Question

In Problems,1-14find the general solution of the given second-order differential equation.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The general solution is$y=c_1\cos 3x+c_2\sin 3x$

Step-by-step solution

Define Second order differential equation

For a second-order differential equation $a{y}^{\text{'}\text{'}}+b{y}^{\text{'}}+cy=0$, where aandb are constants, the auxiliary equation will be of the form $a{m}^2+bm+c=0$.

There will be three forms of the general solution corresponding to three cases that depend on the solution of an auxiliary equation.

1. m₁ and m₂ are real and distinct roots that are possible when $\left(b^2-4ac>0\right)$

2. m₁ and m₂ are real and equal roots that are possible when $\left(b^2-4ac=0\right)$

3. m₁ and m₂ are conjugate complex roots$\left(b^2-4ac<0\right)$

Case I: Distinct Real Roots

The general solution of $ay\text{'}\text{'}+by\text{'}+cy=0$ on the interval $(-\infty ,\infty )$will be $y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$.

Case II: Repeated Real Roots

The general solution will be$y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$.

Case III: Conjugate Complex Roots

If m₁ and m₂ are complex roots, then $m_1=\alpha +i\beta$and $m_2=\alpha -i\beta$, where $\alpha$ and $\beta >0$are real and $i^2=-1$.

The general solution will be

$$\begin{gathered} y=c_1{e}^{\alpha x}\cos \beta x+c_2{e}^{\alpha x}\sin \beta x \\ =e^{\alpha x}\left(c_1\cos \beta x+c_2\sin \beta x\right) \end{gathered}$$

Find the given differential equation

Consider the given differential equation $y\text{'}\text{'}+9y=0$.

Consider $y=e^{mx}$as the solution of the differential equation $y\text{'}\text{'}+9y=0$.

Substitute $y=e^{mx},y\text{'}=m{e}^{mx}$and $y\text{'}\text{'}=m^2{e}^{mx}$ into $y\text{'}\text{'}+9y=0$to obtain the auxiliary equation,

$$\begin{gathered} m^2{e}^{mx}+9e^{mx}=0 \\ {e}^{mx}\left(m^2+9\right)=0 \end{gathered}$$

Solve the auxiliary equation as follows.

$m^2+9=0$

Here,$a=1,b=0\text{and}c=9$

Determine the roots.

$m=\frac{-(-b)\pm \sqrt{b^2-4ac}}{2a}$

Therefore,

$$\begin{gathered} m_1=\frac{-(-b)+\sqrt{b^2-4ac}}{2a} \\ =\frac{-\left(0\right)+\sqrt{-36}}{2\left(1\right)} \\ =\frac{6i}{2\left(1\right)} \\ =0+3i \end{gathered}$$

And,

$$\begin{gathered} m_2=\frac{-(-b)-\sqrt{b^2-4ac}}{2a} \\ =\frac{-\left(0\right)-\sqrt{-36}}{2\left(1\right)} \\ =\frac{-6i}{2\left(1\right)} \\ =0-3i \end{gathered}$$

Compute the solution of the given differential equation

The roots of the auxiliary equation are conjugate complex numbers.

For a differential equation which have complex roots, the roots are $m_1=\alpha +i\beta$and $m_2=\alpha -i\beta$ .

Here, $m_1=0+3i\text{and}{m}_2=0-3i$

On comparison,$\alpha =0\text{and}\beta =3$

By above procedure, compute the general solution to the differential equation.

$$\begin{gathered} y=e^{\alpha x}\left(c_1\cos \beta x+c_2\sin \beta x\right) \\ =e^{0·x}\left(c_1\cos 3x+c_2\sin 3x\right) \\ =c_1\cos 3x+c_2\sin 3x \end{gathered}$$

Thus, the general solution is $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{3}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{3}\mathit{x}$.