Question

Let t^2y''+10ty'+8y=0Find all values of r such that y = tr satisfies the differential equation for t > tr0. If there is more than one correct answer, enter your answers as a comma separated list.

Answer 1

Answer:

The correct question is

Let $t^2y''+10ty'+8y=0$. Find all values of r such that $y = t^r$ satisfies the differential equation for t > 0. If there is more than one correct answer, enter your answers as a comma separated list.

The answer is r = 8, r = 1

Step-by-step explanation:

By substituting     $y = t^r$  into the given equation $t^2y''+10ty'+8y=0$.

That is substituting $y = t^r$ , $y' = rt^{r-1}$, $y'' = r(r-1)t^{r-2}$ into the given equation, we have

$t^2 r(r-1)t^{r-2} +10t rt^{r-1}+8 t^r =0$

Implies that $r(r-1)t^r +10rt^r + 8 t^r =0$

Implies that  $[r(r-1) +10r + 8]t^r = 0$

Implies that  $[r^2 - r +10r + 8]t^r = 0$

Implies that  $[r^2 - 9r + 8]t^r = 0$

Implies that  $r^2 - 9r + 8 = 0$, assuming $t^r \ne 0$  being the assumed non- trivial solution.

The by solving this quadratic equation $r^2 - 9r + 8 = 0$ using factorization method, we have

$r^2$  – r – 8r  + 8 = 0

Implies that r(r – 1) – 8(r – 1) = 0

Implies that (r – 8)(r – 1) = 0

Implies that r – 8 = 0 or r – 1 = 0

Implies that r = 8 or r = 1

Therefore, the value of r that satisfies the given equation is r = 8, r = 1.