 # For what values of r does the function y = Mary Jackson 2021-12-19 Answered
For what values of r does the function $y={e}^{rx}$ satisfy the differential equation $y{}^{″}-4{y}^{\prime }+y=0?$
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Step 1
1) Find ${y}^{\prime }$ and $y{}^{″}$
2) Substitute the expressions for in the given differential equation
3) Solve for r
Step 2
It is given that
$y={e}^{rx}$
Differentiate with respect to x
${y}^{\prime }=\frac{d\left({e}^{rx}\right)}{dx}$
The Chain Rule for differential
$\frac{d\left[f\left[g\left(x\right)\right]\right]}{dx}=\frac{d\left[f\left[g\left(x\right)\right]\right]}{d\left[g\left(x\right)\right]}×\frac{d\left[g\left(x\right)\right]}{dx}$
$\frac{dy}{dx}=\frac{d\left({e}^{rx}\right)}{d\left(rx\right)}×\frac{d\left(rx\right)}{dx}$
${y}^{\prime }={e}^{rx}×\left(r×1×{x}^{1-1}\right)$
Remember that:
1) the derivative of ${e}^{x}$ is ${e}^{x}$ itself
2) the derivative of ${x}^{n}$ is $n{x}^{n-1}$, this is known as power rule
${y}^{\prime }=r{e}^{rx}$
Step 3
Note that:
${y}^{\prime }=ry$
Deffirntiate again, to get
$y{}^{″}=r{y}^{\prime }=r×r{e}^{rx}={r}^{2}{e}^{rx}$
Substitute the expressions for in the following differential equation
$y{}^{″}-4{y}^{\prime }+y=0$
${r}^{2}{e}^{rx}-4r{e}^{rx}+{e}^{rx}=0$
Since ${e}^{rx}$ is never 0, we can divide both sides by it
${r}^{2}-4r+1=0$
add 3 to both sides
${r}^{2}-4r+4=3$
Notice that the LHS is a perfect square
${\left(r-2\right)}^{2}=3$
Take the square root of both sides
$r-2=±\sqrt{3}$
Add 2 to both sides
$r=2±\sqrt{3}$

We have step-by-step solutions for your answer! Thomas White
Step 1
Derivatives.
${y}^{\prime }=r{e}^{rx}$
$y{}^{″}={r}^{2}{e}^{rx}$
Step 2
Substitute into differential equation
${r}^{2}{e}^{rx}-4r{e}^{rx}+{e}^{rx}=0$
Step 3
Factor
${e}^{rx}\left({r}^{2}-4r+1\right)=0$
Step 4
Solve:
${e}^{rx}=0$ or ${r}^{2}-4r+1=0$
${e}^{rx}=0$ does not exist
${r}^{2}-4r+1=0$ must be solved with quadratic formula.
Step 5
$r=\frac{--4±\sqrt{{\left(-4\right)}^{2}-4×1×1}}{2×1}$
$r=\frac{4±\sqrt{16-4}}{2}$
$r=2±\frac{\sqrt{12}}{2}$
$r=2±\sqrt{3}$

We have step-by-step solutions for your answer! nick1337

Step 1
$y=1{e}^{rx}$
${y}^{\prime }=r{e}^{rx}$
${y}^{″}={r}^{2}{e}^{rx}$
${r}^{2}-4r+1=0$
$r=\frac{\left[4±\sqrt{16-4}\right]}{2}$
$r=\left[4±2\sqrt{3}\right]2$
$r=2±\sqrt{3}$

We have step-by-step solutions for your answer!