Answer:
The values of r are -1 and -5.
Step-by-step explanation:
Since it is given that [tex]y=t^r[/tex] satisfy's the differential equation we put the 'y' in the equation and equate to 0
Also we have
[tex]y'=\frac{dt^r}{dt}=rt^(r-1)\\\\y''=\frac{d^{2}t^r}{dt^2}\\\\y''=r(r-1)t^{r-2}[/tex]
Using the values in the above equation we get
[tex]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)[/tex]
Since equation 'i' is a quadratic equation it is solved as under
[tex]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[/tex]
Answer:
The values of r are -1 and -5.
Step-by-step explanation:
