Question

Use the method of Laplace transforms to find a general solution to the given differential equation by assuming $y\left(0\right)=aandy\text{'}\left(0\right)=b$

, whereare arbitrary constants.

$y\text{'}\text{'}-4y\text{'}+3y=0$.

Short answer

The solution to the given differential equation by using Laplace transformation is

$y\left(t\right)=\frac{3a-b}{2}{e}^t+\frac{b-a}{2}{e}^{3t}$

Step-by-step solution

Define Laplace Transform

• The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
• In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Determine the solution of the initial value problem:

Applying the Laplace transform and using its linearity solve as:

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}-4y\text{'}+3y\right\}=0 \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}-4\mathcal{L}\left\{y\text{'}\right\}+3\mathcal{L}\left\{y\right\}=0 \end{gathered}$$

Solve for the Laplace transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\left(0\right)\right]-4\left[sY\left(s\right)-y\left(0\right)\right]+3Y\left(s\right)=0 \\ {s}^{2}Y\left(s\right)-as-b-4sY\left(s\right)-4a+3Y\left(s\right)=0 \\ {s}^{2}Y\left(s\right)-4sY\left(s\right)+3=as+b-4a \\ \left(s^2-4s+3\right)Y\left(s\right)=as+b-4a \end{gathered}$$

Solve as:

$Y\left(s\right)=\frac{as+b-4a}{s^2-4s+3}$

Using partial fractions to simplify as:

$$\begin{gathered} \frac{as+b-4a}{s^2-4s+3}=\frac{as+b-4a}{\left(s-1\right)\left(s-3\right)} \\ =\frac{A}{s-1}+\frac{B}{s-3} \end{gathered}$$

Resolving the above fraction as:

$as+b-4a=A\left(s-3\right)+B\left(s-1\right)$

Using $s=3,1$respectively and solve as:

$$\begin{gathered} s=3:b-a=2B\Rightarrow B=\frac{b-a}{2} \\ s=1:3a-b=2A\Rightarrow A=\frac{3a-b}{2} \end{gathered}$$

Substitute the value and rewrite as:

$Y\left(s\right)=\frac{3a-b}{2\left(s-1\right)}+\frac{b-a}{2\left(s-3\right)}$

Using the inverse Laplace transform to obtain the solution of given differential equation as:

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^1\left\{\frac{3a-b}{2\left(s-1\right)}+\frac{b-a}{2\left(s-3\right)}\right\}\left(t\right) \\ =\frac{3a-b}{2}\mathcal{L}\left\{\frac{1}{s-1}\right\}+\frac{b-a}{2}\mathcal{L}\left\{\frac{1}{s-3}\right\} \\ =\frac{3a-b}{2}{e}^t+\frac{b-a}{2}{e}^{3t} \end{gathered}$$

Therefore, the solution for the initial value problem is$y\left(t\right)=\frac{3a-b}{2}{e}^t+\frac{b-a}{2}{e}^{3t}$