Math Problem Statement

Solve this question of Differential Equation till the end

Solve the second-order linear differential equation: x^2 d^2y/dx^2 + 2x dy/dx - 12y = x^3 log(x).

Solution

We are tasked with solving the given second-order linear differential equation:

x2d2ydx2+2xdydx12y=x3lnx.x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} - 12y = x^3 \ln x.

This is a non-homogeneous Cauchy-Euler equation. Let's solve it step by step.


Step 1: General Solution of the Homogeneous Equation

The associated homogeneous equation is: x2d2ydx2+2xdydx12y=0.x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} - 12y = 0.

We assume a solution of the form yh=xmy_h = x^m, and substitute yhy_h into the homogeneous equation. Calculating derivatives: dyhdx=mxm1,d2yhdx2=m(m1)xm2.\frac{dy_h}{dx} = mx^{m-1}, \quad \frac{d^2y_h}{dx^2} = m(m-1)x^{m-2}.

Substitute these into the equation: x2m(m1)xm2+2xmxm112xm=0.x^2 \cdot m(m-1)x^{m-2} + 2x \cdot mx^{m-1} - 12x^m = 0.

Simplify: m(m1)xm+2mxm12xm=0.m(m-1)x^m + 2mx^m - 12x^m = 0.

Factor out xmx^m (since xm0x^m \neq 0): m(m1)+2m12=0.m(m-1) + 2m - 12 = 0.

Simplify the quadratic equation: m2+m12=0.m^2 + m - 12 = 0.

Factorize: (m+4)(m3)=0.(m+4)(m-3) = 0.

Thus, m=4m = -4 and m=3m = 3.

The general solution of the homogeneous equation is: yh=C1x4+C2x3,y_h = C_1 x^{-4} + C_2 x^3, where C1C_1 and C2C_2 are arbitrary constants.


Step 2: Particular Solution of the Non-Homogeneous Equation

Now consider the full equation: x2d2ydx2+2xdydx12y=x3lnx.x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} - 12y = x^3 \ln x.

We seek a particular solution ypy_p using the method of undetermined coefficients. Since the right-hand side is x3lnxx^3 \ln x, we try: yp=Ax3lnx+Bx3.y_p = Ax^3 \ln x + Bx^3.

Derivatives of ypy_p:

dypdx=A(3x2lnx+x2)+B3x2,\frac{dy_p}{dx} = A \left( 3x^2 \ln x + x^2 \right) + B \cdot 3x^2, d2ypdx2=A(6xlnx+5x)+B6x.\frac{d^2 y_p}{dx^2} = A \left( 6x \ln x + 5x \right) + B \cdot 6x.

Substitute ypy_p and its derivatives into the equation:

[ x^2 \left[ A \left( 6x \ln x + 5x \right) + B \cdot 6x \right] + 2x \left[ A \left( 3x^2 \ln x + x^2 \right) + B \cdot 3x^2 \right] - 12 \left[ A x^3 \ln x + Bx^3 \right] = x^3 \ln x. ]

Simplify each term:

  1. From x2d2ypdx2x^2 \frac{d^2 y_p}{dx^2}: x2A(6xlnx+5x)+x2B(6x)=A(6x3lnx+5x3)+B(6x3).x^2 \cdot A (6x \ln x + 5x) + x^2 \cdot B (6x) = A (6x^3 \ln x + 5x^3) + B (6x^3).

  2. From 2xdypdx2x \frac{dy_p}{dx}: 2xA(3x2lnx+x2)+2xB(3x2)=A(6x3lnx+2x3)+B(6x3).2x \cdot A (3x^2 \ln x + x^2) + 2x \cdot B (3x^2) = A (6x^3 \ln x + 2x^3) + B (6x^3).

  3. From 12yp-12y_p: 12(Ax3lnx+Bx3)=12Ax3lnx12Bx3.-12 \cdot \left( A x^3 \ln x + Bx^3 \right) = -12A x^3 \ln x - 12B x^3.

Combine all terms: [ \big[ 6A + 6A - 12A \big] x^3 \ln x + \big[ 5A + 2A + 6B + 6B - 12B \big] x^3 = x^3 \ln x. ]

Equating coefficients:

  • For x3lnxx^3 \ln x: 6A+6A12A=1    A=16.6A + 6A - 12A = 1 \implies A = \frac{1}{6}.

  • For x3x^3: 5A+2A+6B+6B12B=0    7A=0    B=0.5A + 2A + 6B + 6B - 12B = 0 \implies 7A = 0 \implies B = 0.

Thus, the particular solution is: yp=16x3lnx.y_p = \frac{1}{6} x^3 \ln x.


Step 3: General Solution

The general solution is the sum of the homogeneous and particular solutions: y=yh+yp=C1x4+C2x3+16x3lnx.y = y_h + y_p = C_1 x^{-4} + C_2 x^3 + \frac{1}{6} x^3 \ln x.


Would you like detailed verification of any step? Here are 5 related questions for further exploration:

  1. How do we solve a Cauchy-Euler equation with different forms of non-homogeneous terms?
  2. What if the trial solution ypy_p overlaps with the homogeneous solution?
  3. Why do we assume yh=xmy_h = x^m in solving Cauchy-Euler equations?
  4. How does the method of undetermined coefficients apply to non-polynomial terms?
  5. What are the applications of Cauchy-Euler equations in physics and engineering?

Tip: Always check if the form of the particular solution overlaps with the homogeneous solution. If it does, multiply the trial solution by xx to ensure linear independence.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Cauchy-Euler Equations
Method of Undetermined Coefficients
Homogeneous and Non-Homogeneous Solutions

Formulas

General Cauchy-Euler equation: x^2 d^2y/dx^2 + bx dy/dx + c y = f(x)
Quadratic characteristic equation: m^2 + bm + c = 0

Theorems

Cauchy-Euler Method
Superposition Principle for Linear Differential Equations

Suitable Grade Level

Undergraduate