Question

Solve the given initial value problem

_{$\begin{array}{l}y\text{'}\text{'}\text{'}-4y\text{'}\text{'}+7y\text{'}-6y=0\\ y\left(0\right)=1\\ y\text{'}\left(0\right)=0\\ y\text{'}\text{'}\left(0\right)=0\end{array}$}

Short answer

_{The solution is$y\left(t\right)=e^{2t}-\sqrt{2}{e}^t\sin \sqrt{2}t$}

Step-by-step solution

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.

Solving by basic differentiation:

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

$\begin{array}{l}{r}^3-4r^2+7r-6=0\\ r^3-4r^2+7r-6=(r-2)(r^2-2r+3)=0\\ \\ y\left(t\right)=c_1{e}^{2t}+c_2{e}^t\sin \sqrt{2}t+c_3{e}^t\cos \sqrt{2}t\\ y\text{'}\left(t\right)=2c_1{e}^{2t}+c_2{e}^t\sin \sqrt{2}t+\sqrt{2}{c}_2{e}^t\cos \sqrt{2}t+c_3{e}^t\cos \sqrt{2}t-\sqrt{2}{c}_3{e}^t\sin \sqrt{2}t\\ y\text{'}\text{'}\left(t\right)=4c_1{e}^{2t}+c_2{e}^t\sin \sqrt{2}t+\sqrt{2}{c}_2{e}^t\cos \sqrt{2}t+\sqrt{2}{c}_2{e}^t\cos \sqrt{2}t-\sqrt{2}{c}_2{e}^t\sin \sqrt{2}t+c_3{e}^t\cos \sqrt{2}t-\sqrt{2}{c}_3{e}^t\sin \sqrt{2}t-\sqrt{2}{c}_3{e}^t\sin \sqrt{2}t-\sqrt{2}{c}_3{e}^t\cos \sqrt{2}t\\ \\ y\left(0\right)=1\\ y\text{'}\left(0\right)=0\\ y\text{'}\text{'}\left(0\right)=0\\ \\ y\left(t\right)=e^{2t}-\sqrt{2}{e}^t\sin \sqrt{2}t\end{array}$

Hence, the final answer is:

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