Q45E

Question

Find a particular solution to the non-homogeneous equation $ty'' - (t + 1) y' + y = t^{2} e^{2 t}$ , given that $f (t) = e^{t}$ is a solution to the corresponding homogeneous equation.

Step-by-Step Solution

Verified

Answer

The solution for the given homogeneous equation $ty'' - (t + 1) y' + y = t^{2} e^{2 t}$ is:

$y = c_{1} e^{t} - c_{2} \frac{t + 1}{e^{2 t}}$ .

1Step 1: Finding y

Given differential equation is $ty'' - (t + 1) y' + y = t^{2} e^{2 t}$ then the standard form of the solution is $y'' - \frac{t + 1}{t} y^{'} + \frac{1}{t} y = {te}^{2 t}$ and $f (t) = e^{t}$ is one of the solutions and $p (t) = - \frac{t + 1}{t}$ .

Then $y_{2} = y_{1} (t) \int \frac{e^{- \int p (t)} dt}{y_{1}^{2} (t)} dt$

2Step 2: Simplification

Simplify the above

$= e^{t} \int \frac{e^{\int \frac{1 + t}{t}} dt}{{(e^{t})}^{2}} dt = e^{t} \int \frac{e^{\ln (t^{1}) + t}}{e^{2 t}} dt = e^{t} \times \int \frac{t \times e^{t}}{e^{2 t}} dt = e^{t} \times \frac{(- t - 1) e^{- t}}{e^{2 t}} = - \frac{t + 1}{e^{2 t}}$

So, the solution is $y = c_{1} e^{t} - c_{2} \frac{t + 1}{e^{2 t}}$ .

Q44E

Q46 E

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