒〖p(x)/h(x) dx〗+c
And by using the rules of logarithms, the function y=f(x) could be obtained:
y=e^(-∫▒〖p(x)/h(x) dx〗-c)
Thus, we can conclude that for all homogeneous cases of first order linear ODEs, they can be systematically solved through the separation of variables and integrating on both sides. However, another method is required to solve for inhomogeneous cases, as the variables will no longer be able to be directly separated.
First Order Linear Inhomogeneous ODE
Inhomogeneous cases for first order linear differential equations can also be systematically solved, but the separation of variables method cannot be used for inhomogeneous cases, as the term independent of y b(x) does not allow a common factor of y to be taken out:
h(x) dy/dx+p(x)y+b(x)=0
dy/dx+p(x)/h(x) y=-b(x)/h(x)
At this point, there seems to be no obvious way to find a general solution to this differential equation. The solution is in fact less obvious, as it involves some unintuitive steps. First, nn observation can be made here that the first term and the second term on the LHS have some function y=f(x) and its first derivative dy/dx. Recall that in the product rule for differentiation, the derivative of the product of two function u and v is equal to:
d/dx (u∙v)=u dv/dx+v du/dx
A similarity can be noticed here between the differential equation and the product rule, as the first and second term on the RHS of the product rule equation involves the function v and its first derivative. Thus, the LHS of the differential equation can be expressed as the derivative of the product of two function u and y, where u is some unknown function when multiplied the function y and differentiated would give:
d/dx (u∙y)=u dy/dx+du/dx y
To make the LHS of the differential equation same as the expression above, the unknown function α can be multiplied by the whole differential equation, which would give:
u dy/dx+u p(x)/h(x) y=-u b(x)/h(x)
Since the above expression should be the same as the LHS of the differential equation, the terms can be equated, which when done to the coefficient of the function y, the following equivalence could be obtained:
u p(x)/h(x) =du/dx
Notice the above differential equation is very much so separable, and by applying the separation of variables to the differential equation above, the unknown function α could be found:
∫▒〖1/u du〗=∫▒〖(p(x))/(h(x)) dx〗
ln|u|=∫▒〖(p(x))/(h(x)) dx〗
u=e^∫▒〖(p(x))/(h(x)) dx〗
Because of the equivalence of the LHS of the first order linear differential equation with the RHS in its alternative “product rule” form, the differential equation could be rewritten as:
e^∫▒〖(p(x))/(h(x)) dx〗 dy/dx+e^∫▒〖(p(x))/(h(x)) dx〗 p(x)/h(x) y=d/dx (e^∫▒〖(p(x))/(h(x)) dx〗∙y)=〖-e〗^∫▒〖(p(x))/(h(x)) dx〗 b(x)/h(x)
and when the dx term on the LHS is moved to the RHS, then integrating both sides, the equation becomes:
∫▒d(e^∫▒〖(p(x))/(h(x)) dx〗∙y) =∫▒〖〖-e〗^∫▒〖(p(x))/(h(x)) dx〗 b(x)/h(x) dx〗
Then, by rearranging the equation, the general solution to the differential equation above can be found:
e^∫▒〖(p(x))/(h(x)) dx〗∙y=-∫▒〖e^∫▒〖(p(x))/(h(x)) dx〗 b(x)/h(x) dx-c〗
y=-(∫▒〖e^∫▒〖(p(x))/(h(x)) dx〗 b(x)/h(x) dx+c〗)/e^∫▒〖(p(x))/(h(x)) dx〗
In fact, this class of differential equations have a special name, and they are called exact differential equation. The function u=e^∫▒〖(p(x))/(h(x)) dx〗is called the integrating factor for the differential equation, as it allows the LHS of the differential equation to be rewritten in product form.
To show the application of the integrating factor method to solve for a first order exact equation, consider the differential equation below:
csc(x) dy/dx+x^2 y-sec(x)=0
First, the coefficient of the derivative term must be eliminated for the transformation to the product form, which can be done through rearranging:
dy/dx+(x^2 y)/csc(x) =sec(x)/csc(x)
By using the trigonometric identities csc(x)=1/(sin(x)) and sec(x)=1/(cos(x)), the differential equation becomes:
dy/dx+x^2 sin(x)y=tan(x)
Now, it can be transformed by multiplying both sides by the integrating factor u:
u=e^(∫▒〖x^2 sin(x)〗 dx)
To obtain the function, the integral on the exponent needs to be evaluated:
∫▒〖x^2 sin(x)〗 dx
Similarly, since it is the product of two functions, integration by parts can be used to solve it:
Sign (±)
Differentiate
Integrate
+ t=x^2 v=sin(x)
– dt/dx=2x ∫▒〖v dx〗=-cos(x)
+ (d^2 t)/〖dx〗^2 =2 ∫▒∫▒〖v 〖dx〗^2 〗=-sin(x)
– (d^3 t)/〖dx〗^3 =0 ∫▒∫▒∫▒〖v 〖dx〗^3 〗=cos(x)
When taking the sum of the product of all boxes connected by the arrows, the exponent of the integrating factor becomes:
∫▒〖x^2 sin(x)〗 dx=-x^2 cos(x)+2x sin(x)+2 cos(x)
Substituting the integral back into the integrating factor, the integrating factor u becomes:
u=e^(-x^2 cos(x)+2x sin(x)+2cos(x))
and multiplying it on both sides of the differential equation gives:
e^(-x^2 cos(x)+2x sin(x)+2cos(x)) dy/dx+e^(-x^2 cos(x)+2x sin(x)+2cos(x)) x^2 sin(x)y=e^(-x^2 cos(x)+2x sin(x)+2cos(x)) tan(x)
Now, the LHS can be transformed into its “product rule” form
e^(-x^2 cos(x)+2x sin(x)+2cos(x)) dy/dx+e^(-x^2 cos(x)+2x sin(x)+2cos(x)) x^2 sin(x)y=d/dx(e^(-x^2 cos(x)+2x sin(x)+2 cos(x) ) y)
Finally, equating this to the RHS of the equation, separate the variables and integrating on both sides would give the solution:
d/dx (e^(-x^2 cos(x)+2x sin(x)+2 cos(x) ) y)=e^(-x^2 cos(x)+2x sin(x)+2 cos(x) ) tan(x)
y=(∫▒〖e^(-x^2 cos(x)+2x sin(x)+2 cos(x) ) tan(x) 〗 dx)/e^(-x^2 cos(x)+2x sin(x)+2 cos(x) ) +c
The solution above is extremely complex, and its practical use is highly limited. This proves the point that most differential equations do not have general solutions that can be expressed as elementary functions, or functions that are integrable and closed under a finite number of integral operations. The reason that the function is non-elementary is because of the presence of the terms e^(sin(x)), e^(cos(x)) and e^(〖-x〗^2 cos(x)), none of which are closed under indefinite integration. In this case, as far as the IA concerns, the example differential equation has no closed form general solution.
However, the differential equation below can be solved, and its solution is relatively simple:
x^2 dy/dx+xy=6
To solve this, simply apply the same process to the differential equation, by rearranging it into the standard form for exact equations, and multiply both sides by the integrating factor which would allow it to be expressed in product differential form:
dy/dx+x/x^2 y=6/x^2
dy/dx+1/x y=6/x^2
Then, the integrating factor becomes:
u=e^∫▒〖1/x dx〗=e^(ln|x| )=x
By multiplying it on both sides, LHS of the differential equation becomes:
x dy/dx+x 1/x y=d/dx(xy)
Now, equating the LHS to the RHS of the equation allows the variables to be separated and solved by integrating both sides:
d/dx (xy)=6/x
∫▒d(xy) =∫▒〖6/x dx〗
y=6/x ln(x)+c/x
Which is the general solution to the exact equation theoretically. To show that it really is the solution, if the solution is substituted back into the exact equation, the LHS and the RHS should be equal to each other:
x^2 d/dx (6/x ln(x)+c/x)+x(6/x ln(x)+c/x)=6
The derivative of y is evaluated using the product rule:
d/dx (6/x ln(x)+c/x)=6/x^2 -(6 ln(x))/x^2 -c/x^2 =(6-6 ln(x)-c)/x^2
Substituting the result into the exact equation gives
x^2 ((6-6 ln(x)-c)/x^2 )+x((6 ln(x)+c)/x)=6
By simplifying the terms, LHS=RHS:
6-6 ln(x)-c+6 ln(x)+c=6
6=6
While first order differential equations are capable of modelling a range of phenomenon, such as exponential growth and decay systems, some physical systems require higher order derivatives to be expressed, such as the equations of motion of an object under uniform acceleration, or constant acceleration, which is the second order derivative of its position. To model, the equations of motion of classical mechanical systems, second order differential equations are required.
Second Order Linear Homogeneous Ordinary Differential Equations
Second order linear homogeneous differential equations take on the general form of:
q(x)(d^2 y)/(dx^2 )+h(x)dy/dx+b(x)y=0
Although the equation is homogeneous, the variables cannot be separated, so a different approach is needed to solve the equation. One of the fundamental theorems that is used is that a linear differential equation of order n will have n number of linearly independent solutions, which means that the solutions are not constant multiples of each other.
y_n (x)≠ky_(n±1) (x),k∈R
Also, any linear combination of the linearly independent solutions is also a solution to the differential equation:
〖〖y(x)=c〗_1 y〗_1 (x)+〖c_2 y〗_2 (x)+〖c_3 y〗_3 (x)+⋯〖c_n y〗_n (x)=∑_(k=1)^n▒〖〖c_k y〗_k (x) 〗
Where c_k is some constant multiple of its corresponding linearly independent solution y_k (x).
While this theorem is not entirely intuitive, there are resemblances to how an n^th degree polynomial has n number of roots. Derivations of the theorem will not be shown, as it is far too complex, but it will be taken as a given fact as it is necessary for deriving a solution to the equation of the second order linear differential equation. However, the theorem can be shown to be true when substituting it back into the linear homogeneous ODE.
Since there are n number of linearly independent solutions for an n^th order linear ODE, when applying the theorem above to the case of a second order linear homogeneous differential equation, there must be 2 differential equations to match the 2 linearly independent solutions, and 1 differential equations to match the linear combinations of the linearly independent functions:
The solutions to the ODE (Equation 1):
y(x)=c_1 y_1 (x)+c_2 y_2 (x)
The differential equations for the linearly independent solutions (Equation 2&3):
q(x)(d^2 y_1)/(dx^2 )+h(x)(dy_1)/dx+b(x)y_1=0
q(x)(d^2 y_2)/(dx^2 )+h(x)(dy_2)/dx+b(x)y_2=0
The differential equations for the linearly dependent solution (Equation 4):
q(x)(d^2 y)/(dx^2 )+h(x)dy/dx+b(x)y=0
By substituting Equation 1 into Equation 4, the following can be obtained:
q(x) d^2/(dx^2 ) (c_1 y_1 (x)+c_2 y_2 (x))+h(x) d/dx (c_1 y_1 (x)+c_2 y_2 (x))+b(x)(c_1 y_1 (x)+c_2 y_2 (x))=0
Differentiating the functions on the equation above gives:
q(x) c_1 d^2/(dx^2 ) y_1 (x)+q(x) c_2 〖d^2/(dx^2 ) y〗_2 (x)+h(x) c_1 d/dx y_1 (x)+h(x) c_2 〖d/dx y〗_2 (x)+〖b(x)c〗_1 y_1 (x)+〖b(x)c〗_2 y_2 (x)=0
Note that by rearranging the equation, Equation 2 &3 could be obtained:
c_1 (q(x) d^2/(dx^2 ) y_1 (x)+h(x) d/dx y_1 (x)+b(x)y_1 (x))+c_2 (q(x) d^2/(dx^2 ) y_2 (x)+h(x) d/dx y_2 (x)+b(x) y_2 (x))=0
Since Equation 2&3 are both homogeneous, they equal 0, so
c_1 (0)+c_2 (0)=0
Because the LHS=RHS=0, the linear combination of two linearly independent equation is indeed a solution to the differential equation.
When considering the two linearly independent solutions, if the coefficients are variable, then the solution is not easy to be thought of, as a differential equation of this form is not easy to solve and will involve guesswork to find a solution. However, the differential equation could be simplified if its coefficients were constant and non-zero:
a (d^2 y)/(dx^2 )+b dy/dx+cy=0,a,b,c∈R
By examining the differential equation above, the only type of function that obeys the criteria given by the differential equation above is the exponential function, as it is the only function that gives a constant multiple of itself when differentiated. Thus, it can be hypothesized that the exponential function in the form below is a solution to the differential equation:
y=e^kx
By substituting the proposed solution into the differential equation, one can verify the solution’s validity:
a d^2/(dx^2 )(e^kx)+b d/dx(e^kx)+ce^kx=0
Differentiating all terms:
ak^2 y+bky+cy=0
Note that the common factor of y can be extracted from the equation:
y(ak^2+bk+c)=0
Thus, in order for the equation above to be true, the terms inside the bracket must evaluate to 0, as the function y is non-zero, so the equation is true given that:
ak^2+bk+c=0
And by solving the quadratic for k, k is equal to:
k=(-b±√(b^2-4ac))/2a
Therefore, the set of solutions of the linear second order homogeneous ODE are:
y_1=A_1 e^((-b+√(b^2-4ac))/2a x)
y_2=A_2 e^((-b-√(b^2-4ac))/2a x)
y=y_1+y_2=A_1 e^((-b+√(b^2-4ac))/2a x)+A_2 e^((-b-√(b^2-4ac))/2a x)
The quadratic that when solved gives the value of the constant k is called the auxiliary, or characteristic equation for the second order differential equation, and it allows the constant for the exponential solution to be found, given that the coefficients are constant.
Second order differential equations are especially important in physics, as Newton’s second law, which governs the motion of an object in classical systems is a second order differential equation. Solving this equation gives the motion of the object in the system, and it is what lead to the birth of classical mechanics that is still widely used in many areas of physics and engineering.
Applications of Linear ODEs for Solving the Motion of an Object in Freefall and a Mass on a Spring
Newton’s second law is given by the formula below:
F=m (d^2 x)/(dt^2 )
where x, F and m are the displacement, force and mass of the object respectively. For any given classical system, the first things to identify are any forces acting on the system, and for the case of an object in freefall while neglecting frictional forces, there is a single force acting on the system, which is the gravitational field of some planet g, and m is the mass of the object:
the gravitational field strength is the force exerted on per unit mass, which is inversely proportional to the squared of the radius (r):
g=F/m=((GMm/r^2 ))/m
G is the gravitational constant, M is the mass planet, and m is the mass of the object inside the gravitational field. Since m cancel out, the gravitational field strength becomes:
g=GM/r^2
Note that the gravitational field strength is independent of the mass if the object, which is true because the field strength is simply the acceleration of freefall. In fact, the radius r is actually equal to the displacement x, as the displacement of an object in freefall is radial and one dimensional, so it only falls along the vertical axis. Thus, Newton’s second law can be rewritten as:
GM/r^2 =(d^2 r)/(dt^2 )
Rearranging the equation would then give us:
GM=r^2 (d^2 r)/(dt^2 )
However, the coefficient of the derivative term is not constant, which means the standard method for solving constant coefficient second order linear homogeneous ODEs would not work. Despite this, the equation can be simplified under the assumption that the gravitational field is uniform for sufficiently small values of r. This can be assumed because the gravitational constant G≈6.674×〖10〗^(-11) is so small that as long as the distance between the object and the plant is not too large, the value for the gravitational field g is constant. Then, the differential equation becomes:
(d^2 r)/(dt^2 )=g
This can be solved simply by separating the variables and integrating twice on both sides:
∫▒∫▒〖d^2 r〗=∫▒∫▒〖gdt^2 〗
r=∫▒〖gt+u〗 dt
r=1/2 gt^2+ut+h
The equation above gives the radial displacement of an object from the surface of the planet, depending on some constant c and k. The significance of each of these constants are the initial conditions of the system, which is what each of their value is at time t=0. Since the constant u is obtained by integrating once, and h obtained by integrating twice, they can be found by setting them equal to their respective order of derivative, so:
h=r_(t=0)
u=〖dr/dt〗_(t=0)
r=1/2 gt^2+〖dr/dt〗_(t=0) t+r_(t=0)
Which describes the displacement of an object that is in freefall.
For a mass on an ideal spring, the force exerted on it is directly proportional to the negative displacement of the object, where k is the spring constant:
F=ma=-kx
Which can then be rewritten as:
a=(d^2 x)/(dt^2 )=-kx/m
Note that this is indeed a second order linear homogeneous ordinary differential equation with constant coefficients, so the standard method can be used to find the general solutions of to the differential equation.
(d^2 x)/(dt^2 )+k/m x=0
Solving the auxiliary equation allows us to find the constant σ of the exponential:
σ^2+k/m=0
σ=±i√(k/m)
So, the class of solutions become:
x_1=〖A_1 e〗^(i√(k/m) t)
x_2=〖A_2 e〗^(-i√(k/m) t)
x=〖A_1 e〗^(i√(k/m) t)+〖A_2 e〗^(-i√(k/m) t)
Note that the solution is a complex exponential, which means that the solution has two components to it:
x_1=〖A_1 cos〗(√(k/m) t)+iA_1 sin(√(k/m) t)
x_2=〖A_2 cos〗(-√(k/m) t)+iA_2 "sin" (-√(k/m) t)=〖A_2 cos〗(√(k/m) t)-iA_2 "sin" (√(k/m) t)
x=〖A_1 cos〗(√(k/m) t)+iA_1 sin(√(k/m) t)+A_2 cos(√(k/m) t)-iA_2 "sin" (√(k/m) t)=(A_1+A_2 ) cos(√(k/m) t)+i(A_1-A_2)"sin" (√(k/m) t)
When the mass is in a spring, it is undergoing simple harmonic motion, so it can be inferred that the solution is sinusoidal, as the mass would oscillate back and forth from the equilibrium point. This suggests that either the real or the imaginary component of the complex exponential could be solutions to the ODE, as well as their sum. This can be tested by substituting the sinusoidal solutions into the ODE:
A_1 d^2/(dt^2 ) cos(√(k/m) t)+A_1 k/m cos(√(k/m) t)=0
Differentiating the first term:
〖-A〗_1 k/m cos(√(k/m) t)+A_1 k/m cos(√(k/m) t)=0
Thus, it is true for cosine, and since sine is a phase shift of 90 degrees with respect to cosine, the solution involving sine is also true.
Conclusion
Differential equations are powerful, at modelling dynamic phenomena in nature, but they are also limited in terms of providing solutions to these dynamic phenomena. The complicated nature of differential equations makes a large number of them very difficult to solve, if not impossible, which creates problems for scientists trying to model phenomena using differential equations, due to the exponentially increasing complexity of the solutions. In areas of theoretical physics, such as quantum field theory, scientists use complex partial differential equations to model the interaction of particle fields, but any interaction involving 2 or more particles cause the field equations, which are partial differential equations, to be unsolvable, as the result is an infinite number of solutions. To counter this issue created by these complex differential equations, theoretical physicists use perturbation theory to simplify these complex models so that they become solvable and use physical measurements to determine initial conditions for these differential equations, so that the infinities can be eliminated, a process known as renormalization.
Although differential equations have been around for hundreds of years, many aspects of it remains a mystery, and this raises the question whether the current model of mathematics is not sufficient to describe some aspects of reality modelled by differential equations.