4 Ways to Solve Differential Equations

2024 Author: Samantha Chapman | [email protected]. Last modified: 2023-12-17 09:13

In a course on differential equations, the derivatives studied in an analysis course are used. The derivative is the measure of how much a quantity changes as a second varies; for example, how much the speed of an object changes with respect to time (in comparison to the slope). Such measures of change frequently occur in everyday life. For instance, the law of compound interest states that the rate of accumulation of interest is proportional to the initial capital, given by dy / dt = ky, where y is the sum of the compound interest of the money earned, t is time, and k is a constant (dt is a instant time interval). Although credit card interest is generally compounded daily and reported as the APR, annual percentage rate, a differential equation can be solved to give the instant solution y = c and ^ (kt), where c is an arbitrary constant (the fixed interest rate). This article will show you how to solve common differential equations, especially in mechanics and physics.

Index

Steps

Method 1 of 4: The basics

Step 1. Definition of derivative

The derivative (also referred to as the differential quotient, especially in British English) is defined as the limit of the ratio of the increment of a function (usually y) to the increment of a variable (usually x) in that function, at tend to 0 of the latter; the instantaneous change of one quantity relative to another, such as speed, which is the instantaneous change of distance versus time. Compare the first derivative and the second derivative:

• First derivative - the derivative of a function, example: Speed is the first derivative of distance with respect to time.
• Second derivative - the derivative of the derivative of a function, example: Acceleration is the second derivative of distance with respect to time.

Step 2. Identify the order and degree of the differential equation

L' order of a differential equation is determined by the derivative of the highest order; the degree is given by the highest power of a variable. For example, the differential equation shown in Figure 1 is second order and third degree.

Step 3. Learn the difference between a general or complete solution and a particular solution

A complete solution contains a number of arbitrary constants equal to the order of the equation. To solve a differential equation of order n, you have to calculate n integrals and for each integral you have to introduce an arbitrary constant. For example, in the law of compound interest, the differential equation dy / dt = ky is of first order and its complete solution y = ce ^ (kt) contains exactly one arbitrary constant. A particular solution is obtained by assigning particular values to the constants in the general solution.

Method 2 of 4: Solving 1st Order Differential Equations

It is possible to express a first order and first degree differential equation in the form M dx + N dy = 0, where M and N are functions of x and y. To solve this differential equation, do the following:

Step 1. Check if the variables are separable

The variables are separable if the differential equation can be expressed as f (x) dx + g (y) dy = 0, where f (x) is a function of only x, and g (y) is a function of only y. These are the easiest differential equations to solve. They can be integrated to give ∫f (x) dx + ∫g (y) dy = c, where c is an arbitrary constant. A general approach follows. See Figure 2 for an example.

• Eliminate fractions. If the equation contains derivatives, multiply by the differential of the independent variable.
• Collect all terms containing the same differential into one term.
• Integrate each part separately.
• Simplify the expression, for example, by combining terms, converting logarithms to exponents and using the simplest symbol for arbitrary constants.

Step 2. If the variables cannot be separated, check if it is a homogeneous differential equation

A differential equation M dx + N dy = 0, is homogeneous if the replacement of x and y with λx and λy results in the original function multiplied by a power of λ, where the power of λ is defined as the degree of the original function. If this is your case, please follow the steps below. See Figure 3 as an example.

• Given y = vx, it follows dy / dx = x (dv / dx) + v.
• From M dx + N dy = 0, we have dy / dx = -M / N = f (v), since y is a function of v.
• Hence f (v) = dy / dx = x (dv / dx) + v. Now the variables x and v can be separated: dx / x = dv / (f (v) -v)).
• Solve the new differential equation with separable variables and then use the substitution y = vx to find y.

Step 3. If the differential equation cannot be solved using the two methods explained above, try to express it as a linear equation, in the form dy / dx + Py = Q, where P and Q are functions of x alone or are constants

Note that here x and y can be used interchangeably. If so, continue as follows. See Figure 4 as an example.

• Let y = uv be given, where u and v are functions of x.
• Calculate the differential to get dy / dx = u (dv / dx) + v (du / dx).
• Substitute in dy / dx + Py = Q, to get u (dv / dx) + v (du / dx) + Puv = Q, or u (dv / dx) + (du / dx + Pu) v = Q.
• Determine u by integrating du / dx + Pu = 0, where the variables are separable. Then use the value of u to find v by solving u (dv / dx) = Q, where, again, the variables are separable.
• Finally, use the substitution y = uv to find y.

Step 4. Solve the Bernoulli equation: dy / dx + p (x) y = q (x) y, as follows:

• Let u = y^1-n, so that du / dx = (1-n) y^-n (dy / dx).
• It follows that, y = u^{1 / (1-n)}, dy / dx = (du / dx) y / (1-n), and y = u^{n / (1-n)}.

• Substitute in the Bernoulli equation and multiply by (1-n) / u^{1 / (1-n)}, to give

  du / dx + (1-n) p (x) u = (1-n) q (x).

• Note that we now have a first-order linear equation with the new variable u that can be solved with the methods explained above (Step 3). Once solved, replace y = u^{1 / (1-n)} to get the complete solution.

Method 3 of 4: Solving 2nd Order Differential Equations

Step 1. Check if the differential equation satisfies the form shown in equation (1) in Figure 5, where f (y) is a function of y alone, or a constant

If so, follow the steps described in Figure 5.

Step 2. Solving second order linear differential equations with constant coefficients:

Check if the differential equation satisfies the form shown in equation (1) in Figure 6. If so, the differential equation can be solved simply as a quadratic equation as shown in the following steps:

Step 3. To solve a more general second-order linear differential equation, check if the differential equation satisfies the form shown in equation (1) in Figure 7

If this is the case, the differential equation can be solved by following the following steps. For an example, see the steps in Figure 7.

• Solve equation (1) of Figure 6 (where f (x) = 0) using the method described above. Let y = u be the complete solution, where u is the complementary function for equation (1) in Figure 7.

• By trial and error find a particular solution y = v of equation (1) in Figure 7. Follow the steps below:

  • If f (x) is not a particular solution of (1):

    • If f (x) is of the form f (x) = a + bx, assume that y = v = A + Bx;
    • If f (x) is in the form f (x) = ae^bx, assume that y = v = Ae^bx;
    • If f (x) is in the form f (x) = a₁ cos bx + a₂ sin bx, assume that y = v = A₁ cos bx + A₂ sin bx.
  • If f (x) is a particular solution of (1), assume the above form multiplied by x for v.

  The complete solution of (1) is given by y = u + v.

  Method 4 of 4: Solving Higher Order Differential Equations

  Higher-order differential equations are much more difficult to solve, with the exception of a few special cases:

  

  Step 1. Check if the differential equation satisfies the form shown in equation (1) in Figure 5, where f (x) is a function of x alone, or a constant

  If so, follow the steps described in Figure 8.

  

  Step 2. Solving nth order linear differential equations with constant coefficients:

  Check if the differential equation satisfies the form shown in equation (1) in Figure 9. If so, the differential equation can be solved as follows:

  

  Step 3. To solve a more general n-th order linear differential equation, check if the differential equation satisfies the form shown in equation (1) in Figure 10

  If this is the case, the differential equation can be solved with a method similar to that used to solve second order linear differential equations, as follows:

  Practical Applications

  1. 

     Law of compound interest:

     the speed of the accumulation of interest is proportional to the initial capital. More generally, the rate of change with respect to an independent variable is proportional to the corresponding value of the function. That is, if y = f (t), dy / dt = ky. Solving with the separable variable method, we will have y = ce ^ (kt), where y is the capital accumulating at compound interest, c is an arbitrary constant, k is the interest rate (for example, interest in dollars to one dollar a year), t is time. It follows that time is money.

     • Note that the compound interest law applies in many areas of daily life.

       For example, suppose you want to dilute a saline solution by adding water to reduce its salt concentration. How much water will you need to add and how does the concentration of the solution vary with respect to the speed at which you run the water?

       Let s = the amount of salt in the solution at any given time, x = the amount of water passed into the solution and v = the volume of the solution. The concentration of the salt in the mixture is given by s / v. Now, suppose that a volume Δx leaks out of the solution, so that the amount of salt leaking is (s / v) Δx, hence the change in the amount of salt, Δs, is given by Δs = - (s / v) Δx. Divide both sides by Δx, to give Δs / Δx = - (s / v). Take the limit as Δx0, and you will have ds / dx = -s / v, which is a differential equation in the form of the law of compound interest, where here y is s, t is x and k is -1 / v.

     • 

       Newton's law of cooling '' 'is another variant of the law of compound interest. It states that the cooling rate of a body with respect to the temperature of the surrounding environment is proportional to the difference between the temperature of the body and that of the surrounding environment. Let x = body temperature in excess of the surrounding environment, t = time; we will have dx / dt = kx, where k is a constant. The solution for this differential equation is x = ce ^ (kt), where c is an arbitrary constant, as above. Suppose the excess temperature, x, was first 80 degrees and drops to 70 degrees after one minute. What will it be like after 2 minutes?

       Given t = time, x = temperature in degrees, we will have 80 = ce ^ (k * 0) = c. Furthermore, 70 = ce ^ (k * 1) = 80e ^ k, so k = ln (7/8). It follows that x = 70e ^ (ln (7/8) t) is a particular solution of this problem. Now enter t = 2, you will have x = 70e ^ (ln (7/8) * 2) = 53.59 degrees after 2 minutes.

     • 

       Various layers of the atmosphere with respect to the rise in altitude above sea level In thermodynamics, the atmospheric pressure p above sea level changes in proportion to the altitude h above sea level. Here too it is a variation of the law of compound interest. The differential equation in this case is dp / dh = kh, where k is a constant.

     • 

       In chemistry, the rate of a chemical reaction, where x is the quantity transformed in a period t, is the time rate of change of x. Given a = the concentration at the start of the reaction, then dx / dt = k (a-x), where k is the rate constant. This is also a variation of the law of compound interest where (a-x) is now a dependent variable. Let d (a-x) / dt = -k (a-x), s or d (a-x) / (a-x) = -kdt. Integrate, to give ln (a-x) = -kt + a, since a-x = a when t = 0. Rearranging, we find that the velocity constant k = (1 / t) ln (a / (a-x)).

     • 

       In electromagnetism, given an electric circuit with a voltage V and a current i (amperes), the voltage V undergoes a reduction when it exceeds the resistance R (ohm) of the circuit and the induction L, according to the equation V = iR + L (of / dt), or di / dt = (V - iR) / L. This is also a variation of the law of compound interest where V - iR is now the dependent variable.

  2. 

     In acoustics, a simple harmonic vibration has an acceleration which is directly proportional to the negative value of the distance. Remembering that acceleration is the second derivative of distance, then d ² s / dt ² + k ² s = 0, where s = distance, t = time, and k ² is the measure of acceleration at unit distance. This is the simple harmonic equation, a second order linear differential equation with constant coefficients, as solved in Figure 6, equations (9) and (10). The solution is s = c₁cos kt + c₂sin kt.

     It can be further simplified by establishing c₁ = b sin A, c₂ = b cos A. Substitute them to get b sin A cos kt + b cos A sin kt. From trigonometry we know that sin (x + y) = sin x cos y + cos x sin y, so that the expression is reduced to s = b sin (kt + A). The wave that follows the simple harmonic equation oscillates between b and -b with a period of 2π / k.

     • 

       Spring: let's take an object of mass m connected to a spring. According to Hooke's law, when the spring stretches or compresses by s units with respect to its initial length (also called equilibrium position), it exerts a restoring force F proportional to s, i.e. F = - k²s. According to Newton's second law (force is equal to the product of mass times acceleration), we will have m d ² s / dt ² = - k²s, or m d ² s / dt ² + k²s = 0, which is an expression of the simple harmonic equation.

     • 

       Rear armotizer and spring of a BMW R75 / 5 motorcycle Damped vibrations: consider the vibrating spring as above, with a damping force. Any effect, such as the friction force, which tends to reduce the amplitude of the oscillations in an oscillator, is defined as a damping force. For example, a damping force is provided by a car armotizer. Typically, the damping force, F_d, is roughly proportional to the speed of the object, that is, F_d = - c² ds / dt, where c² is a constant. By combining the damping force with the restoring force, we will have - k²s - c² ds / dt = m d ² s / dt ², based on Newton's second law. Or, m d ² s / dt ² + c² ds / dt + k²s = 0. This differential equation is a second-order linear equation that can be solved by solving the auxiliary equation mr² + c²r + k² = 0, after replacing s = e ^ (rt).

       Solve with the quadratic formula r₁ = (- c² + sqrt (c⁴ - 4 mk²)) / 2 m; r₂ = (- c² - sqrt (c⁴ - 4 mk²)) / 2 m.

       • Over-damping: If c⁴ - 4mk² > 0, r₁ and r₂ they are real and distinct. The solution is s = c₁ and ^ (r₁t) + c₂ and ^ (r₂t). Since c², m, and k² are positive, sqrt (c⁴ - 4mk²) must be less than c², which implies that both roots, r₁ and r₂, are negative, and the function is in exponential decay. In this case, Not an oscillation occurs. A strong damping force, for example, can be given by a high viscosity oil or a lubricant.
       • Critical damping: If c⁴ - 4mk² = 0, r₁ = r₂ = -c² / 2m. The solution is s = (c₁ + c₂t) and ^ ((- c²/ 2m) t). This is also an exponential decay, without oscillation. The slightest decrease, however, in the damping force will cause the object to oscillate once the equilibrium point is exceeded.
       • Underdamping: If c⁴ - 4mk² <0, the roots are complex, given by - c / 2m +/- ω i, where ω = sqrt (4 mk² - c⁴)) / 2 m. The solution is s = e ^ (- (c²/ 2m) t) (c₁ cos ω t + c₂ sin ω t). This is an oscillation damped by the factor e ^ (- (c²/ 2m) t. Since c² and m are both positive, and ^ (- (c²/ 2m) t) will tend to zero as t approaches infinity. It follows that sooner or later the motion will decay to zero.

       Advice

       • Replace the solution in the original differential equation to see that the equation is satisfied. This way you can check if the solution is correct.
       • Note: the inverse of the differential calculus is said integral calculation, which deals with the sum of the effects of continuously changing quantities; for example, the calculation of the distance (compare with d = rt) covered by an object whose instantaneous variations (velocity) in a time interval are known.
       • Many differential equations are not solvable with the methods described above. The above methods, however, are sufficient to solve many common differential equations.