Question

(a) By inspection find a particular solution of $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{10}$

(b) By inspection find a particular solution of$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{x}$

(c) Find a particular solution of$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{x}\mathbf{+}\mathbf{10}$

(d) Find a particular solution of$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{8}\mathit{x}\mathbf{+}\mathbf{5}$

Short answer

(a)$\text{y = 5}$

(b)$y=-2x$

(c)$y=-2x+5$

(d)$y=4x+\frac{5}{2}$

Step-by-step solution

Step 1(a): Derive the equation and find the value of y for y''+2y=10

The simplest method for finding a particular solution is to try a solution of the same form as the right hand side of the non homogeneous equation, i.e. if it's a polynomial of degree n, we try a general polynomial of degree n and see if we can match the coefficients.

Since, the right hand side of the given equation is a constant, let us consider the constant function as the particular solution of the given equation. So, we have $y\text{'}=y\text{'}\text{'}=0$. Substituting these in the equation gives

$\begin{array}{rcl}y\text{'}\text{'}+2y& =& 10\\ 0+2c& =& 10\\ c& =& 5\end{array}$

Hence, by inspection y = 5 is the particular solution.

 Step 2(b): Derive the equation and find the value of y for y''+2y=-4x

The simplest method for finding a particular solution is to try a solution of the same

Now, for this exercise, since, the right hand side of the given equation is a polynomial of degree one, let us consider the function y = Cx as the particular solution of the given equation. So, we have $y\text{'}=c,y\text{'}\text{'}=0$. Substituting these in the equation gives

$\begin{array}{rcl}y\text{'}\text{'}+2y& =& -4x\\ 0+2cx& =& -4x\\ c& =& -2\end{array}$

Hence, by inspection, y = -2x is the particular solution.

Step 3(c): Derive the equation and find the particular solution of y for y''+2y=-4x+10

From part (a), we saw that i$y_{p_1}=5$s the particular solution of

$y\text{'}\text{'}+2y=10$

And from part (b), $y_{p_2}=-2x$is the particular solution of

$y\text{'}\text{'}+2y=-4x$

Therefore, it follows from (13) of Theorem 4.1.7 that the solution of

$y\text{'}\text{'}+2y=-\underset{g_2\left(x\right)}{4x}+\underset{g_1\left(x\right)}{10}$

Is the superposition of $y_{p_2}and{y}_{p1}$, i.e.

$\begin{array}{rcl}y& =& y_{p_2}+y_{p_1}\\ & =& -2x+5\end{array}$

Hence, by inspection, y = -2x + 5 is the particular solution.

Step 4(d): Derive the equation and find the particular solution of y for y''+2y=8x+5

Similarly, the given equation can be written as

$$\begin{gathered} y\text{'}\text{'}+2y=8x+5 \\ =-2(-4x)+\left(\frac{1}{2}\right)\left(10\right) \\ =-2(\underset{g_{2\left(x\right)}}{-4x)}+\left(\frac{1}{2}\right)\underset{g_{1\left(x\right)}}{\left(10\right)} \end{gathered}$$

From part (a), we saw that$y_{p_1}=5$ is the particular solution of

$y\text{'}\text{'}+2y=10$

And from part (b)$y_{p_2}=-2x$ is the particular solution of

$y\text{'}\text{'}+2y=-4x$

Therefore, it follows from (13) of Theorem 4.1.7 that the solution of

$y\text{'}\text{'}+2y=-2(\underset{g_{2\left(x\right)}}{-4x)}+\left(\frac{1}{2}\right)\underset{g_{1\left(x\right)}}{\left(10\right)}$

Is the superposition of $-2y_{p_2}and\frac{1}{2}{y}_{p_1}$, i.e.

$\begin{array}{rcl}y& =& -2y_{p_2}+\frac{1}{2}{y}_{p_1}\\ & =& -2(-2x)+\left(\frac{1}{2}\right)\left(5\right)\\ & =& 4x+\frac{5}{2}\end{array}$

Hence, by inspection, $y=4x+\frac{5}{2}$ is the particular solution.