Math Problem Statement

y double prime + 4 yprime = x

Solution

The given differential equation is:

y+4y=xy'' + 4y' = x

This is a linear non-homogeneous second-order differential equation. Let's solve it step by step.


1. Solve the Homogeneous Equation

The homogeneous part of the equation is: y+4y=0y'' + 4y' = 0

Its characteristic equation is: r2+4r=0r^2 + 4r = 0

Factoring: r(r+4)=0r(r + 4) = 0

So, the roots are: r=0,r=4r = 0, \quad r = -4

The general solution to the homogeneous equation is: yh=C1+C2e4xy_h = C_1 + C_2 e^{-4x}


2. Solve for a Particular Solution

The non-homogeneous equation is: y+4y=xy'' + 4y' = x

We use the method of undetermined coefficients. Assume a particular solution of the form: yp=Ax+By_p = Ax + B

Substitute ypy_p into the differential equation: yp+4yp=0+4A=xy_p'' + 4y_p' = 0 + 4A = x

Equating coefficients of xx and constants:

  • Coefficient of xx: 0=10 = 1 → contradiction, adjust to Ax2+BxAx^2 + Bx. Let us refine the solution approach for a better guess for the particular solution since xx alone contradicts.

Now guess Yp -> Degree adjustmentsSince the right-hand side is xx, we revise our guess for the particular solution.

We let: yp=Ax+By_p = Ax + B

and substitute into the differential equation: y+4y=x.y'' + 4y' = x.

Step-by-Step Substitution:

  1. First derivatives:

    • yp=Ay_p' = A,
    • yp=0y_p'' = 0.
  2. Substitute into the equation: 0+4A=x.0 + 4A = x.

  3. Compare coefficients of xx and constants:

    • Coefficient of xx: There is no match, which suggests the need to revise the guess for ypy_p to a polynomial form.

Correct Guess: Quadratic Form

The minimal polynomial guess should be: yp=Ax2+Bx+C.y_p = Ax^2 + Bx + C.

Now, compute derivatives:

  • yp=2Ax+By_p' = 2Ax + B,
  • yp=2Ay_p'' = 2A.

Substitute these into the equation: 2A+4(2Ax+B)=x.2A + 4(2Ax + B) = x.

Simplify: 2A+8Ax+4B=x.2A + 8Ax + 4B = x.

Equating coefficients:

  1. Coefficient of xx: 8A=1    A=188A = 1 \implies A = \frac{1}{8},
  2. Constant term: 2A+4B=0    2(18)+4B=0    B=1162A + 4B = 0 \implies 2\left(\frac{1}{8}\right) + 4B = 0 \implies B = -\frac{1}{16},
  3. No CC-term adjustment needed since no constant appears.

Particular Solution:

yp=18x2116x.y_p = \frac{1}{8}x^2 - \frac{1}{16}x.


3. General Solution

Combine the homogeneous solution yh=C1+C2e4xy_h = C_1 + C_2e^{-4x} with the particular solution ypy_p: y=C1+C2e4x+18x2116x.y = C_1 + C_2e^{-4x} + \frac{1}{8}x^2 - \frac{1}{16}x.


Let me know if you'd like detailed verification steps or further explanation!

5 Related Questions:

  1. How do we solve non-homogeneous differential equations using other methods, like variation of parameters?
  2. What is the role of the characteristic equation in solving differential equations?
  3. How do we select the correct form of the particular solution for various right-hand sides?
  4. What happens if the particular solution overlaps with the homogeneous solution?
  5. Can this method be applied to higher-order differential equations?

Tip:

Always check if the particular solution conflicts with the homogeneous solution and adjust your guess accordingly (e.g., by multiplying by xx).

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Math Problem Analysis

Mathematical Concepts

Linear Differential Equations
Second-Order Differential Equations
Method of Undetermined Coefficients

Formulas

Characteristic equation: r^2 + 4r = 0
General solution: y = y_h + y_p
Particular solution guess: y_p = Ax^2 + Bx + C

Theorems

Superposition principle for linear differential equations
Method of Undetermined Coefficients

Suitable Grade Level

Undergraduate (Calculus II or Differential Equations)