Q19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problem 19, solve the given initial value problem 

 $\begin{array}{l}y\text{'}\text{'}\text{'}-y\text{'}\text{'}-4y\text{'}+4y=0\\ y\left(0\right)=-4\\ y\text{'}\left(0\right)=-1\\ y\text{'}\text{'}\left(0\right)=-19\end{array}$

Step-by-Step Solution

Answer

The solution is  C $y\left(t\right)=e^t-3e^{2t}-2e^{-2t}$

Step 1: Basic differentiation

The Sum rule says the derivative of a sum of functions is the sum of their derivatives. The Difference rule says the derivative of a difference of functions is the difference of their derivatives.

Step 2: Solving by basic differentiation:

We will do the following question on the basis of basic differentiation ;

 $\begin{array}{l}{r}^3-r^2-4r+4=0\\ r^3-4r^2+7r-6=(r-1)(r-2)(r+2)=0\\ \\ y\left(t\right)=c_1{e}^t+c_2{e}^{2t}+c_3{e}^{-2t}\\ y\text{'}\left(t\right)=c_1{e}^t+2c_2{e}^{2t}-2c_3{e}^{-2t}\\ y\text{'}\text{'}\left(t\right)=c_1{e}^t+4c_2{e}^{2t}+4c_3{e}^{-2t}\\ \\ y\left(0\right)=-4\\ y\text{'}\left(0\right)=-1\\ y\text{'}\text{'}\left(0\right)=-19\\ \\ y\left(t\right)=e^t-3e^{2t}-2e^{-2t}\end{array}$

Hence, the final answer is:

 $y\left(t\right)=e^t-3e^{2t}-2e^{-2t}$