Question

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)

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

Short answer

The general solution of the given systemis given by,

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

From (11),$y\left(x\right)=\cosh x-\left(\tanh 1\right)\sinh x$

Step-by-step solution

Obtain the general solution using (10)

According to (10) the auxiliary equation $m^2-k^2=0$for$y\text{'}\text{'}-k^{2}y=0$

Has distinct real roots $m_1=k and m_2=-k$and so, the general solution of the, DE (Differential Equation) is

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

Now, for the given DE, the auxiliary equation is

$m^2-1=0 \text{or} m^2-{\left(1\right)}^2=0$

It has distinct real roots $m_1=1$and $m_2=-1$and so, the general solution of the given DE is

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

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

$$\begin{gathered} c_1{e}^0+c_2{e}^0=1 \\ {c}_1+c_2=1······\left(1\right) \end{gathered}$$

For the solution, we have the derivative

$y\text{'}=c_1{e}^x-c_2{e}^{-x}$

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

$$\begin{gathered} c_1{e}^1-c_2{e}^{-1}=0 \\ {c}_1=\frac{c_2}{e^2}······\left(2\right) \end{gathered}$$

From (1),

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

Substituting $c_2$and find $c_1=\frac{1}{1+e^2}$

Substituting $c_1$and $c_2$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$y\left(x\right)=\frac{1}{1+e^2} e^x+\frac{e^2}{1+e^2} e^{-x}$

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

$y=c_1\cosh kx+c_2\sinh kx$

Therefore, for the given DE, the general solution is

$y=c_1\cosh x+c_2\sinh x$

Substituting $y\left(0\right)=1$in the above equation gives,

$$\begin{gathered} c_1\cosh 0+c_2\sinh 0=1 \\ {c}_1=1······\left(3\right) \end{gathered}$$

Using $\cosh x=\frac{e^x+e^{-x}}{2}$and $\sinh x=\frac{e^x-e^{-x}}{2}$,

For the solution, we have the derivative

$y\text{'}=c_1\sinh x+c_2\cos h x$

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

$c_1\sinh 1+c_2\cosh 1=0$

From (3),

$$\begin{gathered} \sinh 1+c_2\cosh 1=0 \\ {c}_2=-\frac{\sinh 1}{\cosh 1} \\ {c}_2=-\tanh 1 \end{gathered}$$

Substituting $c_{1 }\text{and} c_2$in the general solution gives

$y\left(x\right)=\cosh x-\left(\tanh 1\right)\sinh x$

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

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

From (11),$y\left(x\right)=\cosh x-\left(\tanh 1\right)\sinh x$