32E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 29 - 32, use the method of Laplace transforms to find a general solution to the given differential equation by   assuming $y\left(0\right)=a and y\text{'}\left(0\right)=b$ a and b are arbitrary constants.

$y\text{'}\text{'}-5y\text{'}+6y=-6t{e}^{2t}$

Step-by-Step Solution

Answer

 

The General solution to the given differential equation is

$y\left(t\right)=\left(3a-b+6\right)e^{2t}+6t{e}^{2t}+3t^2{e}^{2t}+\left(b-2a-6\right)e^{3t}$

Step 1: 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.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$
  

Step 2: Apply Laplace transform:

Applying the Laplace transform and using its linearity we get 

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}-5y\text{'}+6y\right\}=\left\{-6t{e}^{2t}\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}-5\mathcal{L}\left\{y\text{'}\right\}+6\mathcal{L}\left\{y\right\}=-\frac{6}{{(s-2)}^2} \end{gathered}$$

Solve for the transform as:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\left(0\right)\right]-5\left[sY\left(s\right)-y\left(0\right)\right]+6Y\left(s\right)=-\frac{6}{{\left(s-2\right)}^2} \\ {s}^{2}Y\left(s\right)-as-b-5sY\left(s\right)-5a+6Y\left(s\right)=-\frac{6}{{(s-2)}^2} \end{gathered}$$

$$\begin{gathered} s^{2}Y\left(s\right)-5sY\left(s\right)+6Y\left(s\right)=-\frac{6}{{\left(s-2\right)}^2}+as+b-5a \\ \left(s^2-5s+6\right)Y\left(s\right)=\frac{a{s}^3+\left(b-9a\right)s^2+\left(24a-4b\right)s+\left(4b-20a-6\right)}{{\left(s-2\right)}^2} \end{gathered}$$

Solve further as:

$Y\left(s\right)=\frac{a{s}^3+\left(b-9a\right)s^2+\left(24a-4b\right)s+\left(4b-20a-6\right)}{{\left(s-2\right)}^2\left(s^2-5s+6\right)}$

Using partial fractions solve as:

$$\begin{gathered} \frac{a{s}^3+\left(b-9a\right)s^2+\left(24a-4b\right)s+\left(4b-20a-6\right)}{{\left(s-2\right)}^2\left(s^2-5s+6\right)} \\ =\frac{\left\{a{s}^3+\left(b-9a\right)s^2 \\ +\left(24a-4b\right)s+\left(4b-20a-6\right) \\ \right\}}{{\left(s-2\right)}^3\left(s-3\right)} \\ =\frac{A}{s-2}+\frac{B}{{\left(s-2\right)}^2}+\frac{C}{{\left(s-2\right)}^3}+\frac{D}{s-3} \end{gathered}$$

Resolve for the partial fraction as:

$$\begin{gathered} a{s}^3+\left(b-9a\right)s^2+\left(24a-4b\right)s+\left(4b-20a-6\right) \\ =\left\{A{\left(s-2\right)}^2\left(s-3\right)+B\left(s-2\right)\left(s-3\right) \\ +C\left(s-3\right)+Ds-2^3\right\} \\ =\left\{\left(A+D\right)s^3+\left(B-7A-6D\right)s^2 \\ +\left(16A-5B+C+12D\right)s \\ -12A+6B-3C-8D\right\} \end{gathered}$$

Solve for the system of equation as:

$\left\{\begin{array}{l}\begin{array}{l}A+D=a\\ B-7A-6D=b-9a\\ 16A-5B+C+12D=24a-4b\end{array}\\ -12A+6B-3C-8D=4b-20a-6\end{array}\right.\Rightarrow \left\{\begin{array}{l}\begin{array}{l}A=3a-b6\\ B=6\end{array}\\ C=6\\ D=b-2a-6\end{array}\right.$

Step 3: Use Inverse Laplace transform:

$Y\left(s\right)=\frac{3a-b+6}{s-2}+\frac{6}{{(s-2)}^2}+\frac{6}{{(s-2)}^3}+\frac{b-2a-6}{s-3}$

Using the inverse Laplace transform, Obtain the solution of given differential equation,

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{3a-b+6}{s-2}+\frac{6}{{\left(s-2\right)}^2}+\frac{6}{{\left(s-2\right)}^3}+\frac{b-2a-6}{s-3}\right\}\left(t\right) \\ =\left\{\left(3a-b+6\right)\mathcal{L}\left\{\frac{1}{s-2}\right\}+6\mathcal{L}\left\{\frac{1}{{(s-2)}^2}\right\} \\ +3\mathcal{L}\left\{\frac{2}{{(s-2)}^3}\right\}+(b-2a-6)\mathcal{L}\left\{\frac{1}{s-3}\right\}\right\} \\ =\left(3a-b+6\right)e^{2t}+6t{e}^{2t}+3t^2{e}^{2t}+\left(b-2a-6\right)e^{3t} \end{gathered}$$

Therefore, the solution for the differential equation is 

$y\left(t\right)=\left(3a-b+6\right)e^{2t}+6t{e}^{2t}+3t^2{e}^{2t}+\left(b-2a-6\right)e^{3t}$