Math Problem Statement

Berechnen Sie die allgemeine Lösung der folgenden DGL: x^2 y'' - 3x y' + 4y = ln(x), x > 0

Solution

The problem presented is a second-order linear differential equation:

x2y3xy+4y=ln(x)forx>0x^2 y'' - 3x y' + 4y = \ln(x) \quad \text{for} \quad x > 0

This is a non-homogeneous linear differential equation. The general approach to solving such an equation involves two steps:

  1. Solve the homogeneous equation x2y3xy+4y=0x^2 y'' - 3x y' + 4y = 0 to find the complementary solution yc(x)y_c(x).
  2. Find a particular solution yp(x)y_p(x) to the non-homogeneous equation x2y3xy+4y=ln(x)x^2 y'' - 3x y' + 4y = \ln(x).

Let's proceed step by step:

1. Solve the homogeneous equation:

The homogeneous equation is: x2y3xy+4y=0x^2 y'' - 3x y' + 4y = 0

This is a Cauchy-Euler equation. To solve it, assume a solution of the form: y(x)=xry(x) = x^r

Substitute y(x)=xry(x) = x^r into the homogeneous equation: y=rxr1,y=r(r1)xr2y' = r x^{r-1}, \quad y'' = r(r-1) x^{r-2}

Substituting into the homogeneous equation: x2r(r1)xr23xrxr1+4xr=0x^2 \cdot r(r-1) x^{r-2} - 3x \cdot r x^{r-1} + 4x^r = 0 r(r1)xr3rxr+4xr=0r(r-1) x^r - 3r x^r + 4x^r = 0 r(r1)3r+4=0r(r-1) - 3r + 4 = 0 Simplifying: r2r3r+4=0r^2 - r - 3r + 4 = 0 r24r+4=0r^2 - 4r + 4 = 0 (r2)2=0(r-2)^2 = 0 Thus, r=2r = 2 is a double root.

The general solution for the homogeneous equation is then: yc(x)=(C1+C2ln(x))x2y_c(x) = (C_1 + C_2 \ln(x)) x^2 where C1C_1 and C2C_2 are constants.

2. Find a particular solution:

To find a particular solution to the non-homogeneous equation: x2y3xy+4y=ln(x)x^2 y'' - 3x y' + 4y = \ln(x)

We can use the method of variation of parameters or undetermined coefficients. Since the right-hand side is ln(x)\ln(x), we propose a particular solution of the form: yp(x)=Aln(x)+By_p(x) = A \ln(x) + B

Substitute yp(x)y_p(x) into the non-homogeneous equation to determine AA and BB. I will now perform this step.

Let me calculate this.The general solution to the differential equation:

x2y3xy+4y=ln(x)x^2 y'' - 3x y' + 4y = \ln(x)

is:

y(x)=C1x2+C2x2ln(x)+14ln(x)+14y(x) = C_1 x^2 + C_2 x^2 \ln(x) + \frac{1}{4} \ln(x) + \frac{1}{4}

where C1C_1 and C2C_2 are arbitrary constants determined by initial conditions (if given).

This solution consists of the complementary (homogeneous) part: yc(x)=C1x2+C2x2ln(x)y_c(x) = C_1 x^2 + C_2 x^2 \ln(x) and the particular solution: yp(x)=14ln(x)+14y_p(x) = \frac{1}{4} \ln(x) + \frac{1}{4}

Would you like further details on any specific part, or do you have any other questions?

Here are 5 related questions you might want to explore:

  1. How can initial conditions be used to determine C1C_1 and C2C_2?
  2. What method would you use to solve this differential equation if the right-hand side was x3x^3 instead of ln(x)\ln(x)?
  3. Can you explain how variation of parameters works in solving non-homogeneous differential equations?
  4. How does the method of undetermined coefficients differ from variation of parameters?
  5. What are the specific applications of Cauchy-Euler equations in physics and engineering?

Tip: Always check the form of the non-homogeneous term to choose the appropriate method for finding a particular solution (e.g., undetermined coefficients or variation of parameters).

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Non-Homogeneous Linear Differential Equations
Cauchy-Euler Equation
Undetermined Coefficients

Formulas

Cauchy-Euler Equation: x^2 y'' - 3x y' + 4y = ln(x)
General solution for the homogeneous equation y_c(x) = (C1 + C2 ln(x)) x^2
Particular solution y_p(x) = (1/4) ln(x) + (1/4)

Theorems

Cauchy-Euler Theorem
Method of Undetermined Coefficients

Suitable Grade Level

University Level (Calculus and Differential Equations)