91影视

Consider a differential equation as follows.

$\mathrm{f}"-\mathrm{伪蹿}\text{'}+\mathrm{尾蹿}=0$

To find the value of $伪$and $\mathrm{尾}$such that $e^{-\mathrm{x}}\mathrm{and}{\mathrm{e}}^{-5\mathrm{x}}$are the solutions of the differential equation.

Now, substitute the value${\mathrm{e}}^{-\mathrm{x}}$for f in $\mathrm{f}"-\mathrm{伪蹿}\text{'}+\mathrm{尾蹿}=0$as follows.

$$\begin{gathered} \mathrm{f}"-\mathrm{伪蹿}\text{'}+\mathrm{尾蹿}=0 \\ \left({\mathrm{e}}^{-\mathrm{x}}\right)"+\mathrm{伪}\left({\mathrm{e}}^{-\mathrm{x}}\right)\text{'}+\mathrm{尾}\left({\mathrm{e}}^{-\mathrm{x}}\right)=0 \\ \left({\mathrm{e}}^{-\mathrm{x}}\right)\text{'}+{\mathrm{伪别}}^{-\mathrm{x}}+{\mathrm{尾别}}^{-\mathrm{x}}=0 \\ {\mathrm{e}}^{-\mathrm{x}}+{\mathrm{伪别}}^{-\mathrm{x}}+{\mathrm{尾别}}^{-\mathrm{x}}=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\mathrm{e}}^{-\mathrm{x}}\left(1-\mathrm{伪}+\mathrm{尾}\right)=0 \\ {\mathrm{e}}^{-\mathrm{x}}>0,1-\mathrm{伪}+\mathrm{尾}=0 \\ 1-\mathrm{伪}+\mathrm{尾}=0 \end{gathered}$$

Similarly, substitute the value ${\mathrm{e}}^{-5\mathrm{x}}$for f in$\mathrm{f}"-\mathrm{伪蹿}\text{'}+\mathrm{尾蹿}=0$as follows.

$$\begin{gathered} \mathrm{f}"+\mathrm{伪蹿}\text{'}+\mathrm{尾蹿}=0 \\ \left({\mathrm{e}}^{-5\mathrm{x}}\right)"+\mathrm{伪}\left({\mathrm{e}}^{-5\mathrm{x}}\right)\text{'}+\mathrm{尾}\left({\mathrm{e}}^{-5\mathrm{x}}\right)=0 \\ \left(-5{\mathrm{e}}^{-5\mathrm{x}}\right)\text{'}-5{\mathrm{伪别}}^{-5\mathrm{x}}+{\mathrm{尾别}}^{-5\mathrm{x}}=0 \\ 25{\mathrm{e}}^{-5\mathrm{x}}-5{\mathrm{伪别}}^{-5\mathrm{x}}+{\mathrm{尾别}}^{-5\mathrm{x}}=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\mathrm{e}}^{-5\mathrm{x}}\left(25-5\mathrm{伪}+\mathrm{尾}\right)=0 \\ {\mathrm{e}}^{-5\mathrm{x}}>0,25-5\mathrm{伪}+\mathrm{尾}=0 \\ 25-5\mathrm{伪}+\mathrm{尾}=0 \end{gathered}$$

Solve the equations as follows.

$$\begin{gathered} -5\mathrm{伪}+5\mathrm{尾}=0 \\ \frac{-5\mathrm{伪}+\mathrm{尾}=-25}{4\mathrm{尾}=20} \\ \mathrm{尾}=\frac{20}{4} \end{gathered}$$

Simplify further as follows.

$\mathrm{尾}=5$

Substitute the value 5 for$尾$in$1-\mathrm{伪}+\mathrm{尾}=0$as follows.

$$\begin{gathered} 1-\mathrm{伪}+\mathrm{尾}=0 \\ 1-\mathrm{伪}+5=0 \\ -\mathrm{伪}+6=0 \\ \mathrm{伪}=-6 \end{gathered}$$

Substitute the values -6 for $\mathrm{伪}$and 5 for $尾$in localid="1659413102620" $\mathrm{f}"-\mathrm{伪蹿}\text{'}+\mathrm{尾蹿}=0$as follows.

$$\begin{gathered} \mathrm{f}"-\mathrm{伪蹿}\text{'}+\mathrm{尾蹿}=0 \\ \mathrm{f}"-6\mathrm{f}\text{'}+5\mathrm{f}=0 \end{gathered}$$

Thus, the differential equation is $\mathrm{f}"-6\mathrm{f}\text{'}+5\mathrm{f}=0$.