# Top Techniques for Solving Ordinary Differential Equations

April 21, 2023
Jane Lee
United States
Ordinary Differential Equations
With more than 20 years of expertise in the area of differential equations and numerical techniques, Dr. Jane Lee is a famous mathematician. She completed her groundbreaking research on the creation of numerical techniques for resolving differential equations at MIT, where she also earned her Ph.D. in Applied Mathematics.
Science and math both depend on ordinary differential equations (ODEs). Numerous phenomena, ranging from straightforward harmonic motion to intricate biological and physical systems, are described by ODEs. This article covers the best methods for solving ODEs, which are crucial for comprehending and studying complex systems

## Separation of Variables Method:

One effective way for resolving a class of first-order ODEs is the separation of variables method. To use this approach, it is assumed that the differential equation's solution may be written as the product of two functions, one of which only depends on the independent variable and the other of which only depends on the dependent variable.
Consider a differential equation of the first order of the following type:
f(x, y) = dy/dx
When x acts as an independent variable and y acts as the dependent variable. We assume that y may be expressed as the intersection of the two functions u(x) and v(y) to solve this equation using the separation of variables method:
y = u(x) * v(y)
Using y's derivative with respect to x, we can calculate:
u'(x) * v(y) = dy/dx
When these expressions are substituted into the initial differential equation, we obtain:
f(x, u(x) * v(y)) = u'(x) * v(y)
This equation can be rearranged as follows:
f(x, u(x) * v(y)) = u'(x) / u(x) / v(y)
This equation's left and right sides are solely dependent on x and y, respectively. By integrating both sides to their respective variables, we can thus separate the variables:
f(x, u(x) * v(y))) / v(y) dy = u'(x) / u(x) dx
When we simplify these integrals, we obtain:
f(x, u(x) * v(y)) = v(y) dy + C2 = ln |u(x)| + C1 =
where C1 and C2 are integration constants.
By exponentiating both sides of the equation, we can find u(x):
e(f(x, u(x)*v(y)) / v(y) dy + C) |u(x)| =
in where C = C2 - C1.
We can disregard the absolute value and select the sign that meets the initial condition because u(x) can be either positive or negative.
After that, we separate the variables in the initial equation to find v(y):
f(x, y) = dy / v(y) = u(x) dx
When we combine the two sides to the relevant variables, we obtain:
f(x, y) = u(x) dx = dy / v(y)
When we simplify these integrals, we obtain:
f(x, y) = ln |v(y)| + C3 = u(x) dx
where C3 is an integration constant.
By exponentiating both sides of the equation, we can find v(y):
e(f(x, y) / u(x) dx + C') = |v(y)|
where C' is an integration constant.
The generic solution to the differential equation can be expressed as follows.
u(x) * v(y) = C
where C is an integration constant.
In conclusion, a class of first-order ordinary differential equations may be solved effectively using the separation of variables method. It entails assuming that the solution can be expressed as the union of two functions, u(x) and v(y), and then figuring out how to solve for these unions by dividing the variables into their component parts. When the right-hand side of a first-order differential equation has a separable form and can be expressed as the product of functions of x and y, this approach is especially helpful.

## Using Homogeneous Equations

When all terms in an ordinary differential equation (ODE) are homogeneous functions of the dependent variable and its derivatives, the equation is said to be homogeneous. A homogeneous function meets the scaling property, which states that scaling the parameters by a certain amount will also scale the function value by the same amount.
The following is an example of a homogeneous differential equation of order n:
a_(n-1)(x)y(n-1) + a_(n)(x)y(n) +... A_0(x)y = 0 and A_1(x)y' = 0.
where the continuous functions of x are a_n(x), a_(n-1)(x),..., and a_0(x), and the nth derivative of y with respect to x is denoted by y(n).
We can employ the method of substitution to resolve homogeneous differential equations. The differential equation is changed to read y = vx, where v is a new function of x. The derivative of y can be expressed in terms of v and its derivatives using the chain rule:
y' = vx' + v'x
vx'' = 2v'x' + v''x+2v''x2;
vx''' = 3v''x2 + 3v'x' + 3v'x' + v'''x3
...
y(n) = vn(x)xn + an-1vn-1(x)xn+1 +... A0(x)v + av1(x)v
where the original differential equation's coefficients are a0(x), a1(x),..., an-1(x), and an(x).
We get the following results by substituting these expressions into the initial differential equation:
a_n(x)vn(x)xn + a_(n-1)(x)vn(x)xn+... A0(x)v = 0 and + a1(x)vx'
The result of multiplying both sides of the equation by xn is:
The formula is: a_n(x)vn(x) + a_(n-1)(x)vn-1(x)xn-1/xn +... A0(x)v/xn = 0 + a1(x)vx'/xn
The preceding equation can be written as follows because v and its derivatives are continuous functions of x:
a_(n-1)(x)vn(x) + a_(n-2)(x)vn(x) +... A0(x)v = 0 and + a1(x)vx'
This can be resolved using any method for resolving linear differential equations. which is a linear differential equation in v.
After determining v, we may simplify and substitute y = vx to get the solution to the first differential equation:
y = vx
y' = vx' + v'x
vx'' = 2v'x' + v''x+2v''x2;
...
y(n) = vn(x)xn + an-1vn-1(x)xn+1 +... A0(x)v + av1(x)v
As a result, the following is the homogeneous differential equation's general solution:
C1v1(x)xn plus C2v2(x)xn plus yields y. Cnvn(x)x +
When y = vx is substituted for the linear differential equation to give the linearly independent solutions v1(x), v2(x),..., vn(x), and C1, C2,..., Cn are arbitrary constants.

## Using Exact Differential Equations:

The approach of integrating factors can be used to solve a special type of ordinary differential equations (ODEs) called exact differential equations. When the gradient of a scalar function, often referred to as a potential function, can be used to define the differential operator in a differential equation, that equation is said to be exact.
Consider the subsequent differential equation:
N(x,y)dy = 0 and M(x,y)dx = 0.
There exists a scalar function (x,y) such that if this equation is correct:
x = M(x,y), y = N(x,y), etc.
Finding (x,y) and using it to get the differential equation's solution is necessary to solve this equation.
By taking y as a constant and integrating the first equation with regard to x, we can find (x,y):
(x,y) = g(y) + M(x,y)dx
where g(y) is an integration constant that only depends on y. We may distinguish both sides of the equation with regard to y and contrast it with the second equation to find g(y):
"/" = "g(y)/" = "N(x,y)"
This suggests that, with regard to y, g(y) is an antiderivative of N(x,y).
As a result, we have:
(x, y) = g'(y)dy + M(x, y)dx
where every antiderivative of N(x,y) with regard to y is represented by g'(y).
The chain rule now allows us to express d as follows:
M(x,y)dx + N(x,y)dy = d = /x dx + /y dy
With this equation and the initial differential equation in comparison, we obtain:
dφ = 0
This suggests that x, y are constants, let's say C. As a result, we have:
M(x,y)dx plus g'(y,dy) equals C.
This is the precise differential equation's general solution.
Finding a potential function (x,y) may not be simple in practice. If the equation is not precise, we can apply the integrating factors approach to make it precise. To achieve this, the differential equation is multiplied by an appropriate integrating factor, which is a function that ensures the equation is accurate.
We can use the following formula to determine the integrating factor:
P(x)dx: (x) = e
where P(x) is a function of x such that the differential equation is exactly equal to itself when multiplied by (x).
The differential equation can then be multiplied by (x), giving us:
(x)N(x,y)dy = 0 and (x)M(x,y)dx = 0.
The potential function (x,y) and the general solution can be discovered using the method outlined above if this equation is precise. If not, we can continue looking for a good integrating factor until we find an exact equation.

## Power Series Method:

Ordinary differential equations (ODEs) that cannot be resolved using various approaches, such as variable separation or homogeneous equations, can often be resolved using the power series method. By using a power series expansion, or a collection of terms with escalating powers of a variable, the differential equation's solution is expressed using this technique.
Consider the subsequent differential equation:
y''(x), + p(x), + q(x), = 0
where p(x) and q(x) are x-dependent functions. We suppose that the solution y(x) can be represented as a power series expansion with the following form to solve this equation using the power series method:
y(x) = a_nxn = n=0
where a_n are undetermined constants.
This power series expansion is what we get when we substitute it into the differential equation:
[n(n-1)a_nx(n-2) + pa_nx(n-1) + q(x)a_nx(n)] = 0 where n=0.
This equation can be made simpler by grouping terms with the same power of x:
a_0[q(x)a_0] = 0 (n = 0) a_1[(10)a_0 + p(x)a_1] + a_0[q(x)a_1] = 0 (n = 1) a_2[(21)a_1 + p(x)a_2] + a_1[(10)a_1 + p(x)a_2] + a_0[q(x)a_2] = 0 (n = 2) ... p(x)a_(n-1) + a_n[(n(n-1))a_(n-2)] + a_n... + (n 3) a_0[q(x)a_n] = 0
These equations are a system of a_n coefficients linear homogeneous equations. We require the determinant of the coefficient matrix to be zero to arrive at a non-trivial solution. This provides us with the typical equation:
Δ(λ) = λ^2 + p(x)λ + q(x) = 0
where the determinant () and the variable () are present.
This characteristic equation can then be solved to provide the roots _1 and _2. The differential equation's general solution is therefore provided by:
c_1y_1(x) + c_2y_2(x) = y(x).
where y_1(x) and y_2(x) are two linearly independent solutions to the differential equation and c_1 and c_2 are constants.
Finding a series solution around a point x_0 can also be done using the power series method. In this instance, we presume that the answer can be written as a power series expansion of the following form:
y(x) = a_n(x-x_0) = 0
where the center of the power series expansion is denoted by x_0.
We establish a recurrence relation for the coefficients a_n by substituting this power series expansion into the differential equation:
a_n = [p(x_0)a_(n-1) + q(x_0)a_(n-2)] = (-1/n(n-1))
Starting from the known values of a_0 and a_1, this recurrence relation can be utilized to obtain the coefficients a_n iteratively.

## The Laplace Transform Method

The Laplace transform method is an effective method for solving ODEs, especially those that involve discontinuous or nonperiodic functions. The problem is transformed into an algebraic equation using the Laplace transform in this method, making it simpler to solve. By taking the inverse Laplace transform of the outcome, the solution is then obtained.

## Numerical Methods

These are a group of computer-based techniques for roughly approximating the solutions of ODEs. These techniques are especially helpful for complex systems of ODEs or ODEs that cannot be solved analytically. The finite element method, the Runge-Kutta method, and Euler's method are the most often used numerical techniques.

## Green's Function Method:

Green's function method is a potent method for resolving ODEs, especially those with boundary conditions. With this approach, the solution is expressed as a linear combination of Green's functions, or functions that satisfy the ODE provided certain boundary conditions are met. Applying the boundary conditions to the solution yields the coefficients of the linear combination.

## Parameter Variation Method:

For the solution of non-homogeneous ODEs, the modification of the parameters method is an effective method. The method of indeterminate coefficients is another name for it. This approach is based on the idea that the solutions to the related homogeneous ODEs can be combined linearly to describe the solution to a non-homogeneous ODE.
The second-order ODEs of the following kind are particularly well suited to the modification of parameters method:
f(x) = y'' + p(x)y' + q(x)y
If f(x) is a non-homogeneous function, p(x) and q(x) are continuous functions.
Finding the answers to the related homogeneous equation is the first step in utilizing the variation of parameters approach to solve this equation:
y''+p(x)y'+q(x)y = 0
The answers can be stated as follows:
c1y1(x) + c2y2(x) = y1(x)
where y1(x) and y2(x) are the homogeneous equation's linearly independent solutions and c1 and c2 are constants.
Next, we assume that a specific formulation of the non-homogeneous equation's solution is as follows:
u(x)y1(x) plus v(x)y2(x) equals y_p(x).
where the functions to be calculated are u(x) and v(x).
This is what we get when we add it to the non-homogeneous equation:
The equations u'(x)y1(x) + v'(x)y2(x) + u(x)y1'(x) + v(x)y2'(x) + p(x)(u'(x)y1(x) + v'(x)y2(x)) + q(x)(u(x)y1(x) + v(x)y2(x)) = f(x
When we simplify this equation and group-related terms, we obtain:
G(x) = u'(x)y1(x) + v'(x)y2(x).
where (y1(x)y2'(x)-y2(x)y1'(x)) = f(x)/(y1(x)y2'(x)).
Now that we have the inverse of the Wronskian, which is given by: we can solve for u'(x) and v'(x).
W(y1, y2)(x) = y1(x), y2(x), y1'(x)
The result is:
W(y1, y2)(x) = u'(x) = g(x)y2(x)
v'(x) is equal to -g(x)y1(x) / W(y1, y2)(x).
When we combine these equations, we obtain:
g(x)y2(x)/W(y1, y2)(x)(dx + c1) = u(x)
v(x)=-g(x)y1(x)/W(y1, y2)(x)dx + c2
where c1 and c2 are integration constants.
Finally, the following represents the non-homogeneous equation's solution:
c1y1(x) + c2y2(x) = y(x) = y_p(x)
where c1 and c2 are integration constants, and y_p(x) is the specific solution derived using the variation of parameters approach.

## Conclusion:

In conclusion, understanding and analyzing complex systems in mathematics and science rely heavily on the methods for solving ODEs described above. Each method has advantages and disadvantages of its own, so it is best to select one based on the specifics of the issue at hand. If you fully comprehend these methods, you'll be able to approach ODEs with assurance and deduce