Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 13–20, solve the given initial value problem.

${\mathit{y}}^{\mathbf{"}}\mathbf{-}\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{:}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$

Step-by-Step Solution

Answer

The solution is $y\left(t\right)=\frac{4}{3}{e}^t-\frac{1}{3}{e}^{3t}$.

Step 1: Find the solution of the differential equation.

The given differential equation is. y" - 4y' + 3y = 0

The auxiliary equation is;

$\begin{array}{rcl}{r}^2-4r^1+3r^0& =& 0\\ r^2-4r+3& =& 0\\ r^2-r-3r+3& =& 0\\ \left(r-1\right)\left(r-3\right)& =& 0\\ r& =& 1,3\end{array}$

Therefore, the solution is y(t) = c₁e^t + c₂e^3t.

Step 2: Apply initial conditions .

The initial conditions are $y\left(0\right)=1,y^{\text{'}}\left(0\right)=\frac{1}{3}$

Therefore,

$\begin{array}{rcl}y\left(0\right)& =& c_1{e}^0+c_2{e}^0\\ c_1+c_2& =& 1\end{array}$

And 

$\begin{array}{rcl}{y}^{\text{'}}\left(t\right)& =& c_1{e}^t+3c_2{e}^{3t}\\ y^{\text{'}}\left(0\right)& =& c_1{e}^0+3c_2{e}^0\\ c_1+3c_2& =& \frac{1}{3}\end{array}$

Solving for c₁,c₂ then;

$$\begin{gathered} c_1=\frac{4}{3} \\ {c}_2=-\frac{1}{3} \end{gathered}$$

Therefore, the solution is $y\left(t\right)=\frac{4}{3}{e}^t-\frac{1}{3}{e}^{3t}$.