Method of Undetermined Coefficients

In the previous section, we have seen than a general solution of

$$L_n(y) = f (x) \;\; {\mbox{ on }} \;\; I$$ (8.7.6)

can be written in the form

$$y = y_h + y_p,$$

where $y_h$ is a general solution of $L_n(y) = 0$ and $y_p$ is a particular solution of (8.7.6). In view of this, in this section, we shall attempt to obtain $y_p$ for (8.7.6) using the method of undetermined coefficients in the following particular cases of $f(x);$

1. $f(x) = k e^{{\alpha}x}; \;\; k \neq 0, {\alpha}\; {\mbox{ a real constant}}$
2. $f(x) = e^{{\alpha}x}\bigl( k_1 \cos (\beta x) + k_2 \sin (\beta x) \bigr); \;\; k_1, k_2, {\alpha}, \beta \in {\mathbb{R}}$
3. $f(x) = x^m.$

Case I. $f(x) = k e^{{\alpha}x}; \;\; k \neq 0, {\alpha}\; {\mbox{ a real constant.}}$
We first assume that ${\alpha}$ is not a root of the characteristic equation, i.e., $p({\alpha}) \neq 0.$ Note that $L_n(e^{{\alpha}x}) = p({\alpha}) e^{{\alpha}x}.$ Therefore, let us assume that a particular solution is of the form

$$y_p = A e^{{\alpha}x},$$

where $A,$ an unknown, is an undetermined coefficient. Thus

$$L_n(y_p) = A p({\alpha}) e^{{\alpha}x}.$$

Since $p({\alpha}) \neq 0,$ we can choose $A = \displaystyle\frac{k}{p({\alpha})}$ to obtain

$$L_n(y_p) = k e^{{\alpha}x}.$$

Thus, $y_p = \displaystyle\frac{k}{p({\alpha})} e^{{\alpha}x}$ is a particular solution of $L_n(y) = k e^{{\alpha}x}.$

Modification Rule: If ${\alpha}$ is a root of the characteristic equation, i.e., $p({\alpha}) = 0,$ with multiplicity $r,$ (i.e., $p({\alpha}) = p^\prime({\alpha}) = \cdots = p^{(r-1)}({\alpha}) = 0$ and $p^{(r)}({\alpha}) \neq 0$ ) then we take, $y_p$ of the form

$$y_p = A x^r e^{{\alpha}x}$$

and obtain the value of $A$ by substituting $y_p$ in $L_n(y) = k e^{{\alpha}x}.$

EXAMPLE 8.7.1  

1. Find a particular solution of
   $$y^{\prime\prime} - 4 y = 2 e^{x}.$$
   
   Solution: Here $f(x) = 2 e^x$ with $k = 2$ and ${\alpha}= 1.$ Also, the characteristic polynomial, $p({\lambda}) = {\lambda}^2 - 4.$ Note that ${\alpha}= 1$ is not a root of $p({\lambda}) = 0.$ Thus, we assume $y_p = A e^{x}.$ This on substitution gives
   $$A e^x - 4 A e^x = 2 e^x \;\; \Longrightarrow - 3 A e^x = 2 e^x.$$
   So, we choose $A = \displaystyle\frac{-2}{3},$ which gives a particular solution as
   $$y_p = \frac{-2 e^{x}}{3}.$$

2. Find a particular solution of
   $$y^{\prime\prime\prime} -3 y^{\prime\prime} + 3 y^\prime - y = 2 e^{x}.$$
   
   Solution: The characteristic polynomial is $p({\lambda}) = {\lambda}^3 - 3{\lambda}^2 + 3 {\lambda}- 1 = ({\lambda}- 1)^3$ and ${\alpha}= 1.$ Clearly, $p(1) = 0$ and ${\lambda}={\alpha}=1$ has multiplicity $r = 3.$ Thus, we assume $y_p = A x^3 e^{x}.$ Substituting it in the given equation,we have
   

   $$A e^x \left( x^3 + 9 x^2 + 18 x + 6 \right) - 3 A e^x \left( x^3 + 6 x^2 + 6 x\right)$$

   $$+ 3 A e^x \left( x^3 + 3 x^2 \right) - A x^3 e^x = 2 e^x.$$

   
   Solving for $A,$ we get $A = \displaystyle\frac{1}{3},$ and thus a particular solution is $y_p = \displaystyle \frac{x^3 e^{x}}{3}.$
3. Find a particular solution of
   $$y^{\prime\prime\prime} - y^\prime = e^{2x}.$$
   
   Solution: The characteristic polynomial is $p({\lambda}) = {\lambda}^3 - {\lambda}$ and ${\alpha}= 2.$ Thus, using $y_p = A e^{ 2 x},$ we get $A = \displaystyle\frac{1}{p({\alpha})} = \frac{1}{6},$ and hence a particular solution is $y_p = \displaystyle\frac{e^{2x}}{6}.$
4. Solve $y^{\prime\prime\prime} -3 y^{\prime\prime} + 3 y^\prime - y = 2 e^{2x}.$

EXERCISE 8.7.2   Find a particular solution for the following differential equations:

1. $y^{\prime\prime} - 3 y^{\prime} + 2 y = e^{x}.$
2. $y^{\prime\prime} - 9 y = e^{3 x}.$
3. $y^{\prime\prime\prime} -3 y^{\prime\prime} + 6 y^\prime - 4 y = e^{2x}.$

Case II. $f(x) = e^{{\alpha}x}\bigl( k_1 \cos (\beta x) + k_2 \sin (\beta x) \bigr); \;\; k_1, k_2, {\alpha}, \beta \in {\mathbb{R}}$
We first assume that ${\alpha}+ i \beta$ is not a root of the characteristic equation, i.e., $p({\alpha}+ i \beta) \neq 0.$ Here, we assume that $y_p$ is of the form

$$y_p = e^{{\alpha}x}\bigl( A \cos (\beta x) + B \sin (\beta x) \bigr),$$

and then comparing the coefficients of $e^{{\alpha}x} \cos x$ and $e^{{\alpha}x} \sin x$ (why!) in $L_n(y) = f(x),$ obtain the values of $A$ and $B.$

Modification Rule: If ${\alpha}+ i \beta$ is a root of the characteristic equation, i.e., $p({\alpha}+ i \beta) = 0,$ with multiplicity $r,$ then we assume a particular solution as

$$y_p = x^r e^{{\alpha}x}\bigl( A \cos (\beta x) + B \sin (\beta x) \bigr),$$

and then comparing the coefficients in $L_n(y) = f(x),$ obtain the values of $A$ and $B.$

EXAMPLE 8.7.3  

1. Find a particular solution of
   $$y^{\prime\prime} + 2 y^\prime + 2 y = 4 e^{x}\sin x.$$
   
   Solution: Here, ${\alpha}= 1$ and $\beta = 1.$ Thus ${\alpha}+ i \beta = 1 + i,$ which is not a root of the characteristic equation $p({\lambda}) = {\lambda}^2 + 2 {\lambda}+ 2=0.$ Note that the roots of $p({\lambda}) = 0$ are $-1 \pm i.$

   Thus, let us assume $y_p = e^{x}\left( A \sin x + B \cos x\right).$ This gives us

   $$(-4 B + 4 A) e^x \sin x + (4 B + 4 A) e^x \cos x = 4 e^x \sin x.$$
   Comparing the coefficients of $e^x \cos x$ and $e^x \sin x$ on both sides, we get $A - B = 1$ and $A + B = 0.$ On solving for $A$ and $B,$ we get $A = -B = \displaystyle\frac{1}{2}.$ So, a particular solution is $y_p = \displaystyle \frac{e^{x}}{2}\left( \sin x - \cos x \right).$
2. Find a particular solution of
   $$y^{\prime\prime} + y = \sin x.$$
   
   Solution: Here, ${\alpha}= 0$ and $\beta = 1.$ Thus ${\alpha}+ i \beta = i,$ which is a root with multiplicity $r = 1,$ of the characteristic equation $p({\lambda}) = {\lambda}^2 + 1=0.$

   So, let $y_p = {x}\left( A \cos x + B \sin x\right).$ Substituting this in the given equation and comparing the coefficients of $\cos x$ and $\sin x$ on both sides, we get $B = 0$ and $A = -\displaystyle\frac{1}{2}.$ Thus, a particular solution is $y_p = \displaystyle \frac{-1}{2} x \cos x.$

EXERCISE 8.7.4   Find a particular solution for the following differential equations:

1. $y^{\prime\prime\prime} - y^{\prime\prime} + y^\prime - y = e^{x}\cos x.$
2. $y^{\prime\prime\prime\prime} + 2 y^{\prime\prime} + y = \sin x.$
3. $y^{\prime\prime} - 2 y^{\prime} + 2 y = e^{x} \cos x.$

Case III. $f(x) = x^m.$
Suppose $p(0) \neq 0.$ Then we assume that

$$y_p = A_m x^m + A_{m-1} x^{m-1} + \cdots + A_0$$

and then compare the coefficient of $x^k$ in $L_n(y_p) = f(x)$ to obtain the values of $A_i$ for $0 \leq i \leq m.$

Modification Rule: If ${\lambda}= 0$ is a root of the characteristic equation, i.e., $p(0) = 0,$ with multiplicity $r,$ then we assume a particular solution as

$$y_p = x^r \left( A_m x^m + A_{m-1} x^{m-1} + \cdots + A_0\right)$$

and then compare the coefficient of $x^k$ in $L_n(y_p) = f(x)$ to obtain the values of $A_i$ for $0 \leq i \leq m.$

EXAMPLE 8.7.5   Find a particular solution of

$$y^{\prime\prime\prime} - y^{\prime\prime} + y^\prime - y = x^2.$$

Solution: As $p(0) \neq 0,$ we assume

$$y_p = A_2 x^2+ A_1 x + A_0$$

which on substitution in the given differential equation gives

$$- 2 A_2 + (2 A_2 x + A_1) - ( A_2 x^2 + A_1 x + A_0) = x^2.$$

Comparing the coefficients of different powers of $x$ and solving, we get $A_2 = -1, \; A_1 = -2$ and $A_0 = 0.$ Thus, a particular solution is

$$y_p = - ( x^2 + 2 x).$$

Finally, note that if $y_{p_1}$ is a particular solution of $L_n(y)= f_1(x)$ and $y_{p_2}$ is a particular solution of $L_n(y)= f_2(x),$ then a particular solution of

$$L_n(y) = k_1 f_1(x) + k_2 f_2(x)$$

is given by

$$y_p = k_1 y_{p_1} + k_2 y_{p_2}.$$

In view of this, one can use method of undetermined coefficients for the cases, where $f(x)$ is a linear combination of the functions described above.

EXAMPLE 8.7.6   Find a particular soltution of

$$y^{\prime\prime} + y = 2 \sin x + \sin 2x.$$

Solution: We can divide the problem into two problems:

1. $y^{\prime\prime} + y = 2 \sin x.$
2. $y^{\prime\prime} + y = \sin 2x.$

For the first problem, a particular solution (Example 8.7.3.2) is $y_{p_1} = 2 \; \displaystyle \frac{-1}{2} x \cos x = - x \cos x.$

For the second problem, one can check that $y_{p_2} = \displaystyle \frac{-1}{3} \sin (2 x)$ is a particular solution.

Thus, a particular solution of the given problem is

$$y_{p_1} + y_{p_2} = - x \cos x - \frac{1}{3} \sin (2 x).$$

EXERCISE 8.7.7   Find a particular solution for the following differential equations:

1. $y^{\prime\prime\prime} - y^{\prime\prime} + y^\prime - y = 5 e^{x}\cos x+ 10 e^{2x}.$
2. $y^{\prime\prime} + 2 y^\prime + y = x + e^{-x}.$
3. $y^{\prime\prime} + 3 y^\prime - 4 y = 4 e^{x} + e^{4x}.$
4. $y^{\prime\prime} + 9 y = \cos x + x^2 + x^3.$
5. $y^{\prime\prime\prime} - 3 y^{\prime\prime} + 4 y^\prime = x^2 + e^{2x} \sin x.$
6. $y^{\prime\prime\prime\prime} + 4 y^{\prime\prime\prime} + 6 y^{\prime\prime} + 4 y^\prime+ 5 y = 2 \sin x + x^2.$