Exercise on Second-Order Differential Equations 19

We are asked to solve the following differential equation:

$$ y'' - 2y' + y = 6x e^x $$

This is a second-order linear nonhomogeneous differential equation, also known as a complete equation.

To find the general solution, we first solve the associated homogeneous equation $ y_o $, then add to it a particular solution $ y_p $ of the full equation:

$$ y = y_o + y_p $$

The solution process consists of three steps:

1] Solving the Homogeneous Equation
2] Finding a Particular Solution
3] General Solution

1] Solving the Homogeneous Equation

The homogeneous version of the equation is:

$$ y'' - 2y' + y = 0 $$

We solve the characteristic equation using an auxiliary variable $ t $:

$$ t^2 - 2t + 1 = 0 $$

Solving for $ t $:

$$ t = \frac{2 \pm \sqrt{4 - 4(1)(1)}}{2} = \begin{cases} t_1 = 1 \\ \\ t_2 = 1 \end{cases} $$

Since the characteristic equation has a repeated real root, the general solution to the homogeneous equation is:

$$ y_o = c_1 e^x + x c_2 e^x $$

2] Finding a Particular Solution

Next, we determine a particular solution using the method of undetermined coefficients.

$$ y'' - 2y' + y = 6x e^x $$

The nonhomogeneous term $ 6x e^x $ has the form $ P(x)e^{\lambda x} $, with $ P(x) = 6x $ and $ \lambda = 1 $.

Since $ \lambda = 1 $ is a root of the characteristic equation with multiplicity $ m = 2 $, we choose a trial function of the form:

$$ y_p = x^m e^{\lambda x} Q(x) = x^2 e^x (A + Bx) $$

Note. The multiplicity $ m $ of a root $ \lambda $ is the highest power such that the characteristic polynomial can be factored as $$ P(\lambda) = (t - \lambda)^m R(t) $$ In this case, since $ t^2 - 2t + 1 = (t - 1)^2 $, we confirm $ \lambda = 1 $ is a double root.

Given that $ P(x) = 6x $ is a degree-1 polynomial, we let $ Q(x) = A + Bx $:

$$ y_p = x^2 e^x (A + Bx) $$

We now compute the first and second derivatives:

First derivative:

$$ y_p' = D_x[ x^2 e^x (A + Bx) ] $$

Using the product rule:

$$ y_p' = 2x e^x (A + Bx) + x^2 e^x (A + Bx) + x^2 e^x B $$

Second derivative:

$$ y_p'' = D_x[ y_p' ] = D_x[2x e^x (A + Bx)] + D_x[x^2 e^x (A + Bx)] + D_x[x^2 e^x B] $$

Simplifying yields:

$$ y_p'' = e^x \cdot \left[2A + 4x(A + Bx) + 6xB + x^2(A + Bx) + 2x^2B\right] $$

Now substitute $ y_p $, $ y_p' $, and $ y_p'' $ into the original equation:

$$ y'' - 2y' + y = 6x e^x $$

After substitution and simplification, the left-hand side becomes:

$$ e^x \cdot (2A + 6xB) $$

Equating this to the right-hand side $ 6x e^x $, we get:

$$ \begin{cases} 2A = 0 \\ \\ 6B = 6 \end{cases} \Rightarrow \begin{cases} A = 0 \\ \\ B = 1 \end{cases} $$

Explanation. There is no constant term on the right-hand side, so the coefficient of $ e^x $ must be zero: $ 2A = 0 $. The coefficient of $ x e^x $ must match: $ 6B = 6 $.

Substitute $ A = 0 $, $ B = 1 $ into the expression for $ y_p $:

$$ y_p = x^2 e^x (0 + x) = x^3 e^x $$

3] General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:

$$ y = y_o + y_p $$

With:

$$ y_o = c_1 e^x + x c_2 e^x \quad \text{and} \quad y_p = x^3 e^x $$

The final result is:

$$ y = c_1 e^x + x c_2 e^x + x^3 e^x $$

And that completes the solution.