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)

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{5} \end{gathered}$$

Short answer

From 10: $y\left(x\right)=\frac{1}{2}\left(1+\frac{5}{\sqrt{3}}\right)e^{\sqrt{3x}}+\frac{1}{2}\left(1-\frac{5}{\sqrt{3}}\right)e^{-\sqrt{3x}}$

From 11:$y\left(x\right)=\cosh \sqrt{3}x+\frac{5}{\sqrt{3}}\sinh \sqrt{3}x$

Step-by-step solution

Determine the general solution using (10)

We previously concluded that the given system's is general solution then its used the type of matrix form.

$y\text{'}\text{'}-3y=0,$

$y\left(0\right)=1,y\text{'}\left(0\right)=5$

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 IS

localid="1667994797558" $y=c_1{e}^{kx}+c_2{e}^{-kx}$

The give DE the auxiliary equation is$y=c_1{e}^{kx}+c_2{e}^{-kx}$

$m^2-3=0$

Or

$m^2-{\left(\sqrt{3}\right)}^2=0$

Distinct reel roots:

$m_1=\sqrt{3} \text{and} m_2=-\sqrt{3}$so, general solution of the given DE is

$y=c_1{e}^{\sqrt{3x}}+c_2{e}^{-\sqrt{3x}}$

Substituting$y\left(0\right)=1$above equation give,

$$\begin{gathered} c_1{e}^{\sqrt{3}\left(0\right)}+c_2{e}^{-\sqrt{3}\left(0\right)}=1 \\ {c}_1{e}^0+c_2{e}^0=1 \\ {c}_1+c_2=1······\left(1\right) \end{gathered}$$

The solution derivative:

$y\text{'}=\sqrt{3}{c}_1{e}^{\sqrt{3x}}-\sqrt{3}{c}_2{e}^{-\sqrt{3x}}$

Substituting $y\text{'}\left(0\right)=5$above equation gives

$$\begin{gathered} \sqrt{3}{c}_1{e}^{{}^{\sqrt{3}}\left(0\right)\backslash \mathrm{kern}1pt}-\sqrt{3}{c}_2{e}^{{}^{-\sqrt{3}}\left(0\right)}=5 \\ \sqrt{3}{c}_1{e}^0-\sqrt{3}{c}_2{e}^0=5 \\ \sqrt{3}{c}_1-\sqrt{3}{c}_2=5\backslash \mathrm{hfill} \\ {c}_1-c_2=\frac{5}{\sqrt{3}}······\left(2\right) \end{gathered}$$

Solving (1) and (2) gives

$c_1=\frac{1}{2}\left(1+\frac{5}{\sqrt{3}}\right) \text{and} c_1=\frac{1}{2}\left(1-\frac{5}{\sqrt{3}}\right)$

Substitute $c_1 \text{and} c_2$general solution gives

$y\left(x\right)=\frac{1}{2}\left(1+\frac{5}{\sqrt{3}}\right)e^{\sqrt{3}x}+\frac{1}{2}\left(1-\frac{5}{\sqrt{3}}\right)e^{-\sqrt{3}x}$

Determine the general solution using (11)

According to (11), an alternative for the general solution of$y\text{'}\text{'}-k^{2}y=0$

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

given DE general solution is

$y=c_1\cos \sqrt{3}x +c_2\sinh \sqrt{3}x$

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

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

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

The solution have derivative:

$y\text{'}=\sqrt{3}{c}_1\sinh \sqrt{3}x+\sqrt{3c_2}\cosh \sqrt{3}x$

Substituting $y\text{'}\left(0\right)=5$equation gives,

$$\begin{gathered} \sqrt{3}{c}_1\sin \left(\sqrt{3}.0\right)+\sqrt{3}{c}_2\cos \left(\sqrt{3}.0\right) =5 \\ \sqrt{3}{c}_1\sinh 0+\sqrt{3}{c}_2\cosh 0=5 \\ \sqrt{3}{c}_2=5 \\ {c}_2=\frac{5}{\sqrt{3}}······\left(2\right) \end{gathered}$$

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

$y\left(x\right)=\cosh \sqrt{3}x+\frac{5}{\sqrt{3}}\sinh \sqrt{3}x$

Hence, the solutions are

From 10: $y\left(x\right)=\frac{1}{2}\left(1+\frac{5}{\sqrt{3}}\right)e^{\sqrt{3x}}+\frac{1}{2}\left(1-\frac{5}{\sqrt{3}}\right)e^{-\sqrt{3x}}$

From 12:$y\left(x\right)=\cosh \sqrt{3}x+\frac{5}{\sqrt{3}}\sinh \sqrt{3}x$