Differential Equations - Homogeneous Equations with Constant Coefficients

A real linear homogeneous differential equation of $n^\text{th}$ order with constant real coefficients is an equation in the form

$$a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0 \qquad (1)$$

with $a_{0}, a_{1} ,a_{n-1} ,\ldots , a_{n}$ real constants.

Observation:

$$\dfrac{d^ky}{dx^k} e^{rx} = r^k e^{rx}. \qquad (2)$$

Now, substituting $y = e^{rx}$ in $(1)$ and using $(2),$ we obtain

$$a_{n} r^n e^{rx} + a_{n-1} r^{n-1} e^{rx} + \cdots + a_{1} r e^{rx} + a_{0} e^{rx} = 0.$$

Since $e ^ {rx} \neq 0$, we can cancel out this term to obtain the polynomial

$$a_ {n}r^n + a_ {n-1}r^{n-1}+ \cdots + a_ {1} r + a_ {0} = 0. \qquad (3)$$

We see that $y = e^{rx}$ will be the solution of equation $(1)$ precisely when $r$ is a root of this polynomial.

This equation is called the characteristic equation or auxiliary equation of the differential equation $(1).$

Theorem 1 $$$$

If the $n$ roots $r_1, r_2, \ldots, r_{n}$ of the characteristic equation are real and distinct, then

$$y(x) = c_1 e^{r_1x} + c_2e^{r_2x} +\cdots +c_ne^{r_nx}$$

is a general solution of the equation $(1),$ with $c_1, c_2, \ldots, c_n$ constants. $_\square$

Solve $$$$

$$\begin{array}{c}&y'' + 2y' - 8y = 0, &y(0) = 5, &y'(0) = - 2.\end{array}$$

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We need to solve the characteristic equation $r^2 + 2r - 8 = 0.$

Observe that $r = 2, -4$ are the roots of this equation. Then since $y(x) = c_1 e^{2x} + c_2 e^{-4x}$ is the general solution, $y'(x) = 2 c_1 e^{2x} - 4 c_2 e^{-4x}$. Now, using the initial conditions, we have $$y(0) = c_1 + c_2 = 5, y'(0) = 2c_1 - 4c_2 = -2 \Rightarrow c_1 = 3, c_2=2.$$

Thereofore, the desired particular solution is $$y(x) = 3e^{2x} + 2e^{-4x}. \ _\square$$

If there are repeated real roots of the characteristic equation $(3)$, then we cannot generate $n$ linearly independent solutions to equation $(1)$ by the method of Theorem 1. Let us now consider the case that characteristic equation $$a_ {n}r^n + a_ {n-1}r^{n-1}+ \cdots + a_ {1} r + a_ {0} = 0 \qquad (3)$$ has repeated real roots.

Theorem 2 $$$$

If the characteristic equation $(3)$ has a repeated real root $r$ of multiplicity $k,$ then part of the general solution of the differential equation corresponding to $r$ in equation $(1)$ is of the form

$$(c_1 + c_ {2}x + c_ {3}x^2 + \cdots + c_ {k}x^{k-1}) \cdot e^{rx}. \ _\square$$

Find a general solution of $$$$

$$y^{(4)} + 3 \cdot y^{(3)} + 3 \cdot y'' + y' = 0.$$

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The characteristic equation of the differential equation is $r^4 + 3 \cdot r^3 + 3 \cdot r^2 + r= r \cdot (r+1)^3=0$.

It has the single root $r_1 = 0,$ which gives the solution $y_1 = c_1$ to the general solution, and the triple root $(k=3)$ $r_2 = -1,$ which gives $y_2 = (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}.$ Thus, the general solution of the differential equation is $$y(x) = c_1 + (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}. \ _\square$$

Because the coefficients of the differential equation and its characteristic equation are real, any root complex appears in complex conjugate pair $a \pm bi,$ where $a$ and $b$ are real and i = $\sqrt{-1}.$

Theorem 3: (No repeated complex roots) $$$$

If the characteristic equation $(3)$ has a pair of complex roots not repeated $a \pm bi$, then the relevant part to them of the general solution of equation $(1)$ has the form $e^{ax} \cdot (c_1 \cos bx + c_2 \sin bx). \ _\square$

Solve $y'' - 4y' + 5y = 0.$

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The characteristic equation is $r^2 - 4r + 5 = 0,$ whose roots are $2 \pm i.$ Thus, the general solution is $y(x) = e^{2x} \cdot (c_1 \cos x + c_2 \sin x). \ _\square$

Theorem 2 holds for the case of repeated complex roots. If the conjugate pair $a \pm bi$ have multiplicity $k,$ then the relevant part to them of the general solution has the form

$$\begin{aligned} &\left(A_1 + A_ {2}x + A_ {3}x^2 + \cdots + A_ {k}x^{k-1}\right) \cdot e^{(a+bi)x} + \left(B_1 + B_ {2}x + B_ {3}x^2 + \cdots + B_ {k}x^{k-1}\right) \cdot e^{(a-bi)x}\\ &= \sum_{p=0}^{k-1}x^{p} \cdot e^{ax} \cdot (c_{p} \cos bx + d_{p} \sin bx). \qquad (4) \end{aligned}$$

Find a general solution of $$$$

$$y^{(4)} + 4 \cdot y^{(3)} + 12 \cdot y'' + 16y' + 16y = 0.$$

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The characteristic equation is $(r^2 + 2r + 4)^2 = 0,$ whose roots are $-1 \pm i \cdot \sqrt{3}$ of multiplicity $2.$ Therefore, the equation $(4)$ gives the general solution

$$y(x) =e^{-x} \cdot \left(c_1 \cos x\sqrt{3} + d_1 \sin x\sqrt{3}\right) + xe^{-x} \cdot \left(c_2 \cos x\sqrt{3} + d_2 \sin x\sqrt{3}\right). \ _\square$$

$$$$ DIFFERENTIAL EQUATIONS - NO HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS

Let $y_p$ be a particular solution of the no homogeneous equation $$a_{n}(x) \dfrac{d^ny}{dx^n} + a_{n-1}(x) \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1}(x) \dfrac{dy}{dx} + a_{0}(x) y = f(x) \qquad (1)$$ in an open interval $I$ in which the functions $f(x)$ and $a_{i}(x)$ are continuous. Let $y_1, y_2, \ldots , y_n$ be linearly independent solutions of the associated differential homogeneous equation $$a_{n}(x) \dfrac{d^ny}{dx^n} + a_{n-1}(x) \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1}(x) \dfrac{dy}{dx} + a_{0}(x) y = 0. \qquad (*)$$ If $Y$ is a solution of equation $(1)$ in $I,$ then there exists constants $c_1, c_2, \ldots , c_n$ such that $$Y(x) = c_1 y_1(x) + c_2 y_2(x) + \cdots + c_n y_n(x) + y_p(x)$$ for all $x$ in $I.$ $_\square$

Sketch of Proof:
$Y$ solution of $(1)$
$\Rightarrow$ $Y - y_p$ solution of the associated differential equation $(*)$
$\Rightarrow$ there exist $c_1, c_2, \ldots , c_n$ such that $$Y(x) = c_1 y_1(x) + c_2 y_2(x) + \ldots + c_n y_n(x) + y_p(x)$$ for all $x$ in $I.$ $_\square$

Let now the (???) no homogeneous differential equation of $n^\text{th}$ order with constant coefficients $$a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = f(x), \qquad (1)$$ then the general solution of $(1)$ is $$Y = y_c + y_p,$$ where $y_c$ is a general solution of the associated homogeneous equation $$a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0, \qquad (2)$$ and $y_p$ is a simple particular solution of $(1).$

$$$$ UNDETERMINED COEFFICIENTS METHOD

The method of undetermined coefficients is applied when $f(x)$ according to equation $(1)$ is a finite linear combination of products of functions of three types:

1. A polynomial in $x.$
2. An exponential function.
3. $\cos kx$ or $\sin kx.$

Let's see some examples:

Find a particular solution of $$$$

$$y'' + 3y' + 4y = 3x +2.$$

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Here $f(x) = 3x +2.$ Let's seek $y_p(x) = Ax + B.$ Then $y'_p(x) = A$ and $y''_p(x) = 0$, so $y_p$ will satisfy the differential equation $3A + 4(Ax + B) = 3x +2$ if and only if $4A = 3$ and $3A +4B = 2,$ i.e. $A = \frac{3}{4}$ and $B = \frac {-1}{16}.$ Therefore, $$y_p(x) = \frac {3}{4}x - \frac {1}{16}. \ _\square$$

Find a particular solution of $$$$

$$y'' + 4y = 3x^3.$$

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The complementary function of this equation is $y_c(x) = c_1 \cdot \cos 2x + c_2 \cdot \sin 2x$. Here $f(x) = 3x^3$. Let's seek

$$\begin{aligned} y_p(x) &= Ax^3 + Bx^2 + Cx + D \\ y'_p(x) &= 3Ax^2+ 2Bx + C\\ y''_p(x) &= 6Ax + 2B. \end{aligned}$$

Substituting, we have

$$\begin{aligned} y''_p + 4y_p &= (6Ax + 2B) + 4\left(Ax^3 + Bx^2 + Cx + D\right) \\ &= 4Ax^3 + 4Bx^2 + (6A +4C)x + (2B + D) \\ &= 3x^3 \\ \\ \Rightarrow 4A &= 3, ~4B = 0 , ~6A + 4C = 0, ~2B + D = 0 \\ A &= \frac {3}{4}, ~B = 0, ~C = \frac{-9}{8}, ~D = 0\\ \\ \Rightarrow y_p(x) &= \frac{3}{4}x^3 - \frac{9}{8}x. \ _\square \end{aligned}$$

Solve $$$$

$$\begin{array}{c}&y'' - 3y' + 8y = 3e^{-x} - 10 \cos 3x, &y(0) = 1, &y'(0) = 2.\end{array}$$

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The characteristic equation $r^2 - 3r + 2 = 0$ has the roots $r=1$ and $r=2.$ Then the complementary function is $$y_c = c_1 e^{x} + c_2 e^{2x}.$$ Let's seek a particular solution $$\begin{aligned} y_p &= Ae^{-x} + B\cos 3x + C\sin 3x\\ y'_p &= -Ae^{-x} - 3B\sin 3x + 3C\cos 3x\\ y''_p &= Ae^{-x} - 9B\cos 3x - 9C\sin 3x. \end{aligned}$$ Substituing, we have $$\begin{aligned} y''_p - 3y'_p + 2y_p &=6Ae^{-x} + (-7B - 9C)\cos 3x + (9B - 7C)\sin 3x\\ &= 3e^{-x} - 10\cos 3x\\ \\ \Rightarrow 6A &= 3, ~-7B - 9C = -10, ~9B - 7C = 0 \\ A &= \frac{1}{2}, ~B = \frac{7}{13} , ~C = \frac{9}{13}\\ \\ \Rightarrow y_p(x) &= \frac{1}{2}e^{-x} + \frac{7}{13}\cos 3x + \frac{9}{13}\sin 3x. \end{aligned}$$ Then the general solution is $$y(x) = y_c(x) + y_p(x) = c_1e^{x} + c_2e^{2x} + \frac{1}{2}e^{-x} + \frac{7}{13}\cos 3x + \frac{9}{13}\sin 3x.$$ The initial conditions lead to $$\begin{aligned} y(0) &= c_1 + c_2 + \frac{1}{2} + \frac{7}{12} = 1\\ y'(0) &= c_1 + 2c_2 - \frac{1}{2} + \frac{27}{13} = 2 \\ \Rightarrow c_1 &= - \frac{1}{2}, ~c_2 = \frac{6}{13}. \end{aligned}$$ Therefore, the solution sought is $$y(x) = - \frac{1}{2}e^{x} + \frac{6}{13}e^{2x} + \frac{1}{2}e^{-x} + \frac{7}{13}\cos 3x + \frac{9}{13}\sin 3x. \ _\square$$

References

Editorial: PHH Ecuaciones diferenciales elementales y problemas con condiciones en la frontera authors: C.H Edwards, David Penney