Q20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given initial value problem 

 $\begin{array}{l}y\text{'}\text{'}\text{'}+7y\text{'}\text{'}+14y\text{'}+8y=0\\ y\left(0\right)=1\\ y\text{'}\left(0\right)=-3\\ y\text{'}\text{'}\left(0\right)=13\end{array}$

Step-by-Step Solution

Answer

_{The general solution is  $y\left(t\right)=e^{-t}-e^{-2t}+e^{-4t}$}

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 ;

   $\begin{array}{l}{r}^3+7r^2+14r+8=0\\ r^3-4r^2+7r-6=(r+1)(r^2+6r+8)=0\\ \\ y\left(t\right)=c_1{e}^{-t}+c_2{e}^{-2t}+c_3{e}^{-4t}\\ y\text{'}\left(t\right)=-c_1{e}^{-t}-2c_2{e}^{-2t}-4c_3{e}^{-4t}\\ y\text{'}\text{'}\left(t\right)=c_1{e}^{-t}+4c_2{e}^{-2t}+16c_3{e}^{-4t}\\ \\ y\left(0\right)=1\\ y\text{'}\left(0\right)=-3\\ y\text{'}\text{'}\left(0\right)=13\\ \\ y\left(t\right)=e^{-t}-e^{-2t}+e^{-4t}\end{array}$

Hence, the final answer is:

_{$y\left(t\right)=e^{-t}-e^{-2t}+e^{-4t}$}