Question

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

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

Short answer

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

From 11:$y\left(x\right)=\cos h\sqrt{3}x+\frac{5}{\sqrt{3}}\sin h\sqrt{3x}$

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.

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

According to the auxiliary equation role="math" localid="1663918092380" $m^2-k^2=0$for $y^{\text{'}\text{'}}-k^{2}y=0$

Has distinct real roots role="math" localid="1663918242786" $m_1$=k and $m_2=-k$

And so the general solution of the DE is

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

The give DE the auxiliary equation is $$\begin{gathered} y=c_1{e}^{kx}+c_2{e}^{-kx} \\ {m}^2-3=0 \\ or \\ {m}^2-{\left(\sqrt{3}\right)}^2=0 \end{gathered}$$

Distinct reel roots

$m_1=\sqrt{3}$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(0)=1, the above equation gives

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

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

$$\begin{gathered} \sqrt{3}{c}_1{e}^{\sqrt{3}\left(0\right)}-\sqrt{3}{c}_2{e}^{-\sqrt{3}\left(0\right)}=0 \\ \sqrt{3}{c}_1{e}^{°}-\sqrt{3}{c}_2{e}^{°}=5 \\ \sqrt{3}{c}_1-\sqrt{3}{c}_2=5 \\ {c}_1-c_2=\frac{5}{\sqrt{3}}----2 \end{gathered}$$

Solving 1 and 2 gives

$c_1=\frac{1}{2}\left(1+\frac{5}{\sqrt{3}}\right)and{c}_1=\frac{1}{2}\left(1-\frac{5}{\sqrt{3}}\right)$

substitute $c_{1}and{c}_2$ general solution gives

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

 Determine the general solution using (11)

According to (11), an alternative for the general solution of $$\begin{gathered} y^{\text{'}\text{'}}-k^{2}y=0 \\ y=c_1\cos kx+c_2\sin hkx \end{gathered}$$

Given DE general solution is

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

Substituting y(0 ) =1 equation gives

localid="1663931328665" $$\begin{gathered} c_1\left(\cos \sqrt{3.}0\right)+c_2\sin h\left(\sqrt{3.}0\right)=1 \\ {c}_1\left(\cos 0\right)+c_2\sin \left(0\right)=1 \\ {c}_1=1----1 \\ \cos hx=\frac{e^x+e^{-x}}{2}and\sin hx=\frac{e^x-e^{-x}}{2} \end{gathered}$$

The solution have derivative:

$y^{\text{'}}=\sqrt{3}{c}_1\sin h\sqrt{3}x+\sqrt{3}{c}_2\cos h\sqrt{3}x$

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

$$\begin{gathered} \sqrt{3}{c}_1\sin (3.0)+\sqrt{3}{c}_2\cos \sqrt{3}.0=5 \\ \sqrt{3}{c}_1\sin h0+\sqrt{3}{c}_2\cos h0=5 \\ \sqrt{3}{c}_2=5 \\ {c}_2=\frac{5}{\sqrt{3}}----2 \end{gathered}$$

Substitute $c_{1}and{c}_2$in the general solution gives

$y\left(x\right)=\cos h\sqrt{3}x+\frac{5}{\sqrt{3}}\sin h\sqrt{3x}$

Hence, the solutions are From 10:

$\mathit{y}\left(x\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\sqrt{\mathbf{3}}\mathbf{}\mathit{x}\mathbf{+}\frac{\mathbf{5}}{\sqrt{\mathbf{3}}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\sqrt{\mathbf{3}\mathbf{x}}$

From 12:$\mathbf{y}\mathbf{}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{h}\mathbf{}\sqrt{\mathbf{3}}\mathbf{x}\mathbf{+}\frac{\mathbf{5}}{\sqrt{\mathbf{3}}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{h}\sqrt{\mathbf{3}}\mathbf{x}$