Question

In problem 41 and 42 solve the given problem first using the form of the general solution given in (10). Solve again, this time using the from given in (11).

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

Short answer

The general solution of the given system is given by,

$$\begin{gathered} From\left(10\right),y\left(x\right)=\frac{1}{1+e^2}{e}^x+\frac{e^2}{1+e^2}{e}^{-x} \\ From\left(11\right),y\left(x\right)=\cos hx(-\tan h1)\sin hx \end{gathered}$$

Step-by-step solution

Obtain the general solution using (10)

According to the auxiliary equation $m^2-k^2=0fory\text{'}\text{'}-k^{2}y=0$

Has distinct real roots role="math" localid="1664177909524" $m_1=kand{m}_2=-k$and the general solution of the DE (differential equation) is

$y=c_1{e}^x+c_2{e}^{-x}$

Substituting y(0)=1 in the above equation gives

$$\begin{gathered} c_{1}e°+c_{2}e°=1 \\ {c}_1+c_2=1----1 \end{gathered}$$

For the solution, we have the derivative

role="math" localid="1664181242183" $y^{\text{'}}=c_1{e}^x-c_2{e}^{-x}$

Substituting $y\text{'}\left(1\right)=0$in the above equation gives

$$\begin{gathered} c_1{e}^1-c_2{e}^{-1}=0 \\ {c}_1=\frac{c_2}{e^2}----2 \end{gathered}$$

From (1)

role="math" localid="1664182172837" $$\begin{gathered} 1-c_2=\frac{c_2}{e^2} \\ {c}_2=c_2\left(1+\frac{1}{e^2}\right) \\ {c}_2=\frac{e^2}{1+e^2} \end{gathered}$$

Substituting $c_1$and role="math" localid="1664182253475" $c_2$in the general solution gives

$y\left(x\right)=\frac{1}{1+e^2}{e}^x+\frac{e^2}{1+e^2}{e}^{-x}$

Hence the solution is $\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{2}}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{2}}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

Obtain general solution using (11)

Now, according to (11), an alternative form for the general solution of $y\text{'}\text{'}-k^{2}y=0$is $y=c_1\cos hkx+c_2\sin hkx$

Therefore, for the given DE, the general solution is $y=c_1\cos hx+c_2\sin hx$

Substituting y(0)=1 in the above equation gives,

$$\begin{gathered} c_1\mathrm{co}h0+c_2\sin h0=1 \\ {c}_1=1----3 \end{gathered}$$

Using $\cos hx=\frac{e^x+e^{-x}}{2}$and $\sin hx=\frac{e^x-e^{-x}}{2}$

For the solution, we have the derivative

$y=c_1\sin hx+c_2\cos hx$

Substituting $y\text{'}\left(1\right)=0$in the above equation gives

$c_1\sin h1+c_2\cos h1=0$

From (3)

$$\begin{gathered} \sin h1+c_2\cos h1=0 \\ {c}_2=-\frac{\sin h1}{\cos h1} \\ {c}_2=-\tan h1 \end{gathered}$$

Substituting $c_1$and $c_2$in the general solution gives

$y\left(x\right)=\cos hx-\left(\tan h1\right)\sin hx$

Therefore,the general solution of the given system is given by

From (10) $\mathit{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{2}}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{2}}}\mathit{e}$

From (11)$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{h}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{(}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{h}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{h}\mathbf{}\mathbf{x}$