Q23E

Question

In Problems 22 through 25, use a 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.

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

Step-by-Step Solution

Verified

Answer

The general solution is $y (t) = c_{1} t^{2} + c_{2} t^{3} + (- t + \frac{t^{- 2}}{2}) t^{2} + (\ln | t | - \frac{t^{- 3}}{3}) t^{3}$ .

1Step 1: Find a particular solution.

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

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

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

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

2Step 2: Evaluate v 1    and    v 2 .

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

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

Now integrate the above result.

$\begin{matrix} v_{1} (t) = \int \frac{(t + 1)}{[e^{t}]} dt \\ = - (t + 2) e^{- 1} \end{matrix}$

3Step 3: Determine v ' 2 and v 2

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