For what values of r does the function y = $e^{r x}$ $e^(rx)$ satisfy the differential equation 5y'' + 14y' − 3y = 0?

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For what values of r does the function y = $e^{r x}$ $e^(rx)$ satisfy the differential equation 5y'' + 14y' − 3y = 0?

a) For what values of r does the function y = $e^{r x}$ $e^(rx)$ satisfy the differential equation 5y'' + 14y' − 3y = 0? (Enter your answers as a comma-separated list.)

r = ???

(b) If $r_{1}$ $r_1$ and $r_{2}$ $r_2$ are the values of r that you found in part (a), show that every member of the family of functions y = $a e_{1}^{r_{1} x} + b e_{2}^{r_{2} x}$ $ae^(r_1x) + be^(r_2x)$ is also a solution. (Let $r_{1}$ $r_1$ be the larger value and $r_{2}$ $r_2$ be the smaller value.)

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Answer 1 · 2017-06-29T05:57:10

(a) r = $\frac{1}{5}$ $1/5$ , -3

(b) Shown in explanation.

Explanation:

(b) We replace y with $a e_{1}^{r_{1} x} + b e_{2}^{r_{2} x}$ $ae^(r_1 x) + be^(r_2 x)$ . First, we must figure out the values for both y' and y''.

y = $a e^{\frac{x}{5}} + b e^{- 3 x}$ $ae^(x/5) + be^(-3x)$
y' = $\frac{1}{5} a e^{\frac{x}{5}} - 3 b e^{- 3 x}$ $1/5 ae^(x/5) - 3 be^(-3x)$
y'' = $\frac{1}{25} a e^{\frac{x}{5}} - 9 b e^{- 3 x}$ $1/25 ae^(x/5) - 9 be^(-3x)$

Now we can insert the values in our differential equation 5y'' + 14y' − 3y = 0.

$5 [\frac{1}{25} a e^{\frac{x}{5}} - 9 b e^{- 3 x}] + 14 [\frac{1}{5} a e^{\frac{x}{5}} - 3 b e^{- 3 x}] - 3 [a e^{\frac{x}{5}} + b e^{- 3 x}] = 0$ $5[1/25 ae^(x/5) - 9 be^(-3x)] + 14[1/5 ae^(x/5) - 3 be^(-3x)] − 3[ae^(x/5) + be^(-3x)] = 0$

$a [\frac{1}{5} e^{\frac{x}{5}} + \frac{14}{5} e^{\frac{x}{5}} - 3 e^{\frac{x}{5}}] + b [45 e^{- 3 x} - 42 e^{- 3 x} - 3 e^{- 3 x}] = 0$ $a[1/5 e^(x/5) + 14/5 e^(x/5) - 3 e^(x/5)] + b[45 e^(-3x) - 42 e^(-3x) - 3 e^(-3x)] = 0$

$a [3 e^{\frac{x}{5}} - 3 e^{\frac{x}{5}}] + b [45 e^{- 3 x} - 42 e^{- 3 x} - 3 e^{- 3 x}] = 0$ $a[3 e^(x/5) - 3 e^(x/5)] + b[45 e^(-3x) - 42 e^(-3x) - 3 e^(-3x)] = 0$

$a [3 e^{\frac{x}{5}} - 3 e^{\frac{x}{5}}] + b [3 e^{- 3 x} - 3 e^{- 3 x}] = 0$ $a[3 e^(x/5) - 3 e^(x/5)] + b[3 e^(-3x) - 3 e^(-3x)] = 0$

a[0] + b[0] = 0

0 = 0

Since the solutions are both 0, it means that every member of the family of functions in y = $a e_{1}^{r_{1} x} + b e_{2}^{r_{2} x}$ $ae^(r_1 x) + be^(r_2 x)$ is a solution to 5y'' + 14y' − 3y = 0.

Answer 2 · 2017-06-29T06:02:26

$r = \frac{1}{5}$ $r=1/5$ or $- 3$ $-3$ ; For details please see below.

Explanation:

(a) As $y = e^{r x}$ $y=e^(rx)$

$y'=(dy)/(dx)=re^(rx)$

and $y''=r^2e^(rx)$

Hence $5y'' + 14y' − 3y = 0$ can be written as

$5r^2e^(rx)+14re^(rx)-3e^(rx)-0$

or $5r^2+14r-3=0$

or $5r^2+15r-r-3=0$

or $(5r-1)(r+3)=0$

i.e. $r=1/5$ or $-3$

(b) Consider the two solutions as $r_1$ and $r_2$ , then we have $5r_1^2+14r_1-3=0$ and $5r_2^2+14r_2-3=0$

As $y=ae^(r_1x)+be^(r_2x)$ , then

$y'=ar_1e^(r_1x)+br_2e^(r_2x)$

and $y''=ar_1^2e^(r_1x)+br_2^2e^(r_2x)$

$:.$ $5y''+14y'−3y=5ar_1^2e^(r_1x)+5br_2^2e^(r_2x)+14ar_1e^(r_1x)+14br_2e^(r_2x)-3ae^(r_1x)-3be^(r_2x)$

= $ae^(r_1x)(5r_1^2+14r_1-3)+be^(r_2x)(5r_2^2+14r_2-3)$

= $0$

Hence $y=ae^(r_1x)+be^(r_2x)$ is also a solution of $5y''+14y'−3y=0$

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