Question

Let t^2y″+7ty′+5y=0.

Answer 1

Answer:

The values of r are -1 and -5.

Step-by-step explanation:

Since it is given that $y=t^r$ satisfy's the differential equation we put the 'y' in the equation and equate to 0

Also we have

$y'=\frac{dt^r}{dt}=rt^(r-1)\\\\y''=\frac{d^{2}t^r}{dt^2}\\\\y''=r(r-1)t^{r-2}$

Using the values in the above equation we get

$t^2\times r(r-1)t^{r-2}+7t\times r\times t^{r-1}+5t^{r}=0\\\\r(r-1)t^r+7rt^r+5t^r=0\\\\r(r-1)+7r+5=0.........(i)$

Since equation 'i' is a quadratic equation it is solved as under

$r^2-r+7r+5=0\\\\r^2+6r+5=0\\\\r=\frac{-6\pm \sqrt{6^2-4\times 1\times 5} }{2}\\\\r_1=-1\\\\r_2=-5$

Answer 2

Answer:

The values of r are -1 and -5.

Step-by-step explanation: