"Determine the values of r for which the..." was answered by students like you

Cyrus Munoz

Open question

2022-08-19

Determine the values of r for which the given differential equation has solutions of the form

y = t^{r} f or t > 0 . t^{2} y^{″} + 4 t y^{'} + 2 y = 0

Dale Rose

Beginner2022-08-20Added 7 answers

Plug

t^{r}

into the differential equation. Recall that

\frac{d}{d t} t^{r} = r t^{r - 1}

and

\frac{d^{2}}{{d t}^{2}} t^{r} = \frac{d}{d t} r t^{r - 1} = r (r - 1) t^{r - 2}

t^{2} y^{″} + 4 t y^{'} + 2 y = 0

t^{2} [r (r - 1) t^{r - 2}] + 4 t [r t^{r - 1}] + 2 t^{r} = 0

r (r - 1) t^{r} + 4 r t^{r} + 2 t^{r} = 0

Since t>0 we can divide both sides by

t^{r}

r (r - 1) t^{r} + 4 r t^{r} + 2 t^{r} = 0

r(r-1)+4+2=0

r^{2} - r + 4 r + 2 = 0

r^{2} + 3 r + 2 = 0

Solve this polynomial for r. This implies that

t^{r}

is a solution only if r=-1 or r=-2.

r^{2} + 3 r + 2 = 0

(r + 1) (r + 2) = 0 \Rightarrow r = - 1, r = - 2

Result:
r=-1 and r=-2

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