Solving the Differential Equation 16y'' 8y' y x^2 - 3x Using Variation of Parameters

In differential equations, the method of variation of parameters (also known as the method of undetermined coefficients) is a powerful technique for solving second-order non-homogeneous linear differential equations. This article will guide you through the process of solving the equation 16y'' 8y' y x^2 - 3x using this method. The problem is framed within a larger context of differential equations and the applications of these techniques in various fields of mathematics and engineering.

Introduction to the Characteristic Equation

First, we start with the characteristic equation of the homogeneous part:

16m^2 8m 1 0

This equation can be factored to reveal its roots:

(4m 1)^2 0

Hence, the characteristic roots are:

root -1/4 (repeated root)

The Homogeneous Solution

Given the characteristic roots, the homogeneous solution (yh) can be written as:

yh(x) C1 * e^(-x/4) C2 * x * e^(-x/4)

where C1 and C2 are constants to be determined later.

Setting Up the Particular Solution

To find the particular solution (yp), we introduce new functions u(x) and v(x), and set:

y u(x) * e^(-x/4) v(x) * x * e^(-x/4)

By differentiating y, we arrive at:

y' e^(-x/4) * u' x * e^(-x/4) * v' - (1/4) * e^(-x/4) * u - (1/4) * x * e^(-x/4) * v

And by differentiating again:

y'' e^(-x/4) * u'' - (1/2) * e^(-x/4) * u' - (1/4) * e^(-x/4) * u e^(-x/4) * v'' - (1/2) * e^(-x/4) * v' - (1/4) * x * e^(-x/4) * v' - (1/4) * e^(-x/4) * v (1/4) * x * e^(-x/4) * v''

Substituting y and y'' into the original equation, we eliminate terms to obtain:

16 * [e^(-x/4) * u'' - (1/2) * e^(-x/4) * u' - (1/4) * e^(-x/4) * u e^(-x/4) * v'' - (1/2) * e^(-x/4) * v' - (1/4) * x * e^(-x/4) * v'] - 8 * [e^(-x/4) * u' x * e^(-x/4) * v' - (1/4) * e^(-x/4) * u - (1/4) * x * e^(-x/4) * v] [u * e^(-x/4) v * x * e^(-x/4)] x^2 - 3x

Solving for u' and v'

To facilitate the solution, we impose a condition on u and v:

u' * e^(-x/4) v' * x * e^(-x/4) 0

This simplifies to:

u' v' * x 0

From which we can obtain v' in terms of u' and substitute back into the equation:

u' v' * x 0

v' -u' / x

Substituting v' into y' and solving the remaining differential equations, we get:

u' -x^2 - 3x / 16

v' x^2 3x / 4

Integrating these equations, we find:

u -[1/4 * (x^3 - 9x^2 18x - 72)] * e^(-x/4)

v [1/4 * (x^2 - 5x 20)] * e^(-x/4)

Therefore, the particular solution yp is:

yp u * e^(-x/4) v * x * e^(-x/4)

The Complete Solution

Combining the homogeneous (yh) and particular (yp) solutions, we obtain the general solution:

y yh yp

y C1 * e^(-x/4) C2 * x * e^(-x/4) 16x^2 - 13x - 72

Conclusion

In this article, we have demonstrated the method of variation of parameters to solve the non-homogeneous differential equation 16y'' 8y' y x^2 - 3x. The solution process involves careful manipulation of differential equations, integration, and the application of the method of undetermined coefficients. This method is particularly useful for solving a wide range of non-homogeneous differential equations and can be adapted to various scenarios in mathematics, physics, and engineering. Understanding this method enhances your problem-solving skills in differential equations and provides a solid foundation for advanced topics in the field.

References

[1] Zill, D. G. (2013). A First Course in Differential Equations with Modeling Applications (10th ed.). Brooks/Cole.

[2] Apostol, T. M. (1992). Calculus - One and Several Variables (5th ed.). Wiley.