Question

Solve the given initial value problem. $\mathbf{y}\text{'}\text{'}\text{'}\mathbf{-}\mathbf{4}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{7}\mathbf{y}\text{'}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}$$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}1\mathbf{,}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$$\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution of the given initial value$\mathrm{y}\text{'}\text{'}\text{'}-4\mathrm{y}\text{'}\text{'}+7\mathrm{y}\text{'}-6\mathrm{y}=0$ is$\mathrm{y}\left(\mathrm{t}\right)=\sqrt{2}{\mathrm{e}}^{\mathrm{t}}\left(\sin \right(\sqrt{2}\mathrm{t}\left)\right)+{\mathrm{e}}^{2\mathrm{t}}$ when$\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0$ and$\mathrm{y}\text{'}\text{'}\left(0\right)=0$ .

Step-by-step solution

Differentiate the value of y.

Given differential equation is $\mathrm{y}\text{'}\text{'}\text{'}-4\mathrm{y}\text{'}\text{'}+7\mathrm{y}\text{'}-6\mathrm{y}=0$

Let $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$

Therefore,

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}$,$\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}^2{\mathrm{e}}^{\mathrm{rt}}$ and$\mathrm{y}\text{'}\text{'}\text{'}={\mathrm{r}}^3{\mathrm{e}}^{\mathrm{rt}}$

Then the auxiliary equation is ${\mathrm{r}}^3-4{\mathrm{r}}^2+7\mathrm{r}-6=0$.

Now${\mathrm{r}}^3-4{\mathrm{r}}^2+7\mathrm{r}-6=(\mathrm{r}-2)({\mathrm{r}}^2-2\mathrm{r}+3)$

Finding the general solution

Now we have to find the roots of ${\mathrm{r}}^2-2\mathrm{r}+3.$

$\begin{array}{c}\mathrm{r}=\frac{2\pm \sqrt{2^2-4\times 1\times 3}}{2\times 1}\\ \mathrm{r}=\frac{2\pm \sqrt{4-12}}{}\\ \mathrm{r}=\frac{2\pm \sqrt{-8}}{}\\ \mathrm{r}=\frac{2\pm 2\sqrt{2}\mathrm{i}}{}\\ \mathrm{r}=1\pm \sqrt{2}\mathrm{i}\end{array}$

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left({\mathrm{c}}_1\cos (\sqrt{2}\mathrm{t})+{\mathrm{c}}_2\sin (\sqrt{2}\mathrm{t})\right)+{\mathrm{c}}_3{\mathrm{e}}^{2\mathrm{t}}$.

Substituting the values of y(0)=1,y'(0)=0 and   y''(0)=0

$\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{e}}^0\left({\mathrm{c}}_1\cos (\sqrt{2}\times 0)+{\mathrm{c}}_2\sin (\sqrt{2}\times 0)\right)+{\mathrm{c}}_3{\mathrm{e}}^{2\times 0}\\ {\mathrm{c}}_1+{\mathrm{c}}_3=1...\left(1\right)\end{array}$

And

$\mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}({\mathrm{c}}_1\cos \sqrt{2}\mathrm{t}+{\mathrm{c}}_2\sin \sqrt{2}\mathrm{t})+\sqrt{2}{\mathrm{e}}^{\mathrm{t}}(-{\mathrm{c}}_1\sin \sqrt{\mathrm{t}}+{\mathrm{c}}_2\cos \sqrt{\mathrm{t}})+2{\mathrm{c}}_3{\mathrm{e}}^{2\mathrm{t}}$

Then,

$\begin{array}{l}\mathrm{y}\text{'}\text{'}\left(0\right)={\mathrm{e}}^0{\mathrm{c}}_1\cos (\sqrt{2}\times 0)+{\mathrm{c}}_2\sin (\sqrt{2}\times 0)+\sqrt{2}{\mathrm{e}}^0(-{\mathrm{c}}_1\sin (\sqrt{2}\times 0)+{\mathrm{c}}_2\cos (\sqrt{2}\times 0))+2{\mathrm{c}}_3{\mathrm{e}}^{2\times 0}\\ {\mathrm{c}}_1+\sqrt{2}{\mathrm{c}}_2+2{\mathrm{c}}_3=0...\left(2\right)\end{array}$

And

${\mathrm{y}}^{\text{'}\text{'}\text{'}}={\mathrm{e}}^{\mathrm{t}}{\mathrm{c}}_1\cos (\sqrt{2}\mathrm{t})+{\mathrm{c}}_2\sin (\sqrt{2}\mathrm{t})+\sqrt{2}{\mathrm{e}}^{\mathrm{t}}(-{\mathrm{c}}_1\sin \sqrt{\mathrm{t}}+{\mathrm{c}}_2\cos \sqrt{\mathrm{t}})+\sqrt{2}{\mathrm{e}}^{\mathrm{t}}(-{\mathrm{c}}_1\sin \sqrt{\mathrm{t}}+{\mathrm{c}}_2\cos \sqrt{\mathrm{t}})+2{\mathrm{e}}^{\mathrm{t}}(-{\mathrm{c}}_1\cos (\sqrt{2}\mathrm{t})-{\mathrm{c}}_2\sin (\sqrt{2}\mathrm{t}))+4{\mathrm{c}}_3{\mathrm{e}}^{2\mathrm{t}}$

Then,

$\begin{array}{l}{\mathrm{y}}^{\text{'}\text{'}\text{'}}\left(0\right)={\mathrm{e}}^0({\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)+\sqrt{2}{\mathrm{e}}^0(-{\mathrm{c}}_1\sin 0+{\mathrm{c}}_2\cos 0)+\sqrt{2}{\mathrm{e}}^0(-{\mathrm{c}}_1\sin 0+{\mathrm{c}}_2\cos 0)\\ +2{\mathrm{e}}^0(-{\mathrm{c}}_1\cos \left(0\right)-{\mathrm{c}}_2\sin \left(0\right))+4{\mathrm{c}}_3{\mathrm{e}}^0)\\ -{\mathrm{c}}_1+2\sqrt{2}{\mathrm{c}}_2+4{\mathrm{c}}_3=0...\left(3\right)\end{array}$

On solving the equations, we get:

${\mathrm{c}}_1=0,{\mathrm{c}}_2=-\sqrt{2}$and ${\mathrm{c}}_3=1$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\sqrt{2}{\mathrm{e}}^{\mathrm{t}}\left(\sin \right(\sqrt{2}\mathrm{t}\left)\right)+{\mathrm{e}}^{2\mathrm{t}}$.