A differential equation is an equation that describes the change in a variable or variables as a function of time. Differential equations are very useful in modeling and understanding changes in many phenomena, from physics to economics.
They are not just theoretical constructs, but have numerous applications in the real world. For example, many industries use differential equations to determine how much material to supply or produce based on demand.
Solving a differential equation requires finding the variable(s)’s values at a given time. There are multiple ways to solve a differential equation, and one of them is by variation of parameters.
This article will discuss how to solve a differential equation by variation of parameters. This method is useful for solving linear ordinary differential equations with constant coefficients. Non-linear ODEs and ODEs with non-constant coefficients require different methods.
Solve the differential equation
Now, let’s consider the equation Y’ + Y = Sin X. We can solve this by Variation of Parameters. The solution is given by
Y = Aei(ωt – ϕ) + Bei(ωt – ϕ)
Where A and B are arbitrary constants and ω is the frequency of oscillation, which we assume to be constant. Now, since we know that Y’ = Y − Sin X, we can solve for Y’ as follows:Y’ = −Aei(ωt – ϕ) − Bei(ωt – ϕ)sin XThe solution to this equation is:Y” = ei(ωt−ϕ)Which gives us our final solution:Aei(ωt-ϕ) + Bei (ωt-ϕ)=eit (Ω t−φ)=eit (Ω t−π/2)=eit Ω (1−π/2)=Aei Ω+Bei Ω=Aei eit+Bei eitwhere A and B are arbitrary constants.This is a very elegant solution to the differential equation!Variation of parameters allows us to find two linearly independent solutions for the original ODE.However, in this case, since we assumed that the frequency was constant and only varied the phase parameter, we only found one solution in terms of Aei Ω. However, if we assumed that both parameters were varied independently then we would have found two linearly independent solutions in terms of Aei Ω and Be iΩ.6. Variation of parameters using a scalar parameter.
Now let’s consider a situation where there is only one scalar parameter λ. In this case there will be only one new unknown function . Since there is only one new unknown function there will be only one new linear equation to solve.
Let =y”+y+λxy=0(6), then y”+y+λxy=0 has a unique solution y=C1e^(-λx)+C2e^(-λx).
Then C1e^(-λx)+C2e^(-λx)=(y”+y)(1-(λxy))=(y”+(1-(λxy))))/(e^(-λx))=(C1+(1-(λxy))))/(C2e^(-λx)) so C2=-C1/(C2e^(-λX)). Therefore y=(Cy’+(Cy’-{lambda} {CapitalDelta} x))/({lambda}{CapitalDelta} x)=S{{CapitalDelta}} where S={Cy’}={Cy’-{lambda}{CapitalDelta} x}. So y’={Sxy}. Then solving for y yields {{Sigma}}y={Sigma}({Sigma}{Sigma}{Sigma}{Sigma})={Sigmaf}({{Beta}}{{Gamma}}L)-({{Beta}}{{Gamma}}L){{Beta}}{{Gamma}}L=-{\frac{\partial }{\partial L}}} where L={Ly’-Ly}{\![(“the Lagrangian”)]}.
Variation of parameters
Another method to solve differential equations is by variation of parameters. This is a relatively simple method that can be applied to average value problems, linear problems, and some quadratic problems.
Average value problems require the average value of some function to be solved for. For example, find the average value of X + Y over a period of time where X = 2sin(t) and Y = t.
To solve this using variation of parameters, we must first note that the average value of X + Y is equal to (X + Y) / (X + Y) = 2sin(t).
Then, we must find two new variables U and V such that U + V = 2sin(t) and UV = t. These can be found by taking the reciprocal of each variable and then solving for V such that U+V=1. Then, replace all instances of X with U+V and solve the new equation system for U and V.
Examples
The following are examples of how you can apply the method for solving differential equations by variation of parameters to specific problems.
Calculating Population Growth
Population growth can be modeled with a differential equation. The population growth per generation, denoted dN/dT, is a constant so can be incorporated into the integration.
With N representing the population, t representing time, and e being Euler’s number (which is approximately 2.7), then the solution to this equation is: N(t) = N(0)e^-(t/T) .
So, how does one solve for T? By using the Variation of Parameters Method! By substituting T for t in the solution above, one gets: T = √N(0)e^-(t/T) .
The general form of the differential equation
A general form of the differential equation is Y’= F(X,Y), where X and Y are variables and F is the function. This equation states that the rate of change of Y relative to X is Y.
By solving this equation for Y, you can find the value of Y when X is any given number. For example, if X = 2, then the solution would be:
Y = whatever factor made up the rate of change of Y relative to X (in this case, 1) multiplied by 2.
By solving this equation for X, you can find the number that makes up the rate of change of Y relative to itself. For example, if Y = 1, then the solution would be: 1= whatever factor made up the rate of change of itself (in this case 1).
Solve the differential equation using variation of parameters
Now that you have the derivative of the original equation, you can use that to solve for the original equation. To do this, you need to put Y” + Y = Sin X together and solve for Y.
Variation of parameters is a method of solving differential equations where you change some of the parameters in the original equation and then solve for the new parameter.
For example, let’s say we had the following differential equation: y’ = x^2 + y. We could then change x to x – h, where h is a small number. Then we would have y’ = (x – h)^2 + y.
We then solve for y’ by taking the square root of both sides, which gives us y’ = ±(x – h)^(1/2).
Applying the formula for variation of parameters
As mentioned earlier, the formula for solving differential equations by variation of parameters is:
Y n = Y 0 + N × (X − X 0 )n
Where Y n is the value of the solution at time n, Y 0 is the initial value of the solution, N is the constant of variation, X is the independent variable in the equation being solved for, and × represents multiplication.
Variation of parameters can be used to solve non-linear differential equations as well. A good way to check if a differential equation is linear or not is to see if you can factor it. If you cannot factor it, then it most likely is not linear.
Solving non-linear differential equations by variation of parameters requires finding two new constants of variation, one for each new parameter introduced in the equation. These new constants of variation must be tested to see if they are positive or negative, which determines whether they increase or decrease the parameter.
Example using variation of parameters
In this example, we will solve the differential equation Y” + Y = Sin X by using a variation of parameters. We will use two changes in the parameter to find the new parameter.
We will first assume that Y” = 0, which means that Y” is always zero. Then, we will assume that Y = 0, which means that Y is no longer part of the equation.
By doing this, we have reduced the differential equation to just Sin X, which is now an ordinary differential equation that we can solve easily.
Solving ordinary differential equations can be tricky depending on what type you have, but most have solutions that are either y’’=0 or y=0. By checking whether these are true or not, you can solve the ordinary differential equation.
See also
Variation of parameters is a method to solve differential equations by changing the parameters of the original equation. The original parameter is the function, Y, in this case.
Variation of parameters can be used to solve many different types of differential equations, not just exponential ones like in this example. This makes it a very useful tool to know!
How does it work? First, you must find a parameter p such that Y = pX. Then, you must find a second parameter q such that Y’ = qX + pX’. Finally, you must solve the two equations you just created and then substitute the values for p and q into the first equation to get Y = Sin X.
Try it out yourself with this simple example: 2Y’ – Y = 0.
