Question

(a) For what values of r does the function y = erx satisfy the differential equation 5y'' + 34y' − 7y = 0? (Enter your answers as a comma-separated list.) r = ___.b) If r1 and r2 are the values of r that you found in part (a), show that every member of the family of functions y = aer1x + ber2x is also a solution. (Let r1 be the larger value and r2 be the smaller value.)

Answer 1

Answer:

a) r=-7 and r=1/5

b) See proof below

Step-by-step explanation:

Let y(x) be the function

$\large y(x)=e^{rx}$

Then

$\large y'(x)=re^{rx}\\y''(x)=r^2e^{rx}$

a)

Replacing these values in the differential equation

5y'' + 34y' − 7y = 0

we get

$\large 5r^2e^{rx}+34re^{rx}-7e^{rx}=0$

dividing both sides by $\large e^{rx}$ we get the quadratic equation

$\large 5r^2+34r-7=0$

whose solutions are r = -7 and r=1/5

b)

We must show that for any constants a, b, the function

$\large y(x)=ae^{-7x}+be^{x/5}$

 

is also a solution of the differential equation

We have

$\large y'(x)=-7ae^{-7x}+\frac{be^{x/5}}{5}\\\\y''(x)=49ae^{-7x}+\frac{be^{x/5}}{25}$

So

$\large 5y'' + 34y'-7y=\\\\ =5 \left( 49ae^{-7x}+\frac{be^{x/5}}{25}\right)+34\left( -7ae^{-7x}+\frac{be^{x/5}}{5}\right)-7\left( ae^{-7x}+be^{x/5}\right)=\\\\=245ae^{-7x+}\frac{be^{x/5}}{5}-238ae^{-7x}+\frac{34be^{x/5}}{5}-7ae^{-7x}-7be^{x/5}=\\\\=ae^{-7x}(245-238-7)+be^{x/5}(1/5+34/5-7)=0$

Hence

$\large y(x)=ae^{-7x}+be^{x/5}$

is also a solution for any values of a and b.

Answer 2

a) Roots of the differential equation are $r_{1} = \frac{1}{5}$ and $r_{2} = -7$.

b) Any expression of the form $y = A\cdot e^{r_{1}\cdot x }+B\cdot e^{r_{2}\cdot x}$ is a solution of this equation, since they belong to the family of solutions of the form $y = C_{1}\cdot e^{\frac{1}{5}\cdot x }+C_{2}\cdot e^{-7\cdot x}$.

Analysis of a homogeneous second order differential equation

a) The differential equation is equivalent to $y'' +\frac{34}{5}\cdot y'-\frac{7}{5} \cdot y = 0$, whose solution is $y = C_{1}\cdot e^{\frac{1}{5}\cdot x }+C_{2}\cdot e^{-7\cdot x}$, where $C_{1}$ and $C_{2}$ are constants. In this case, we notice that roots are $r_{1} = \frac{1}{5}$ and $r_{2} = -7$. $\blacksquare$

b) Since the family of solutions are functions of the form $y = C_{1}\cdot e^{\frac{1}{5}\cdot x }+C_{2}\cdot e^{-7\cdot x}$ and $r_{1} = \frac{1}{5}$ and $r_{2} = -7$. Then, we conclude that any expression of the form $y = A\cdot e^{r_{1}\cdot x }+B\cdot e^{r_{2}\cdot x}$ is a solution of this equation. $\blacksquare$

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