Recap from last class#

For the 2nd order, linear, constant coefficients, homogeneous differential equation $y'' + ay' + by = 0$ where \(a^2-4b \[\begin y(x) &= e^x>[A\sin(\omega x) + B\cos(\omega x)] && \omega = \sqrta^2> \end\]

where $A$ and $B$ are our arbitrary constants. All components of the solution are real.

Example#

$y’’ + 2y’ + 3y = 0 $ ; $ \hspacey(0) = 2 $ and $ y’(0) = -3$\

(249)# \[\begin \lambda^2 + 2\lambda + 3 = 0\\ a = 2, \hspace b = 3 \end\]

Since $a^2 - 4b = -8$ , we have complex roots

(250)# \[\begin \lambda &= -\fraca \pm i\sqrt\\ &= -\fraca \pm i \omega \implies \omega = \sqrt\\ \lambda &= -1 \pm i\sqrt \end\]

Great, so we know our general solution is:

To find $A$ and $B$ , we will also need an expression for $y'(x)$

(252)# \[\begin y'(x) &= -e^[A\cos(\sqrtx) + B\sin(\sqrtx)] + + e^[-\sqrt A\sin(\sqrtx) + \sqrt B\cos(\sqrtx)]\\ &= e^[(\sqrt B - A) \cos(\sqrtx) - (\sqrtA + B)\sin(\sqrtx)] \end\]

Using Initial Conditions:
(253)# \[\begin y(0) = 2 &= e^[A\cos(0) + B\sin(0)]\\ 2&=A \end\] (255)# \[\begin \therefore y(x) = e^[2 \cos(\sqrtx) -\frac<\sqrt>\sin(\sqrtx)] \implies \text \end\]
$\underline>$ : Homogeneous, linear $2^\circ$ ODEs with constant coefficients
(256)# \[\begin y'' + ay' + by = 0 \end\]

Three types of solutions, arising from three cases for the characteristic equation $\lambda^2 + a\lambda + b = 0$

Two real and different roots

(257)# \[\begin y(x) = c_1 ^ <\lambda_1x>+ c_2 e^ <\lambda_2x>&& \text \lambda = \frac(-a \pm \sqrt) \end\]

Real, double roots

(258)# \[\begin y(x) = c_1 e^ <\lambda x>+ c_2 x e^ <\lambda x>\end\]

Complex roots

In each case, we have a basis of $\underline>$ linearly independent solutions (for $2^$ order). $\rightarrow$ in case there’s just one, this necessitates that $\lambda_1 \neq \lambda_2$ , which is the definition of the case

Non-homogeneous Linear $2^\circ$ ODEs (sec 2.8)#
(260)# \[\begin y'' + p(x) y' + q(x) y = r(x) \end\]

We build upon our knowledge of solving the homogenous case to solve non-homogeneous case.
$\underline>$ :

(261)# \[\begin y(x) = y_H(x) + y_P(x) \end\]
where $y_H(x) = c_1y_1(x) + c_2y_2(x)$ , which is the “homogeneous solution” for $r(x)\equiv 0$ and $y_P(x)$ is the “particular solution” which accounts for non-homogeneous part and has no arbitrary constants (we need only two for a $2^\circ$ ODE)

$\underline>$ : $c_1$ and $c_2$ must be found using $y(x) = y_H(x) + y_P(x)$ , not just $y_H(x)$

We will do an example first and then describe the general method to find $y_P(x)$

Example#

First, solve the homogeneous case:

(262)# \[\begin y'' + 4y = 0 \implies \text$> \end\]
(263)# \[\begin \lambda ^ 2 + 4 = 0 \implies \lambda = \pm 2i\\ a = 0, b = 4 \implies \omega = \sqrta^2> = 2 \end\]

$\implies$ for imaginary roots, we know the solution is:

Now, I will tell you that

(265)# \[\begin y_P(x) &= \frace^x\\ \therefore y(x) &= y_H(x) + \frace^x\\ y'(x) &= y_H'(x) + \frace^x\\ y''(x) &= y_H''(x) + \frace^x \end\]

Plugging these into original non-homogeneous equation:

(266)# \[\begin y_H'' + \frace^x + 4(y_H + \frace^x) = e^x\\ y_H'' + \frace^x + 4y_H + \frace^x = e^x\\ y_H'' + 4y_H + e^x = e^x && \implies \text y_P(x) = \frace^x \text \end\]

Now we need the particular solution to the general solution of non-homogeneous equation using initial conditions. 1.

(267)# \[\begin y(0) = \frac &= c_1 \sin(2\cdot 0) + c_2 \cos (2\cdot 0) + \frac e^0\\ \frac &= c_2 + \frac\\ c_2 &= \frac \end\]

We need expression for $y'(x)$ for $2^$ I.C.

(268)# \[\begin y'(x) &= 2c_1\cos(2x) - 2c_2 \sin(2x) + \frace^x\\ y'(0) = \frac &= 2c_1 \cos(2\cdot 0) - 2c_2 \sin(2\cdot 0) + \frace^0\\ \frac &= 2c_1 + \frac\\ c_1 &= \frac \end\]
(269)# \[\begin \therefore y(x) = \frac \sin(2x) + \frac \cos(2x) + \frac e^x \implies \text \end\]
Method of Undetermined Coefficients (sec 2.9)#
(270)# \[\begin y'' + ay' + by = r(x) \end\]

If $r(x)$ is a function s.t. $r'(x)$ is like $r(x)$ , e.g. $e^<\lambda x>$ , $\sin(\omega x)$ , $\cos(\omega x)$ , $x^n$ … we can choose $y_P(x)$ based on $r(x)$ \

Rules (from textbook)#

$K_nx^n + K_x^ + . + K_1x + K_0$

$k\cos(\omega x)$ or $k\sin(\omega x)$

$K\cos(\omega x) + M\sin(\omega x)$

Here are some additional rules; we’ll see why these are important later:

Basic Rule. If $r(x)$ is one of the functions in the first column in Table 2.1, choose in the same line and determine its undetermined coefficients by substituting $y_p$ and its derivatives into the differential equation.

Modification Rule. If a term in your choice for $y_p$ happens to be a solution of the homogeneous ODE corresponding to (4), multiply this term by x (or by $x^2$ if this solution corresponds to a double root of the characteristic equation of the homogeneous ODE).

Sum Rule. If is a sum of functions in the first column of Table 2.1, choose for the sum of the functions in the corresponding lines of the second column.

Given $y'' + ay' + by = r(x) \hspace (*)$

Solve $y'' + ay' + by = 0$ to get $y_H(x) = c_1y_1 + c_2y_2$

Then choose $y_P(x)$ based on $r(x)$

(271)# \[\begin y(x) = c_1y_1 + c_2y_2 + y_P\\ y'(x) = c_1y_1' + c_2y_2' + y_p'\\ y''(x) = c_1y_1'' + c_2y_2'' + y_p'' \end\]

Plug back into $(*)$ to check:

(272)# \[\begin (c_1y_1'' + c_2y_2'' + y_P'') + a(c_1y_1' + c_2y_2' + y_P') + b(c_1y_1 + c_2y_2 + y_P) = r(x)\\ c_1(y_1'' + ay_1' + by_1) + c_2(y_2'' + ay_2' + by_2) + y_P'' + ay_P' + by_P = r(x) \end\]

Therefore, $y_P'' + ay_P' + by_P = r(x)$ and $y_P$ is a linear solution to the problem (which is why only some functions work)

Example: $y'' + 2y' - 3y = 4e^$ #
(273)# \[\begin y(x) = y_H(x) + y_P(x) \end\]

Find $y_H(x)$

(275)# \[\begin \therefore y_H(x) = c_1e^x + c_2 e^ \end\]

Choose $y_P(x)$ based on $r(x)$
$\implies$ Since $r(x) = 4e^$ , let’s pick $y_p(x) = Ke^$ , where $K$ is the undetermined coefficient (not arbitrary)

Find $K$ from non-homogeneous ODE

(276)# \[\begin y_P'' + 2y_p' - 3y_p = 4e^\\ y_p' = 2Ke^\\ y_P'' = 4Ke^ \end\] (277)# \[\begin \implies 4Ke^ + 4Ke^ - 3Ke^ = 4e^\\ 5K = 4\\ K = \frac \end\]

Write down general solution

(278)# \[\begin y(x) = y_H(x) + y_P(x)\\ y(x) = c_1e^x + c_2e^ + \frace^ \end\]
Still need 2 IC’s to find $c_1$ and $c_2$

In-class problem#

Consider the differential equation $ $y''+y=0.001x^2$ $ The homogeneous solution is $ $y_h=A\cos x+B\sin x$ $, as we saw in class last week. Find a particular solution to this differential equation using the MoUC rules above.

Non-homogeneous Linear $2^\circ$ ODEs Continued#

$\underline>$ : If you choose a $y_P(x)$ that is not linearly independent of $y_1$ and $y_2$ , things will go wrong.

Example: $y'' - 4y = 4e^$ #

Solve $y'' - 4y = 0$

Choose $y_P(x)$ based on $r(x)$
Since $r(x) = 4e^$ , choose $y_P(x) = Ke^$ as before.
SPOILER: $Ke^$ is not linearly independent of $y_1(x) = e^$

Find $K$ from non-homogeneous ODE

(280)# \[\begin y_P'' - 4y_P = 4e^\\ y_P' = 2Ke^\\ y_P'' = 4Ke^\\ \implies 4Ke^ - 4Ke^ = 4e^\\ 0 = 4e^ \hspace ?? \end\]

If $y(p)$ isn’t linearly independent, it can’t be absorbed into $y_H(x)$ .

(281)# \[\begin y(x) &= c_1e^ + c_2e^ + Ke^\\ &= (c_1 + K)e^ + c_2 e^ \implies \text \end\]

Solution: If $y_P(x)$ is linearly dependent on $y_1(x)$ or $y_2(x)$ , multiply $y_P(x)$ by $x$ to obtain a linearly independent new $y_P(x)$ based on Reduction of Order.
Therefore, choose $y_P(x) = Kxe^$ . Then,

(282)# \[\begin y_P'(x) &= Ke^ + 2Kxe^\\ y_P''(x) &= 2Ke^ + 2Ke^ + 4Kxe^\\ &= 4Ke^ + 4Kxe^ \end\]

Find $K$

(283)# \[\begin y_P'' - 4y_P = 4e^\\ 4Ke^ + 4Kxe^ - 4Kxe^ = 4e^\\ 4K = 4\\ K=1\\ \therefore y_P(x) = xe^ \end\]

General Solution

(284)# \[\begin y(x) &= y_H(x) + y_P(x)\\ y(x) &= c_1e^ + c_2e^ + xe^ \end\]

Another place to be careful:
$\implies$ Double roots yields a homogeneous solution

(285)# \[\begin y_H(x) = c_1e^ <\lambda x>+ c_2xe^ <\lambda x>\end\]
$\implies$ Again, you must be careful to select $y_P(x)$ s.t. it is linearly independent.
$\implies$ if $r(x) \propto e^ <\lambda x>$ or $xe^<\lambda x>$ , then $y_P(x) = Kx^2e^<\lambda x>$ (Obtained from a second reduction of order)

More complex example: $y'' + 2y' -3y = -3x^2 + \sin(x)$ #

Solve the homogeneous ODE

(286)# \[\begin y'' + 2y' - 3y = 0\\ \lambda^2 + 2\lambda -3 = 0\\ (\lambda+3)(\lambda-1) = 0 && \implies \lambda_1 = 1, \hspace \lambda_2 = -3\\ y_H(x) = c_1e^x + c_2e^ \end\]

Choose $y_P(x)$ based on $r(x)$
$\underline>$ : If $r(X)$ is a sum of wo functions, then $y_P(x)$ should be a sum of the corresponding functions.

(287)# \[\begin r(x) &= -3x^2 + \sin(x)\\ \therefore y_P(x) &= Ax^2 + Bx + C + K\sin(x) + M\cos(x) \end\]

Find undetermined coefficients from ODE

(288)# \[\begin y_P' &= 2Ax + B + K\cos(x) - M\sin(x)\\ y_P'' &= 2A - K\sin(x) - M\cos(x) \end\]

Plugging into ODE:

(289)# \[\begin 2A - K\sin(x) - M\cos(x) + 2[2Ax + B + K\cos(x) - M\sin(x)] - 3[Ax^2 + Bx + C + K\sin(x) + M\cos(x)] = -3x^2 + \sin(x) \end\]

Rearrange to collect ‘like’ terms

(290)# \[\begin -3Ax^2 + (4A - 3B)x + (2A + 2B - 3C) + (-K -2M -3K)\sin(x) + (-M + 2K -3M)\cos(x) = -3x^2 + \sin(x) \end\]

Determine the values of the coefficients on each $f(x)$ that will solve the equation

$-3Ax^2 = -3x^2 \implies A = 1$

$(4A-3B)x - 0$ because $r(x)$ has no ‘ $x$ ’ term . Hence, $4-3B = 0 \implies B = \frac$

$2A + 2B - 3C = 0 \implies 2+\frac- 3C = 0 \implies C=\frac$

$-K -2M -3K = 1 \implies 2M = -4K -1 \implies M = -2K -\frac$

(291)# \[\begin -M + 2K -3M = 0\\ 2K = 4M\\ K = 2(-2K-\frac) = -4K-1\\ \implies K = -\frac\\ \implies M = -\frac \end\]
(292)# \[\begin \therefore y_P(x) = -3x^2 + \fracx + \frac-\frac(2\sin x + \cos x) \end\]

Write down the general solution

(293)# \[\begin y(x) &= y_H(x) + y_P(x)\\ &= c_1e^x + c_2e^ + y_P(x) \end\]
Second Order Differential Equations

Method of Undetermined Coefficient and Variations of Parameters

By Zachary Ulissi
© Copyright 2022.

Recap from last class#

Example#

Non-homogeneous Linear \(2^\circ\) ODEs (sec 2.8)#

Example#

Method of Undetermined Coefficients (sec 2.9)#

Rules (from textbook)#

Example: \(y'' + 2y' - 3y = 4e^\) #

In-class problem#

Non-homogeneous Linear \(2^\circ\) ODEs Continued#

Example: \(y'' - 4y = 4e^\) #

More complex example: \(y'' + 2y' -3y = -3x^2 + \sin(x)\) #