Use power series to solve the differential equation. $$y ^ { t } = 4 x y .$$
A)$y ( x ) = c e ^ { - \frac { 4 x ^ { 2 } } { 2 } }$
B) $y ( x ) = c e ^ { \frac { 4 x ^ { 2 } } { 2 } }$
C) $y ( x ) = e ^ { - \frac { x ^ { 2 } } { 2 } }$
D) $y ( x ) = c e ^ { 4 x }$
E)$y ( x ) = 4 e ^ { \frac { x ^ { 2 } } { 2 } }$

Question

Quizplus · Accepted Answer

We can write the given differential equation as
$$y'=4xy.$$ 
Let us assume that $y$ can be written as a power series about $x=0$ as 
$$y(x)=\sum_{n=0}^\infty a_nx^n.$$ 
Differentiating $y$ we get 
$$y'(x)=\sum_{n=1}^\infty na_nx^{n-1}.$$ 
Using these in the differential equation, we get 
$$\sum_{n=1}^\infty na_nx^{n-1}=4x\sum_{n=0}^\infty a_nx^n.$$ 
Rearranging terms, we get 
$$\sum_{n=1}^\infty na_nx^{n-1}=4\sum_{n=0}^\infty a_nx^{n+1}.$$ 
Comparing coefficients of $x^n$ on both sides, we get 
$$na_n=4a_{n-2}$$ 
for $n\geq 2$ and $0a_0=0,$ and $a_1$ is arbitrary. 
Using $a_0=0,$ we get $a_2=0,$ $a_4=0,$ $a_6=0,$ and so on. 
Using $a_1=c,$ we get $a_3=\frac{4}{2\cdot 1}c=2c,$ $a_5=\frac{4}{4\cdot 2}2c=\frac{c}{2},$ $a_7=\frac{4}{6\cdot 3}\frac{c}{2}=\frac{c}{9},$ and so on. 
Thus, we have 
$$y(x)=c\left(1+2x^3+\frac{1}{2}x^5+\frac{1}{9}x^7+\cdots\right)=ce^{\frac{4x^2}{2}}.$$ 
Therefore, the best choice is B.

Use Power Series to Solve the Differential Equation yt=4xy.y ^ { t } = 4 x y .yt=4xy.

Use Power Series to Solve the Differential Equation $y ^ { t } = 4 x y .$