# -b.1.3.19 Determine the values of r for t^2y''+4ty'+2y = 0

• MHB
• karush
In summary, the values of $r$ for which the given differential equation has solutions of the form $y = t^r$ for $t > 0$ are $r = -1, -2$ for the first equation and $r = 1, 4$ for the second equation. These values are found by substituting the given form for $y$ and its derivatives into the equation and solving for $r$. There may be some typos or minor errors in the given work, but the final values are accurate.
karush
Gold Member
MHB
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$\textsf{Determine the values of$r$for which the given differential equation has solutions of the form$ y = t^r$for$t > 0 $}$
$t^2y''+4ty'+2y = 0$
$\color{red}{r=-1,-2}$
$t^2y''-4ty'+4y=0$
$\color{red}{r=1,4}$
ok the book answers are in red, and obviously this is from a factored quadradic but I couldn't an example of the steps.
probably just a couple!

Last edited:
karush said:
$\textsf{Determine the values of$r$for which the given differential equation has solutions of the form$ y = t^r$for$t > 0 $}$
$t^2y''+4ty'+2y = 0$
$\color{red}{r=-1,-2}$
$t^2y''-4ty'+4y=0$
$\color{red}{r=1,4}$
ok the book answers are in red, and obviously this is from a factored quadradic but I couldn't an example of the steps.
probably just a couple!

Okay, if we are to consider a solution of the form:

$$\displaystyle y=t^r$$

Then:

$$\displaystyle y'=rt^{r-1}$$

$$\displaystyle y''=r(r-1)t^{r-2}$$

And thus the ODE becomes:

$$\displaystyle t^2r(r-1)t^{r-2}+4trt^{r-1}+2t^r=0$$

Or:

$$\displaystyle r(r-1)t^{r}+4rt^{r}+2t^r=0$$

$$\displaystyle t^r\left(r(r-1)+4r+2\right)=0$$

Now, if we exclude the trivial solution $$y\equiv0$$ this leaves:

$$\displaystyle r(r-1)+4r+2=0$$

$$\displaystyle r^2+3r+2=0$$

$$\displaystyle (r+1)(r+2)=0$$

Thus:

$$\displaystyle r\in\{-2,-1\}$$

Can you do the second one?

given
$t^2y''-4ty'+4y=0$
let
$\displaystyle y=t^r, \quad y''=r(r-1)t^{r-2}, \quad y''=r(r-1)t^{r-2}$
then
$t^2(r(r-1)t^{r-2}-4rt^{r-1}+4t^r=0$
simplify
$(r(r-1)t^r-4rt^{r-1}+4t^r=r(r-1)-4r+4=r^2-5r+4=0$
factor
(r-1)(r-4)=0
thus
$r\in{1,4}$typos maybe

There are some typos/issues with your work. See if you can spot them...:)

$\displaystyle y=t^r, \quad y'=rt^{r-1}, \quad y''=r(r-1)t^{r-2}$
then
$t^2(r(r-1)t^{r-2})-4t(rt^{r-1})+4(t^r)=0$
then simplify
$(r - 4) (r - 1) t^r=0$
so $r\in\{1,4\}$

Last edited:

## 1. What does the equation "-b.1.3.19 Determine the values of r for t^2y''+4ty'+2y = 0" represent?

The equation represents a second-order linear differential equation, where the values of r need to be determined in order to solve for the function y(t).

## 2. What is the significance of the values of r in this equation?

The values of r determine the type of solution for the differential equation. If r is a real number, the solution will be a combination of exponential and polynomial functions. If r is a complex number, the solution will involve trigonometric functions.

## 3. How do you determine the values of r for this equation?

In order to determine the values of r, we can use the characteristic equation r^2 + 4r + 2 = 0. By solving this quadratic equation, we can find the two values of r that will satisfy the original differential equation.

## 4. Can you give an example of a solution for this equation?

One possible solution for this equation is y(t) = e^(-2t)cos(t). This solution satisfies the differential equation and has the values of r = -2 ± i.

## 5. What are the applications of this type of differential equation?

This type of differential equation is commonly used in physics and engineering to model systems that involve oscillations or vibrations, such as a mass-spring system or an electrical circuit with an inductor and capacitor.

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