Solve the differential equation $$10 y ^ { tt } - x y ^ { t } - y = 0$$
A) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } 2 ^ { n } n ! }$, where $a _ { 0 }$ is an arbitrary constant
B) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } n ! } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { ( n + 1 ) ! }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants
C) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } 2 ^ { n } n ! } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { 10 ^ { n } \cdot 1 \cdot 3 \cdot 5 \cdot 7 \cdots ( 2 n + 1 ) }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants
D) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } 2 ^ { n + 1 } n ! } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { 1 \cdot 3 \cdot 5 \cdot 7 \cdots ( 2 n + 1 ) }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants
E) $y = a _ { 0 } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } \cdot 1 \cdot 3 \cdot 5 \cdot 7 \cdots ( 2 n + 1 ) }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants

Question

Solve the differential equation $$10 y ^ { tt } - x y ^ { t } - y = 0$$ 
A) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } 2 ^ { n } n ! }$, where $a _ { 0 }$ is an arbitrary constant
B) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } n ! } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { ( n + 1 ) ! }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants
C) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } 2 ^ { n } n ! } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { 10 ^ { n } \cdot 1 \cdot 3 \cdot 5 \cdot 7 \cdots ( 2 n + 1 ) }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants
D) $y = a _ { 0 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } 2 ^ { n + 1 } n ! } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n + 1 } } { 1 \cdot 3 \cdot 5 \cdot 7 \cdots ( 2 n + 1 ) }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants
E) $y = a _ { 0 } + a _ { 1 } \sum _ { n = 0 } ^ { \infty } \frac { x ^ { 2 n } } { 10 ^ { n } \cdot 1 \cdot 3 \cdot 5 \cdot 7 \cdots ( 2 n + 1 ) }$, where $a _ { 0 }$ and $a _ { 1 }$ are arbitrary constants

Quizplus · Accepted Answer

We can start by assuming the solution to be of the form $$y = \sum_{n=0}^{\infty}a_nx^n$$Then we have $$y' = \sum_{n=0}^{\infty}na_nx^{n-1}$$and $$y'' = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$$Substituting these expressions into the given differential equation, we get $$10\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} - \sum_{n=0}^{\infty}xa_nx^{n-1} - \sum_{n=0}^{\infty}a_nx^n = 0$$Simplifying, rearranging and collecting coefficients of like powers of x, we get $$10a_2 = a_0 \quad\quad (n=0)$$$$10(n+2)(n+1)a_{n+2} - (n+1)a_n - a_{n-2} = 0 \quad\quad (n\geq1)$$The first equation gives $$a_2 = \frac{1}{5}a_0$$Using this to find $a_4$ in the second equation, we get $$a_4 = \frac{1}{7\cdot10}a_0$$Similarly, we can find $$a_6 = \frac{1}{9\cdot10^2\cdot2}a_0$$and in general, $$a_{2n} = \frac{1}{(2n+1)\cdot10^n\cdot2^n}a_0$$Now, let's look at the odd-powered terms. Using the second equation with $n=0$, we get $$10a_2-a_0 = 0 \Rightarrow a_2 = \frac{1}{10}a_0$$Using this to find $a_3$ in the second equation, we get $$a_3 = \frac{1}{3\cdot10}a_0$$Similarly, we can find $$a_5 = \frac{1}{5\cdot10^2\cdot3}a_0$$and in general, $$a_{2n+1} = \frac{1}{(2n+2)(2n+1)\cdot10^n\cdot1\cdot3\cdot5\cdot7\cdots(2n+1)}a_0$$Therefore, the general solution is $$y = a_0\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n+1)\cdot10^n\cdot2^n}+a_1\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+2)(2n+1)\cdot10^n\cdot1\cdot3\cdot5\cdot7\cdots(2n+1)}$$which matches choice (C).

Solve the Differential Equation 10ytt−xyt−y=010 y ^ { tt } - x y ^ { t } - y = 010ytt−xyt−y=0

Solve the Differential Equation $10 y ^ { tt } - x y ^ { t } - y = 0$