Given that y1(t)=cost is a solution to y''-y'+y=sint and y2(t)=e2t3 is a solution to role="math" localid="1654926813168" y''-y'+y=e2t. Use the superposition principle to find solutions to the following differential equations: (a) y''-y'+y=5sint(b) y''-y'+y=sint-3e2t(c) y''-y'+y=4sint+18e2t

y(t)=5costy(t)=cost-e2ty(t)=4cost+6e2t

Q1E - Fundamentals Of Differential Equations And Boundary Value Problems

Q1E

Question

Given that $y_{1} (t) = cost$ is a solution to $y'' - y' + y = sint$ and $y_{2} (t) = \frac{e^{2 t}}{3}$ is a solution to $y'' - y' + y = e^{2 t}$ . Use the superposition principle to find solutions to the following differential equations:

$(a) y'' - y' + y = 5 sint$

$(b) y'' - y' + y = sint - 3 e^{2 t}$

$(c) y'' - y' + y = 4 sint + 18 e^{2 t}$

Step-by-Step Solution

Verified

Answer

$y (t) = 5 cost$
$y (t) = cost - e^{2 t}$
$y (t) = 4 cost + 6 e^{2 t}$

1Step 1: Write the given equation

Given that $y_{1} (t) = cost$ is a solution to $y'' - y' + y = sint$ and $y_{2} (t) = \frac{e^{2 t}}{3}$ is a solution to $y'' - y' + y = e^{2 t} .$

2Step 2: Use the superposition principle to find solutions.

We need to find solutions to the following differential equation.

$y'' - y' + y = 5 sint$

Using the method of the superposition principle, for any constants $c_{1}$ and $c_{2}$ the function;

$y (t) = c_{1} y_{1} (t) + c_{2} y_{1} (t) \Rightarrow y (t) = c_{1} (cost) + c_{2} (\frac{e^{2 t}}{3})$ is a solution to the differential equation.

Write the $5 sint$ as a linear combination of $sint$ and $e^{2 t}$ .

Thus, superposition is,

$\Rightarrow 5 (sint) + 0 (e^{2 t})$

The coefficients of the above equation are,

$\begin{matrix} c_{1} = 5 \\ c_{2} = 0 \end{matrix}$

Substituting the value of $c_{1}$ and $c_{2}$ , we get:

$\begin{matrix} y (t) = 5 (cost) + 0 (\frac{e^{2 t}}{3}) \\ y (t) = 5 cost \end{matrix}$

Therefore, the solution of a differential equation,

$y (t) = 5 cost$

3Step 3: Use the superposition principle to find solutions.

To solutions to the following differential equation;

$y'' - y' + y = sint - 3 e^{2 t}$

According to the method of the superposition principle, for any constants $c_{1}$ and $c_{2}$ the function

$y (t) = c_{1} y_{1} (t) + c_{2} y_{1} (t) \Rightarrow y (t) = c_{1} (cost) + c_{2} (\frac{e^{2 t}}{3})$ is a solution to the differential equation.

Write the $sint - 3 e^{2 t}$ as a linear combination of $sint$ and $e^{2 t}$ .

Thus, superposition is,

$\Rightarrow 1 (sint) - 3 (e^{2 t})$

The coefficients of the above equation are,

$\begin{matrix} c_{1} = 1 \\ c_{2} = - 3 \end{matrix}$

Substitute the value of $c_{1}$ and $c_{2}$ in the equation (3),

Hence, the solution of the differential equation,

$\begin{matrix} y (t) = 1 (cost) - 3 (\frac{e^{2 t}}{3}) \\ y (t) = cost - e^{2 t} \end{matrix}$

4Step 4: Use the superposition principle to find solutions.

To find solutions to the following differential equation;

$y'' - y' + y = 4 sint + 18 e^{2 t}$

According to the method of the superposition principle, for any constants $c_{1}$ and $c_{2}$ the function

$y (t) = c_{1} y_{1} (t) + c_{2} y_{1} (t) \Rightarrow y (t) = c_{1} (cost) + c_{2} (\frac{e^{2 t}}{3})$ is a solution to the differential equation.