Solve The Given Differential Equation By Undetermined Coefficients:${ Y^{\prime \prime} - 8 Y^{\prime} + 20 Y = 100 X^2 - 91 X E^x }$

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Introduction

Differential equations are a fundamental concept in mathematics, and they have numerous applications in various fields such as physics, engineering, and economics. One of the methods used to solve differential equations is the method of undetermined coefficients. This method is used to solve non-homogeneous linear differential equations with constant coefficients. In this article, we will discuss how to solve the given differential equation using the method of undetermined coefficients.

The Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to solve non-homogeneous linear differential equations with constant coefficients. The general form of a non-homogeneous linear differential equation with constant coefficients is:

any(n)+an−1y(n−1)+⋯+a1y′+a0y=f(x){ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y^{\prime} + a_0 y = f(x) }

where an,an−1,…,a1,a0a_n, a_{n-1}, \ldots, a_1, a_0 are constants, and f(x)f(x) is a function of xx. The method of undetermined coefficients involves assuming a particular solution of the form:

yp=∑i=0ncixi{ y_p = \sum_{i=0}^{n} c_i x^i }

where cic_i are constants to be determined. The particular solution ypy_p is then substituted into the differential equation, and the coefficients cic_i are determined by equating the coefficients of like powers of xx on both sides of the equation.

Solving the Given Differential Equation

The given differential equation is:

y′′−8y′+20y=100x2−91xex{ y^{\prime \prime} - 8 y^{\prime} + 20 y = 100 x^2 - 91 x e^x }

To solve this differential equation using the method of undetermined coefficients, we first assume a particular solution of the form:

yp=c0+c1x+c2x2+c3x3+c4x4+c5x5ex{ y_p = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + c_4 x^4 + c_5 x^5 e^x }

where c0,c1,c2,c3,c4,c5c_0, c_1, c_2, c_3, c_4, c_5 are constants to be determined. We then substitute this particular solution into the differential equation and equate the coefficients of like powers of xx on both sides of the equation.

Finding the Coefficients

We first find the coefficients of the terms involving x2x^2 and x3x^3 on the right-hand side of the differential equation. We have:

100x2−91xex=100x2−91xex{ 100 x^2 - 91 x e^x = 100 x^2 - 91 x e^x }

Comparing the coefficients of x2x^2 on both sides of the equation, we get:

20c2=100{ 20 c_2 = 100 }

Solving for c2c_2, we get:

c2=5{ c_2 = 5 }

Next, we compare the coefficients of x3x^3 on both sides of the equation. We have:

0=0{ 0 = 0 }

This means that there is no term involving x3x^3 on the right-hand side of the differential equation. Therefore, we set c3=0c_3 = 0.

Finding the Coefficients of the Exponential Term

We now find the coefficients of the terms involving exe^x on the right-hand side of the differential equation. We have:

−91xex=−91xex{ -91 x e^x = -91 x e^x }

Comparing the coefficients of xexx e^x on both sides of the equation, we get:

20c5=−91{ 20 c_5 = -91 }

Solving for c5c_5, we get:

c5=−9120{ c_5 = -\frac{91}{20} }

Finding the Coefficients of the Constant Term

We now find the coefficients of the constant term on the right-hand side of the differential equation. We have:

0=0{ 0 = 0 }

This means that there is no constant term on the right-hand side of the differential equation. Therefore, we set c0=0c_0 = 0.

Finding the Coefficients of the Linear Term

We now find the coefficients of the linear term on the right-hand side of the differential equation. We have:

0=0{ 0 = 0 }

This means that there is no linear term on the right-hand side of the differential equation. Therefore, we set c1=0c_1 = 0.

The Particular Solution

We have now found all the coefficients of the particular solution. We have:

yp=5x2−9120x5ex{ y_p = 5 x^2 - \frac{91}{20} x^5 e^x }

The General Solution

The general solution of the differential equation is the sum of the complementary solution and the particular solution. The complementary solution is the solution of the homogeneous differential equation:

y′′−8y′+20y=0{ y^{\prime \prime} - 8 y^{\prime} + 20 y = 0 }

The complementary solution is:

yc=c1e4x+c2e2x{ y_c = c_1 e^{4x} + c_2 e^{2x} }

where c1c_1 and c2c_2 are arbitrary constants. Therefore, the general solution of the differential equation is:

y=yc+yp=c1e4x+c2e2x+5x2−9120x5ex{ y = y_c + y_p = c_1 e^{4x} + c_2 e^{2x} + 5 x^2 - \frac{91}{20} x^5 e^x }

Conclusion

In this article, we have discussed how to solve the given differential equation using the method of undetermined coefficients. We have assumed a particular solution of the form:

yp=c0+c1x+c2x2+c3x3+c4x4+c5x5ex{ y_p = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + c_4 x^4 + c_5 x^5 e^x }

and determined the coefficients c0,c1,c2,c3,c4,c5c_0, c_1, c_2, c_3, c_4, c_5 by equating the coefficients of like powers of xx on both sides of the equation. We have found that the particular solution is:

yp=5x2−9120x5ex{ y_p = 5 x^2 - \frac{91}{20} x^5 e^x }

and the general solution of the differential equation is:

y=c1e4x+c2e2x+5x2−9120x5ex{ y = c_1 e^{4x} + c_2 e^{2x} + 5 x^2 - \frac{91}{20} x^5 e^x }

Introduction

In our previous article, we discussed how to solve differential equations using the method of undetermined coefficients. In this article, we will answer some of the most frequently asked questions about solving differential equations with undetermined coefficients.

Q: What is the method of undetermined coefficients?

A: The method of undetermined coefficients is a technique used to solve non-homogeneous linear differential equations with constant coefficients. It involves assuming a particular solution of the form:

yp=∑i=0ncixi{ y_p = \sum_{i=0}^{n} c_i x^i }

where cic_i are constants to be determined.

Q: What are the steps involved in solving a differential equation using the method of undetermined coefficients?

A: The steps involved in solving a differential equation using the method of undetermined coefficients are:

  1. Assume a particular solution of the form:

yp=∑i=0ncixi{ y_p = \sum_{i=0}^{n} c_i x^i }

  1. Substitute the particular solution into the differential equation.

  2. Equate the coefficients of like powers of xx on both sides of the equation.

  3. Solve for the constants cic_i.

  4. The particular solution is then added to the complementary solution to obtain the general solution.

Q: What is the complementary solution?

A: The complementary solution is the solution of the homogeneous differential equation:

y′′−8y′+20y=0{ y^{\prime \prime} - 8 y^{\prime} + 20 y = 0 }

The complementary solution is:

yc=c1e4x+c2e2x{ y_c = c_1 e^{4x} + c_2 e^{2x} }

where c1c_1 and c2c_2 are arbitrary constants.

Q: How do I determine the constants cic_i?

A: The constants cic_i are determined by equating the coefficients of like powers of xx on both sides of the equation. For example, if the differential equation is:

y′′−8y′+20y=100x2−91xex{ y^{\prime \prime} - 8 y^{\prime} + 20 y = 100 x^2 - 91 x e^x }

then we would equate the coefficients of x2x^2 on both sides of the equation to obtain:

20c2=100{ 20 c_2 = 100 }

Solving for c2c_2, we get:

c2=5{ c_2 = 5 }

Q: What is the general solution of a differential equation?

A: The general solution of a differential equation is the sum of the complementary solution and the particular solution. For example, if the differential equation is:

y′′−8y′+20y=100x2−91xex{ y^{\prime \prime} - 8 y^{\prime} + 20 y = 100 x^2 - 91 x e^x }

then the general solution is:

y=c1e4x+c2e2x+5x2−9120x5ex{ y = c_1 e^{4x} + c_2 e^{2x} + 5 x^2 - \frac{91}{20} x^5 e^x }

where c1c_1 and c2c_2 are arbitrary constants.

Q: What are some common mistakes to avoid when solving differential equations using the method of undetermined coefficients?

A: Some common mistakes to avoid when solving differential equations using the method of undetermined coefficients are:

  • Not assuming a particular solution of the correct form.
  • Not equating the coefficients of like powers of xx on both sides of the equation.
  • Not solving for the constants cic_i correctly.
  • Not adding the particular solution to the complementary solution to obtain the general solution.

Conclusion

In this article, we have answered some of the most frequently asked questions about solving differential equations with undetermined coefficients. We have discussed the method of undetermined coefficients, the steps involved in solving a differential equation using this method, and some common mistakes to avoid. We hope that this article has been helpful in clarifying any confusion about solving differential equations with undetermined coefficients.