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