For problems 13-17 find a particular solution of the nonhomogeneous equation, given that the functions y1(x) and y2(x) are linearly independent solutions of the corresponding homogeneous equation. x^2y''+xy'-4y=x(x+x^3)

y_g(x) = C1*x^2 + C2*x^-2 +  x^4 / 12

Step-by-step explanation:

Given:-

- The following second order ODE :

                                  x^2y''+xy'-4y=x*(x+x^3)  

Find:-

Find a particular solution of the nonhomogeneous equation    

Solution:-

- First note that the ODE given is a Cauchy Euler ODE. The order of derivative of independent and dependent variables are similar. The general form of Cauchy Euler ODE is:

                           a*x^n y^(n) + b*x^n-1 y^(n-1) + c*x^n-2 y^(n-2) + ... + d*y = f(x)

- We will use the following Auxiliary Equation to find the complementary solutions - Solving Homogeneous part of ODE.

                           am*(m-1) + bm + c = 0

Where, a,b,c are constants such that:

                            x^2y'' + xy' - 4y = 0

                            a = 1 , b = 1 , c = -4

- Solve the Auxiliary equation for (m) as follows:

                            m*(m-1) + m - 4 = 0

                            m^2 - 4 = 0

                            m = +/- 2   ( Real and distinct roots )

- The complementary solutions to the Real and distinct roots from Auxiliary Equation is:

                            yc(x) = y1(x) + y2(x)

                            yc(x) = C1*x^2 + C2*x^-2   ( Complementary Solution ).

- Now for the non-homogeneous part of ODE. The function f(x) is defined as:

                            f(x) = x*( x + x^3 ) = x^2 + x^4

- We see that (x^2) term is common to both f(x) and complementary solution yc(x). So when we develop a particular solution, we have to make sure that the solution is independent from complementary solution. If not we multiply the particular solution with (x^n). Where n is the smallest possible integer for which the solution is independent. So in our case ( Using undetermined Coefficient method ) :

                           y_p (x) = A*x^4 + B*x^3 + C*x^2 + D*x + E

- To make the solution independent we multiply y_p by (x^3) where n = 3.

                          y_p (x) = A*x^7 + B*x^6 + C*x^5 + D*x^4 + E*x^3

- Take first and second derivatives of the y_p(x) as follows:

                          y'_p(x) = 7A*x^6 + 6B*x^5 + 5C*x^4 + 4D*x^3 + 3E*x^2

                          y''_p(x) = 42Ax^5 + 30Bx^4 + 20Cx^3 + 12Dx^2 + 6Ex

- Substitute y_p(x) , y'_p(x) and y''_p(x) into the ODE given:

                     42Ax^7 + 30Bx^6 + 20Cx^5 + 12Dx^4 + 6Ex^3

              +       7Ax^7  + 6B*x^6  + 5C*x^5 + 4D*x^4 + 3E*x^3

              -      ( 4Ax^7 +  4B*x^6 +  4C*x^5 +  4D*x^4 + 4E*x^3 )

                       45Ax^7  + 32Bx^6 + 21Cx^5 + 12Dx^4 + 5Ex^3  

            45Ax^7  + 32Bx^6 + 21Cx^5 + 12Dx^4 + 5Ex^3 = x^2 + x^4

- Compare the coefficients:

                                           A = B = C = E = 0

                                           D = 1 / 12.

The particular solution is:

                                          y_p(x) = x^4 / 12

- The general solution is as follows:

                                     y_g(x) = yc(x) + y_p(x)

                                     y_g(x) = C1*x^2 + C2*x^-2 +  x^4 / 12

The particular solution to the differential equation

x²y'' + xy' - 4y = x(x + x³)

is

y_p = (1/12)x^4 - x²/2 - x/3

Step-by-step explanation:

Given the differential equation:

x²y'' + xy' - 4y = x(x + x³)(1)

First, we solve the homogeneous part of (1)

x²y'' + xy' - 4y = 0(2)

Let x = e^z

=>z = lnx

Let D = d/dz

dz/dx = (1/x)

dy/dx = (dy/dz).(dz/dx)

= (1/x)(dy/dz)

dy/dz = xdy/dx = xy' = Dy

d²y/dx² = (-1/x²)(dy/dz) + (1/x)(d²y/dx²)(dz/dx)

= (1/x²)(d²y/dx² - dy/dz) = (1/x²)(D² - D)y

Using these, (2) becomes

(D² - D)y + Dy - 4y = 0

(D² - 4)y = 0

The auxiliary equation is

m² - 4 = 0

(m - 2)(m + 2) = 0

m1 = 2, m2 = -2

The complementary function is

y = C1e^(2z) + C2e^(-2z)

But z = lnx

y_c = C1x² + C2/x² (3)

Now we solve (1) using the method of undetermined coefficients.

The nonhomogeneous part is

x(x + x³) = x² + x^4

So, we assume a particular solution of the form

y_p = Ax^4 + Bx³ + Cx² + Dx + E

y'_p = 4Ax³ + 3Bx² + 2Cx + D

y''_p = 12Ax² + 6Bx + 2C

Using these in (1)

x²y''_p + xy'_p - 4y_p = x²(12Ax² + 6Bx + 2C) + x(4Ax³ + 3Bx² + 2Cx + D) - 4(Ax^4 + Bx³ + Cx² + Dx + E)

= x² + x^4

12Ax^4 + 6Bx³ + 2Cx + 4Ax^4 + 3Bx³ + 2Cx² + Dx - 4Ax^4 - 4Bx³ - 4Cx² - 4Dx - 4E = x² + x^4

Comparing the coefficients of various powers of x, we have

12A + 4A - 4A = 1

12A = 1

=> A = 1/12

6B + 3B - 4B = 0

5B = 0

=> B = 0

2C - 4C = 1

-2C = 1

=> C = -1/2

2C + D - 4D = 0

2C - 3D = 0

2C = 3D

2(-1/2) = 3D

=> D = -1/3

-4E = 0

=> E = 0

(A, B, C, D, E) = (1/12, 0, -1/2, -1/3, 0)

y_p = Ax^4 + Bx³ + Cx² + Dx + E

= (1/12)x^4 - (1/2)x² - (1/3)x

The general solution is

y = y_c + y_p

= C1x² + C2/x² + (1/12)x^4 - x²/2 - x/3