Question

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

$\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

$y=e^{\frac{3}{4}·x}\left(c_1\cos \left(\frac{\sqrt{23}}{4}\right)x+c_2\sin \left(\frac{\sqrt{23}}{4}\right)x\right)$

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 m₁and m₂ are complex roots, then $m_1=\alpha +i\beta$ and $m_2=\alpha -i\beta$, where $\alpha and\beta >0$ and 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 $2y^{\text{'}\text{'}}-3y^{\text{'}}+4y=0$.

Consider $y=e^{mx}$as the solution of the differential equation $2y^{\text{'}\text{'}}-3y^{\text{'}}+4y=0$.

Substitute $y=e^{mx},y\text{'}=m{e}^{mx}$and $y\text{'}\text{'}=m^2{e}^{mx}$into $2y^{\text{'}\text{'}}-3y^{\text{'}}+4y=0$ to obtain the auxiliary equation.

$$\begin{gathered} 2m^2{e}^{mx}-3m{e}^{mx}+4e^{mx}=0 \\ {e}^{mx}\left(2m^2-3m+4\right)=0 \end{gathered}$$

For any real $x,e^{mx}\ne 0$. Therefore, $2m^2-3m+4=0$.

Here, $a=2,b=-3\text{and}c=4$

Determine the roots.

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

Therefore,

$$\begin{gathered} m_1=\frac{-\left(b\right)+\sqrt{b^2-4ac}}{2a} \\ =\frac{-(-3)+\sqrt{-23}}{2\left(2\right)} \\ =\frac{3+i·\sqrt{23}}{4} \\ =\frac{3}{4}+i\frac{\sqrt{23}}{4} \end{gathered}$$

And,

$$\begin{gathered} m_2=\frac{-\left(b\right)+\sqrt{b^2-4ac}}{2a} \\ =\frac{-(-3)-\sqrt{-23}}{2\left(2\right)} \\ =\frac{3-i·\sqrt{23}}{4} \\ =\frac{3}{4}-i\frac{\sqrt{23}}{4} \end{gathered}$$

Compute the solution of the given differential equation

Therefore, the solutions to the auxiliary equation are $\left(\frac{3}{4}+i\frac{\sqrt{23}}{4},\frac{3}{4}-i\frac{\sqrt{23}}{4}\right)$. The roots of the auxiliary equation are conjugate complex numbers.

For a differential equation which have complex roots, the roots are localid="1668401526780" role="math" $m_1=\alpha +i\beta and{m}_2=\alpha -i\beta$.

Here, $m_1=\frac{3}{4}+i\frac{\sqrt{23}}{4}$and$m_2=\frac{3}{4}-i\frac{\sqrt{23}}{4}$

On comparison, $\alpha =\frac{3}{4}and\beta =\frac{\sqrt{23}}{4}$

Therefore, the general solution to the differential equation is,

$$\begin{gathered} y=e^{\alpha x}\left(c_1\cos \beta x+c_2\sin \beta x\right) \\ =e^{\frac{3}{4}x}\left(c_1\cos \left(\frac{\sqrt{23}}{4}\right)x+c_2\sin \left(\frac{\sqrt{23}}{4}\right)x\right) \end{gathered}$$

Thus, the general solution is$\mathit{y}\mathbf{=}{\mathit{e}}^{\frac{\mathbf{3}}{\mathbf{4}}\mathbf{·}\mathbf{x}}\left(c_{1}cos\left(\frac{\sqrt{23}}{4}\right)x+c_{2}sin\left(\frac{\sqrt{23}}{4}\right)x\right)$