Short Answer

Solve the given initial value problem
_{$\begin{array}{l} y''' - 4 y'' + 7 y' - 6 y = 0 \\ y (0) = 1 \\ y' (0) = 0 \\ y'' (0) = 0 \end{array}$}

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 $\begin{array}{l} r^{3} - 4 r^{2} + 7 r - 6 = 0 \\ r^{3} - 4 r^{2} + 7 r - 6 = (r - 2) (r^{2} - 2 r + 3) = 0 \\ y (t) = c_{1} e^{2 t} + c_{2} e^{t} \sin \sqrt{2} t + c_{3} e^{t} \cos \sqrt{2} t \\ y' (t) = 2 c_{1} e^{2 t} + c_{2} e^{t} \sin \sqrt{2} t + \sqrt{2} c_{2} e^{t} \cos \sqrt{2} t + c_{3} e^{t} \cos \sqrt{2} t - \sqrt{2} c_{3} e^{t} \sin \sqrt{2} t \\ y'' (t) = 4 c_{1} e^{2 t} + c_{2} e^{t} \sin \sqrt{2} t + \sqrt{2} c_{2} e^{t} \cos \sqrt{2} t + \sqrt{2} c_{2} e^{t} \cos \sqrt{2} t - \sqrt{2} c_{2} e^{t} \sin \sqrt{2} t + c_{3} e^{t} \cos \sqrt{2} t - \sqrt{2} c_{3} e^{t} \sin \sqrt{2} t - \sqrt{2} c_{3} e^{t} \sin \sqrt{2} t - \sqrt{2} c_{3} e^{t} \cos \sqrt{2} t \\ y (0) = 1 \\ y' (0) = 0 \\ y'' (0) = 0 \\ y (t) = e^{2 t} - \sqrt{2} e^{t} \sin \sqrt{2} t \end{array}$

Hence, the final answer is:

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Solve the given initial value problem
_{$\begin{array}{l} y''' - 4 y'' + 7 y' - 6 y = 0 \\ y (0) = 1 \\ y' (0) = 0 \\ y'' (0) = 0 \end{array}$}

Step by Step Solution

Step 1: Basic differentiation

Step 2: Solving by basic differentiation:

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Solve the given initial value problem y'''−4y''+7y'−6y=0y(0)=1y'(0)=0y''(0)=0

Step by Step Solution

Step 1: Basic differentiation

Step 2: Solving by basic differentiation:

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Solve the given initial value problem
_{$\begin{array}{l} y''' - 4 y'' + 7 y' - 6 y = 0 \\ y (0) = 1 \\ y' (0) = 0 \\ y'' (0) = 0 \end{array}$}