91Ó°ÊÓ

A particular solution and a fundamental solution set are given for a nonhomogeneous equation and its corresponding homogeneous equation.

(a) Find a general solution to the nonhomogeneous equation.

(b) Find the solution that satisfies the specified initial condition.

$$\begin{gathered} {\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{-}\mathbf{lnx}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{>}\mathbf{0}\mathbf{;} \\ \mathbf{y}\left(1\right)\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{;} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{lnx}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left\{x,  \mathrm{xlnx},  x{\left(\mathrm{lnx}\right)}^2\right\} \end{gathered}$$

use the annihilator method to determinethe form of a particular solution for the given equation.$u\text{'}\text{'}-5u\text{'}+6u=cos2x+1$

Solve the given initial value problem

_{$\begin{array}{l}y\text{'}\text{'}\text{'}−4y\text{'}\text{'}+7y\text{'}−6y=0\\ y\left(0\right)=1\\ y\text{'}\left(0\right)=0\\ y\text{'}\text{'}\left(0\right)=0\end{array}$}

Find a general solution for the given linear system using the elimination method of Section 5.2.

$\begin{array}{l}\frac{d^{2}x}{d{t}^2}−x+5y=0\\ 2x+\frac{d^{2}y}{d{t}^2}+2y=0\end{array}$

use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}+6y\text{'}+8y=e^{3x}-sinx$

Find a general solution for the given

linear system using the elimination method of Section 5.2.

$\begin{array}{l}\frac{d^{3}x}{d{t}^3}−x+\frac{dy}{dt}+y=0\\ \frac{dx}{dt}−x+y=0\end{array}$

use the annihilator method to determinethe form of a particular solution for the given equation$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{x}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}$

Let\({\bf{L}}\left[{\bf{y}}\right]{\bf{=y'''+y'+ xy,}}\,\,\,\,\,\,\,{{\bf{y}}_{\bf{1}}}\left({\bf{x}}\right){\bf{=sinx,}}\)and\({{\bf{y}}_{\bf{2}}}\left({\bf{x}}\right){\bf{=x}}\).Verifythat\({\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]\left( {\bf{x}} \right){\bf{=xsinx,}}\)and\({\bf{L}}\left[ {{{\bf{y}}_{\bf{2}}}} \right]\left( {\bf{x}} \right){\bf{ = }}{{\bf{x}}^{\bf{2}}}{\bf{ + 1}}\). Then use the superposition principle (linearity) to find a solution to the differential equation:

(a) \({\bf{L}}\left[ {\bf{y}} \right]{\bf{ = 2xsinx - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 1}}\)

(b) \({\bf{L}}\left[ {\bf{y}} \right]{\bf{ = 4}}{{\bf{x}}^{\bf{2}}}{\bf{ + 4 - 6xsinx}}\)

use the annihilator method to determinethe form of a particular solution for the given equation. $\mathit{θ}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{θ}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{x}}$

Let$P\left(r_1\right)=a_n{r}_1^n+a_{n−1}{r}_1^{n−1}+...+a_1{r}_1+a_0$ be a polynomialwith real coefficients$a_n,....,a_0$ . Prove that if r₁ is azero of$P\left(r\right)$ , then so is its complex conjugate r₁. [Hint:Show that$\stackrel{¯}{P\left(r\right)}=P\left(\stackrel{¯}{r}\right)$ , where the bar denotes complexconjugation.]