Question

To find the general solution of the given system.

10.$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\mathit{X}\mathbf{+}\left(\begin{array}{c}t\\ e^{4t}\end{array}\right)$

Short answer

The general solution for the given matrix is$X=C_3\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_4\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-t-1\\ \frac{1}{2}{e}^{4t}\end{array}\right)$

Step-by-step solution

Definition for non-homogeneous system.

• For a nonhomogeneous system of linear first-order differential equations it can be shown that the general solution of X'=AX + F(t), where A is an n*n matrix of constants, is
• $\mathbf{X}=e^{\mathbf{A}t}\mathbf{C}+e^{\mathbf{A}t}{\int }_{t_0}^t{e}^{-\mathbf{A}s}\mathbf{F}\left(s\right)ds.$
  
• Since the matrix exponential e^At is a fundamental matrix, it is always nonsingular and e^-As = (e^As)^-1.

Calculate the Matrix exponential eAt

Given that,

$X\text{'}=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)X+\left(\begin{array}{c}\mathrm{t}\\ {\mathrm{e}}^{4\mathrm{t}}\end{array}\right)$

By using

localid="1664345886240" $\mathbf{X}=e^{\mathbf{A}t}\mathbf{C}+e^{\mathbf{A}t}{\int }_{t_0}^t{e}^{-\mathbf{A}s}\mathbf{F}\left(s\right)ds$

To find the general solution of the given system. Now, let

localid="1664345893711" $A=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$

By using

localid="1664345899031" $e^{At}=I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$

Compute e^At . So,

A²= A.A

localid="1664345908171" $$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right) \end{gathered}$$

A³ = A².A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 8\end{array}\right) \end{gathered}$$

A⁴= A².A²

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 16\end{array}\right) \end{gathered}$$

width="91">Ak=1002k for k=1,2,3,4.....

Therefore, using (2), we have

$e^{At}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)t+\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\frac{t^2}{2!}+\left(\begin{array}{cc}1& 0\\ 0& 8\end{array}\right)\frac{t^3}{3!}$$=\left(\begin{array}{cc}1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}& 0\\ 0& 1+2t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}\end{array}\right)$

The power series in the first and second rows of the matrix represent, respectively, e^t and e^2t, therefore,

$=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)$

Find the General solution of given matrix.

Now, using (1), we have

$$\begin{gathered} X=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right){\int }_{t_0}^t\left(\begin{array}{cc}{e}^{-s}& 0\\ 0& e^{-2s}\end{array}\right)\left(\begin{array}{c}s\\ e^{4s}\end{array}\right)ds \\ =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right){\int }_0^t\left(\begin{array}{c}s{e}^{-s}\\ e^{2s}\end{array}\right) \end{gathered}$$

To integrate

$\int s{e}^{-s}ds$

In the first row, we will use integration by parts:

$$\begin{gathered} u=sdv=e^{-s} \\ du=1v=-e^{-s} \end{gathered}$$

where,

$$\begin{gathered} \int s{e}^{-s}ds=uv-\int vdu \\ =-s{e}^{-8}+\int e^{-s}ds \\ =-s{e}^{-8}-e^{-s} \end{gathered}$$

Then,

$$\begin{gathered} X=C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right){{\left(\begin{array}{c}-s{e}^{-s}-e^{-x}\\ \frac{1}{2}{e}^{2s}\end{array}\right)}^t}_0 \\ =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}-t{e}^{-s}-e^{-t}+1\\ \frac{1}{2}{e}^{2t}-\frac{1}{2}\end{array}\right) \end{gathered}$$

$$\begin{gathered} =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}-t{e}^{-t}-e^{-t}+1\\ \frac{1}{2}{e}^{2t}-\frac{1}{2}\end{array}\right) \\ =C_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_2\left(\begin{array}{c}0\\ 1\end{array}\right)e^t+\left(\begin{array}{c}-t-1+e^t\\ \frac{1}{2}{e}^{4t}-\frac{1}{2}{e}^{2t}\end{array}\right) \\ =\left(\begin{array}{c}1\\ 0\end{array}\right)e^t[C_1+1]+\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}[C_2-\frac{1}{2}]+\left(\begin{array}{c}-t-1\\ \frac{1}{2}{e}^{4t}\end{array}\right) \end{gathered}$$

Let ${\mathbf{C}}_3={\mathbf{C}}_1+1$and ${\mathbf{C}}_4={\mathbf{C}}_2-\frac{1}{2}$,

$=C_3\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_4\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-t-1\\ \frac{1}{2}{e}^{4t}\end{array}\right)$

Therefore, the general solution for the given matrix is

$X=C_3\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_4\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-t-1\\ \frac{1}{2}{e}^{4t}\end{array}\right)$