Question

Solve the ordinary dierential equation

y'' + 2y' + 5y = 4e^-x + 17 sin 2x:

Answer 1

Given the differential equation

$$y'' + 2y' + 5y = 4e^{-x} + 17 \sin 2x$$

We can rewrite the equation as

$$\frac{d^{2} y(x)}{d x^{2}}+2 \frac{d y(x)}{d x}+5 y(x)=4 e^{-x}+17 \sin (2 x)$$

The general solution will be the sum of the complementary solution and particular solution. We thus find the complementary solution by solving

$$\frac{d^{2} y(x)}{d x^{2}}+2 \frac{d y(x)}{d x}+5 y(x)=0$$

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda.$  

Substitute $y(x)=e^{\lambda x}$ into the differential equation:

$$\frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)+2 \frac{d}{d x}\left(e^{\lambda x}\right)+5 e^{\lambda x}=0$$

Substitute:

$$\frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)=\lambda^{2} e^{\lambda x} \text{ and } \frac{d}{d x}\left(e^{\lambda x}\right)=\lambda e^{\lambda x}$$

$$\lambda^{2} e^{\lambda x}+2 \lambda e^{\lambda x}+5 e^{\lambda x}=0$$

Factor out $e^{\lambda x} :$

$$\left(\lambda^{2}+2 \lambda+5\right) e^{\lambda x}=0$$

Since $e^{\lambda x} \neq 0$ for any finite $\lambda$ , the zeros must come from the polynomial:

$$\lambda^{2}+2 \lambda+5=0$$

Solve for $\lambda$ :

$$\lambda=-1+2 i \text{ or } \lambda=-1-2 i$$

$\text{The roots } \lambda=-1 \pm 2 i \text{ give } y_{1}(x)=c_{1} e^{(-1+2 i) x}, y_{2}(x)=c_{2} e^{(-1-2 i) x} \text{ as solutions, }\\ \text{ where } c_{1} \text{ and } c_{2} \text{ are arbitrary constants. } \text{ }\\ \text{The general solution is the sum of the above solutions: }\\ y(x)=y_{1}(x)+y_{2}(x)=c_{1} e^{(-1+2 i) x}+c_{2} e^{(-1-2 i) x} \text{Apply Euler's identity } e^{\alpha+i \beta}=e^{\alpha} \cos (\beta)+i e^{\alpha} \sin (\beta):\\ y(x)=c_{1}\left(e^{-x} \cos (2 x)+i e^{-x} \sin (2 x)\right)+c_{2}\left(e^{-x} \cos (2 x)-i e^{-x} \sin (2 x)\right)\\ \text{ }\\ \text{Regroup terms:}\\ y(x)=\left(c_{1}+c_{2}\right) e^{-x} \cos (2 x)+i\left(c_{1}-c_{2}\right) e^{-x} \sin (2 x)\\ \text{Redefine } c_{1}+c_{2} \text{ as } c_{1} \text{ and } i\left(c_{1}-c_{2}\right) \text{ as } c_{2}, \text{ since these are arbitrary constants: }\\ y(x)=c_{1} e^{-x} \cos (2 x)+c_{2} e^{-x} \sin (2 x)\\ \text{}\\ \text{Determine the particular solution to } \frac{d^{2} y(0)}{d x^{2}}+2 \frac{d y(x)}{d x}+5 y(x)=4 e^{-x}+17 \sin (2 x)\\ \text{by the method of undetermined coefficients: }$

The particular solution will be the sum of the particular solutions to

$$\frac{d^{2} y(x)}{d x^{2}}+2 \frac{d y(x)}{d x}+5 y(x)=4 e^{-x} \text{ and } \frac{d^{2} y(x)}{d x^{2}}+2 \frac{d y(x)}{d x}+5 y(x)=17 \sin (2 x).$$

The particular solution to

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

is of the form:

$$y_{p_{1}}(x)=a_{1} e^{-x}$$

The particular solution to:

$$\frac{d^{2} y(x)}{d x^{2}}+2 \frac{d y(x)}{d x}+5 y(x)=17 \sin (2 x)$$

is of the form:

$$y_{p_{2}}(x)=a_{2} \cos (2 x)+a_{3} \sin (2 x)$$

$\text{Sum } y_{p_{1}}(x) \text{ and } y_{p_{2}}(x) \text{ to obtain } y_{p}(x): \\y_{p}(x)=y_{p_{1}}(x)+y_{p_{2}}(x)=a_{1} e^{-x}+a_{2} \cos (2 x)+a_{3} \sin (2 x)\\ \text{}\\ \text{Solve for the unknown constants } a_{1}, a_{2}, \text{ and } a_{3} :\\ \text{Compute } \frac{d y_{p}(x)}{d x} :\\ \begin{aligned} \frac{d y_{p}(x)}{d x} &=\frac{d}{d x}\left(a_{1} e^{-x}+a_{2} \cos (2 x)+a_{3} \sin (2 x)\right) \\ &=-a_{1} e^{-x}-2 a_{2} \sin (2 x)+2 a_{3} \cos (2 x) \end{aligned}$

$\begin{aligned} \text { Compute } & \frac{d^{2} y_{p}(x)}{d x^{2}}: \\ \frac{d^{2} y_{p}(x)}{d x^{2}} &=\frac{d^{2}}{d x^{2}}\left(a_{1} e^{-x}+a_{2} \cos (2 x)+a_{3} \sin (2 x)\right) \\ &=a_{1} e^{-x}-4 a_{2} \cos (2 x)-4 a_{3} \sin (2 x) \end{aligned}$

$\text{Substitute the particular solution } y_{p}(x) \text{ into the differential equation: }\\ \frac{d^{2} y_{p}(x)}{d x^{2}}+2 \frac{d y_{p}(x)}{d x}+5 y_{p}(x)=4 e^{-x}+17 \sin (2 x)\\ a_{1} e^{-x}-4 a_{2} \cos (2 x)-4 a_{3} \sin (2 x)+2\left(-a_{1} e^{-x}-2 a_{2} \sin (2 x)+2 a_{3} \cos (2 x)\right)+ 5\left(a_{1} e^{-x}+a_{2} \cos (2 x)+a_{3} \sin (2 x)\right)=4 e^{-x}+17 \sin (2 x)\\ \text{}\\ \text{Simplify:}\\ 4 a_{1} e^{-x}+\left(a_{2}+4 a_{3}\right) \cos (2 x)+\left(-4 a_{2}+a_{3}\right) \sin (2 x)=4 e^{-x}+17 \sin (2 x)\\ \text{}\\ \text{Equate the coefficients of } e^{-x} \text{ on both sides of the equation: }\\ 4 a_{1}=4\\ \text{}\\ \text{Equate the coefficients of } \cos (2 x) \text{ on both sides of the equation:}\\ a_{2}+4 a_{3}=0\\ \text{}\\ \text{Equate the coefficients of } \sin (2 x) \text{ on both sides of the equation:}\\ -4 a_{2}+a_{3}=17$

Solving the system, we get:

$$a_{1}=1\\ a_{2}=-4\\ a_{3}=1\\$$

$\text{}\\ \text{Substitute } a_{1}, a_{2}, \text{ and } a_{3} \text{ into } y_{p}(x)=e^{-x} a_{1}+\cos (2 x) a_{2}+\sin (2 x) a_{3}\\$

$$y_{p}(x)=e^{-x}-4 \cos (2 x)+\sin (2 x)$$

The general solution is:

$\begin{aligned} &y(x)=y_{c}(x)+y_{p}(x)= \\ &e^{-x}-4 \cos (2 x)+\sin (2 x)+c_{1} e^{-x} \cos (2 x)+c_{2} e^{-x} \sin (2 x) \end{aligned}$

which is the required answer.