A Two-Part Strategy

Big Idea

The general solution to the differential equation

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + cy = f(x)$

is $y = y_{\mathrm P} + y_{\mathrm C}$ . For convenience, we use $\mathcal D$ –notation to obtain

$\displaystyle (a \mathcal D^2 + b \mathcal D + c)y = f(x)$

We compute $y_{\mathrm C}$ by setting the right-hand side to zero:

$\displaystyle (a \mathcal D^2 + b \mathcal D + c)y = 0.$

Then $y_{\mathrm C}$ will take one of three forms. We compute $y_{\mathrm P}$ by using inverse- $\mathcal D$ operator techniques to evaluate

$\displaystyle y_{\mathrm P} = \frac{1}{a \mathcal D^2 + b \mathcal D + c}(f(x)).$

Questions

Question 1. Evaluate $\displaystyle \frac{1}{\mathcal D^2 + 2}(1 - e^{2x} + \sin(2x))$ .

(Click for Solution)

Solution. By the linearity of the inverse- $\mathcal D$ operator,

$\begin{aligned}\frac{1}{\mathcal D^2 + 2}(1 - e^{2x} + \sin(2x)) &= \frac{1}{\mathcal D^2 + 2}(1) - \frac{1}{\mathcal D^2 + 2}(e^{2x}) + \frac{1}{\mathcal D^2 + 2}(\sin(2x)) \\ &= \frac{1}{0^2 + 2} \cdot 1 - \frac{1}{2^2 + 2} \cdot e^{2x} + \frac{1}{-2^2 + 2} \cdot \sin(2x) \\ &= \frac{1}{2} - \frac{1}{6} \cdot e^{2x} - \frac{1}{2} \cdot \sin(2x). \end{aligned}$

Question 2. Solve the differential equation

$\displaystyle y'' - 4y' + 4y = e^{2x} + 4.$

(Click for Solution)

Solution. Rewriting in $\mathcal D$ -notation,

$(\mathcal D^2 - 4 \mathcal D + 4)y = e^{2x} + 4.$

The general solution is given by $y = y_{\mathrm P} + y_{\mathrm C}$ . To obtain the complementary function $y_{\mathrm C}$ , we solve the equation

$(\mathcal D^2 - 4 \mathcal D + 4)y = 0.$

Since the characteristic equation $m^2 - 4m + 4 = 0$ has real and repeated auxiliary roots $m = 2$ , the general solution is given by

$y_{\mathrm C} = (C_1 x + C_2) e^{2x}.$

To evaluate $y_{\mathrm P}$ , we first use linearity to compute

$\begin{aligned} y_{\mathrm P} &= \frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(e^{2x} + 4) \\ &= \underbrace{\frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(e^{2x})}_{y_1} + \underbrace{\frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(4)}_{y_2}. \end{aligned}$

To shorten presentation, the general solution is $y = y_1 + y_2 + y_{\mathrm C}$ . For the simpler term $y_2$ ,

$\displaystyle y_2 = \frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(4) = \frac 1{0^2 + 4 \cdot 0 + 4} \cdot 4 = 1.$

For the challenging term $y_1$ , we first replace each $\mathcal D$ with $\mathcal D + 2$ to obtain

$\begin{aligned} (\mathcal D + 2)^2 - 4 (\mathcal D + 2) + 4 &= (\mathcal D^2 + 4\mathcal D + 4) - (4\mathcal D + 8) + 4 = \mathcal D^2. \end{aligned}$

Therefore,

$\begin{aligned} y_1 &= \frac{1}{\mathcal D^2 - 4 \mathcal D + 4}(e^{2x}) \\ &= e^{2x} \cdot \frac{1}{\mathcal D^2} (1) \\ &= e^{2x} \cdot \frac 1{\mathcal D}\left(\frac 1{\mathcal D} (1)\right) \\ &= e^{2x} \cdot \frac 1{\mathcal D}(x) = e^{2x} \cdot \frac{x^2}{2}. \end{aligned}$

Consolidating all results,

$y = y_1 + y_2 + y_{\mathrm C} = (\frac{1}{2} x^2 e^{2x} + 1) + (C_1 x + C_2) e^{2x}.$

Question 3. Solve the differential equation

$\displaystyle y'' + 4y' + 4y = e^{-2x} + e^{2x}.$

(Click for Solution)

Solution. Rewriting in $\mathcal D$ -notation,

$(\mathcal D^2 + 4 \mathcal D + 4)y = e^{-2x} + e^{2x}.$

The general solution is given by $y = y_{\mathrm P} + y_{\mathrm C}$ . To obtain the complementary function $y_{\mathrm C}$ , we solve the equation

$(\mathcal D^2 + 4 \mathcal D + 4)y = 0.$

Since the characteristic equation $m^2 + 4m + 4 = 0$ has real and repeated auxiliary roots $m = -2$ , the general solution is given by

$y_{\mathrm C} = (C_1 x + C_2) e^{-2x}.$

To evaluate $y_{\mathrm P}$ , we first use linearity to compute

$\begin{aligned} y_{\mathrm P} &= \frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(e^{-2x} + e^{2x}) \\ &= \underbrace{\frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(e^{-2x})}_{y_1} + \underbrace{\frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(e^{2x})}_{y_2}. \end{aligned}$

To shorten presentation, the general solution is $y = y_1 + y_2 + y_{\mathrm C}$ . For the simpler term $y_2$ ,

$\displaystyle y_2 = \frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(e^{2x}) = \frac 1{2^2 + 4 \cdot 2 + 4} \cdot e^{2x} = \frac{1}{16} e^{2x}.$

For the challenging term $y_1$ , we first replace each $\mathcal D$ with $\mathcal D - 2$ to obtain

$\begin{aligned} (\mathcal D - 2)^2 + 4 (\mathcal D - 2) + 4 &= (\mathcal D^2 - 4\mathcal D + 4) + (4\mathcal D - 8) + 4 = \mathcal D^2. \end{aligned}$

Therefore,

$\begin{aligned} y_1 &= \frac{1}{\mathcal D^2 + 4 \mathcal D + 4}(e^{-2x}) \\ &= e^{-2x} \cdot \frac{1}{\mathcal D^2} (1) \\ &= e^{-2x} \cdot \frac 1{\mathcal D}\left(\frac 1{\mathcal D} (1)\right) \\ &= e^{-2x} \cdot \frac 1{\mathcal D}(x) = e^{-2x} \cdot \frac{x^2}{2}. \end{aligned}$

Consolidating all results,

$y = y_1 + y_2 + y_{\mathrm C} = \frac{1}{2} x^2 e^{-2x} + \frac{1}{16} e^{2x} + (C_1 x + C_2) e^{-2x}.$

—Joel Kindiak, 18 Apr 25, 1440H

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