$\text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. }$
A) $y = a _ { 0 } + e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant
B) $y = a _ { 0 } e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant
C) $y = a _ { 0 } e ^ { x }$, where $a _ { 0 }$ is an arbitrary constant
D) $y = 3 + a _ { 0 } e ^ { z }$, where $a _ { 0 }$ is an arbitrary constant
E) $y = a _ { 0 } + 3 e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant

Question

$\text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. }$ 
A) $y = a _ { 0 } + e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant
B) $y = a _ { 0 } e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant
C) $y = a _ { 0 } e ^ { x }$, where $a _ { 0 }$ is an arbitrary constant
D) $y = 3 + a _ { 0 } e ^ { z }$, where $a _ { 0 }$ is an arbitrary constant
E) $y = a _ { 0 } + 3 e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant

Quizplus · Accepted Answer

Assume that $y = \sum_{n=0}^\infty a_nx^n$ is a power series solution to the differential equation. Then, we have $$\begin{aligned} y' &= \sum_{n=1}^\infty na_nx^{n-1} \ y'' &= \sum_{n=2}^\infty n(n-1)a_nx^{n-2} \end{aligned}$$ Plugging these expressions into the differential equation, we get $$\sum_{n=2}^\infty n(n-1)a_nx^{n-2} - 3\sum_{n=0}^\infty a_nx^n=0.$$ Shifting the index of summation in the first term, we get $$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n} - 3\sum_{n=0}^\infty a_nx^n=0.$$ Equating the coefficient of each power of $x$, we get the recursive equation $$a_{n+2} = \frac{3}{(n+2)(n+1)}a_n$$ with initial conditions $a_0$ and $a_1$. Notice that the recursive equation does not depend on the value of $a_1$. Therefore, the solution to the differential equation is the power series $$\sum_{n=0}^\infty a_nx^n = a_0 \sum_{n=0}^\infty \frac{(3x)^n}{n!} = a_0e^{3x}.$$ Thus, the best choice is (B) $y = a_0e^{3x}$, where $a_0$ is an arbitrary constant.

Use a power series to solve the differential equation yt−3y=0. \text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. } Use a power series to solve the differential equation yt−3y=0.

$\text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. }$