Q20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

y" - 4y' + 4y = 0 : y(1) = 1, y'(1) =1

Step-by-Step Solution

Answer

The solution is y(t) = 2e^2t-2 - te^3t+2.

Step 1: Find the solution of the differential equation.

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

The auxiliary equation is r² - 4r + 4 = 0

Find the roots of the auxiliary equation;

$\begin{array}{rcl}{r}^2-4r+4& =& 0\\ r& =& \frac{4\pm \sqrt{{(-4)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\\ r& =& \frac{4\pm \sqrt{16-16}}{2}\\ r& =& 2,2\end{array}$

Therefore the solution is y(t) = c₁e^2t + c₂te^2t.

Step 2: Apply initial conditions .

The initial conditions are y(1) = 1, y'(t) = 1.

Therefore,

$\begin{array}{rcl}y\left(1\right)& =& c_1{e}^2+c_2\left(1\right)e^2\\ c_1{e}^2+c_2\left(1\right)e^2& =& 1\\ c_1+c_2& =& e^{-2}\end{array}$

And

$\begin{array}{rcl}y\text{'}\left(t\right)& =& 2c_1{e}^{\left(2t\right)}+c_2(2t{e}^{\left(2t\right)}+e^{2t})\\ y\text{'}\left(1\right)& =& 2c_1{e}^2+c_2\left(2\right(1)e^2+e^2)\\ 2c_1{e}^2+c_2\left(3e^2\right)& =& 1\\ 2c_1+3c_2& =& e^{-2}\end{array}$

Solving for c₁,c₂ then;

c₁ = 2e^-2

c₂ = -e²

Therefore, the solution is $y\left(t\right)=2e^{-2}{e}^{2t}-e^{2}t{e}^{3t}\Rightarrow y\left(t\right)=2e^{2t-2}-t{e}^{3t+2}$.