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Problem 1

In Exercises 1 through \(7,\) find the general solution of the system \(X^{\prime}=A X\) for the given matrix A. In each case check on the linear independence of solutions by examining the Wronskian. $$ A=\left(\begin{array}{rr} 8 & -3 \\ 16 & -8 \end{array}\right) $$

Problem 1

In Exercises 1 through \(5,\) replace the given equation by a system of first- order equations. \(y^{\prime \prime}-6 y^{\prime}+8 y=x+2\).

Problem 1

Find the matrix requested given the following matrices: \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right) \quad B=\left(\begin{array}{rr}2 & 0 \\ 1 & -1\end{array}\right) \quad C=\left(\begin{array}{rr}1 & -1 \\ 1 & 2\end{array}\right)\) \(\quad A+2 B\).

Problem 1

Find the general solution of the system \(X^{\prime}=A X\) for the given \(m a-\) \(\operatorname{trix} A\). $$ A=\left(\begin{array}{rr} 4 & 5 \\ -4 & -4 \end{array}\right) $$

Problem 2

In Exercises 1 through \(7,\) find the general solution of the system \(X^{\prime}=A X\) for the given matrix A. In each case check on the linear independence of solutions by examining the Wronskian. $$ A=\left(\begin{array}{rr} 1 & 0 \\ -2 & 2 \end{array}\right) $$

Problem 2

Solve the system \(X^{\prime}=A X\). $$A=\left(\begin{array}{ll}4 & -9 \\ 4 & -8\end{array}\right)$$

Problem 2

Find the general solution of the system \(X^{\prime}=A X\) for the given \(m a-\) \(\operatorname{trix} A\). $$ A=\left(\begin{array}{rr} 4 & 1 \\ -8 & 8 \end{array}\right) $$

Problem 2

In Exercises 1 through \(5,\) replace the given equation by a system of first- order equations. \(y^{\prime \prime}+4 y^{\prime}+4 y=e^{x}\).

Problem 2

Find the matrix requested given the following matrices: \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right) \quad B=\left(\begin{array}{rr}2 & 0 \\ 1 & -1\end{array}\right) \quad C=\left(\begin{array}{rr}1 & -1 \\ 1 & 2\end{array}\right)\) \(\quad A+2 B\). \(2 A+C\).

Problem 3

Find the matrix requested given the following matrices: \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right) \quad B=\left(\begin{array}{rr}2 & 0 \\ 1 & -1\end{array}\right) \quad C=\left(\begin{array}{rr}1 & -1 \\ 1 & 2\end{array}\right)\) \(\quad A+2 B\).. \(A B+2 I\).

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