Q25E

Question

In Problems 22 through 25, use variation of parameters to find a general solution to the differential equation given that the functions $y_{1}$ and $y_{2}$ are linearly independent solutions to the corresponding homogeneous equation for t> 0.

$ty'' + (5 t - 1) y' - 5 y = t^{2} e^{- 5 t}; y_{1} = 5 t - 1, y_{2} = e^{- 5 t}$

Step-by-Step Solution

Verified

Answer

The general solution is $y (t) = c_{1} (5 t - 1) + c_{2} e^{- 5 t} - \frac{e^{- 5 t}}{125} (5 t - 1) + (- \frac{t^{2}}{10} + \frac{t}{25}) e^{- 5 t}$ .

1Step 1: Find a particular solution.

Given the differential equation is $ty'' + (5 t - 1) y' - 5 y = t^{2} e^{- 5 t}$

And $y_{1} = 5 t - 1, y_{2} = e^{- 5 t}$

$y_{h} (t) = c_{1} (5 t - 1) + c_{2} e^{- 5 t}$

The particular solution is $y_{p} = v_{1} (t) (5 t - 1) + v_{2} (t) e^{- 5 t} .$

2Step 2: Evaluate v 1    and    v 2

Here $y_{p} = v_{1} (t) (5 t - 1) + v_{2} (t) e^{- 5 t}$

$\begin{matrix} v_{1}' = \frac{- f (t) y_{2} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} \\ = \frac{- t^{2} e^{- 5 t} . e^{- 5 t}}{t [(5 t - 1) (- 5 e^{- 5 t}) - 5 . e^{- 5 t}]} \\ = \frac{e^{- 5 t}}{25} \end{matrix}$

Now integrate the above result, we get:

$\begin{array}{l} v_{1} (t) = \int \frac{e^{- 5 t}}{25} dt \\ = - \frac{e^{- 5 t}}{125} \end{array}$

3Step 3: Determine v 2 ' and v 2

$\begin{matrix} v_{2}' = \frac{f (t) y_{1} (t)}{a [y_{1} (t) y'_{2} (t) - y'_{1} (t) y_{2} (t)]} \\ = \frac{- t^{2} e^{- 5 t} (5 t - 1)}{t [(5 t - 1) (- 5 e^{- 5 t}) - 5 . e^{- 5 t}]} \\ = - \frac{5 t - 1}{25} \end{matrix}$

Integrate the above result.

$\begin{array}{l} v_{2} (t) = \int - \frac{5 t - 1}{25} dt \\ = - \frac{t^{2}}{10} + \frac{t}{25} \end{array}$

Therefore a particular solution is $y_{p} = - \frac{e^{- 5 t}}{125} (5 t - 1) + (- \frac{t^{2}}{10} + \frac{t}{25}) e^{- 5 t}$

Thus, the general solution is:

$\begin{matrix} y (t) = y_{h} (t) + y_{p} (t) \\ y (t) = c_{1} (5 t - 1) + c_{2} e^{- 5 t} - \frac{e^{- 5 t}}{125} (5 t - 1) + (- \frac{t^{2}}{10} + \frac{t}{25}) e^{- 5 t} \end{matrix}$

Q24E

Q1E

Other exercises in this chapter

Q23E

In Problems 22 through 25, use a variation of parameters to find a general solution to the differential equation given that the functions y1 and y2are linearly

View solution

Q24E

In Problems 22 through 25, use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2

View solution

Q1E

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions

View solution

Q2E

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions

View solution