MHB Do we have to write them separately?

[evinda]

Hello! (Wave)The differential equation $y''+xy=0$ is given.

I want to find the solution of the differential equation, using the power series method.

That's what I have tried:We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$.Then:

$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$

$$y''(x)= \sum_{n=1}^{\infty} (n+1) n a_{n+1} x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n$$Thus:$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n+ x \sum_{n=0}^{\infty} a_n x^n=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=0}^{\infty} a_n x^{n+1}=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=1}^{\infty} a_{n-1} x^n=0 \\ \Rightarrow 2a_2+\sum_{n=1}^{\infty} \left[ (n+2) (n+1) a_{n+2}+ a_{n-1}\right] x^n=0$$So it has to hold:

$$a_2=0 \\ (n+2) (n+1) a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots$$For $n=1$: $3 \cdot 2 \cdot a_3+ a_0=0 \Rightarrow a_3=-\frac{a_0}{6}$

For $n=2$: $4 \cdot 3 \cdot a_4+a_1=0 \Rightarrow a_4=-\frac{a_1}{12}$

For $n=3$: $5 \cdot 4 \cdot a_5+a_2=0 \Rightarrow a_5=0$

For $n=4$: $6 \cdot 5 \cdot a_6+a_3=0 \Rightarrow 30 a_6-\frac{a_0}{6}=0 \Rightarrow a_6=\frac{a_0}{6 \cdot 30}=\frac{a_0}{180}$

For $n=5$: $7 \cdot 6 \cdot a_7+ a_4=0 \Rightarrow 7 \cdot 6 \cdot a_7-\frac{a_1}{12}=0 \Rightarrow a_7=\frac{a_1}{12 \cdot 42}$Will it be as follows?$$a_{3k+2}=0$$

$$a_{3k}=(-1)^k \frac{a_0}{(3k)!} \prod_{i=0}^{k-1} (3i+1)$$$$a_{3k+1}=(-1)^k \frac{a_1}{(3k+1)!} \prod_{i=0}^{k-1} (3i+2)$$

If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ? (Thinking)

 

[Sudharaka]

Hello! (Wave)The differential equation $y''+xy=0$ is given.

I want to find the solution of the differential equation, using the power series method.

That's what I have tried:We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$.Then:

$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$

$$y''(x)= \sum_{n=1}^{\infty} (n+1) n a_{n+1} x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n$$Thus:$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n+ x \sum_{n=0}^{\infty} a_n x^n=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=0}^{\infty} a_n x^{n+1}=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=1}^{\infty} a_{n-1} x^n=0 \\ \Rightarrow 2a_2+\sum_{n=1}^{\infty} \left[ (n+2) (n+1) a_{n+2}+ a_{n-1}\right] x^n=0$$So it has to hold:

$$a_2=0 \\ (n+2) (n+1) a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots$$For $n=1$: $3 \cdot 2 \cdot a_3+ a_0=0 \Rightarrow a_3=-\frac{a_0}{6}$

For $n=2$: $4 \cdot 3 \cdot a_4+a_1=0 \Rightarrow a_4=-\frac{a_1}{12}$

For $n=3$: $5 \cdot 4 \cdot a_5+a_2=0 \Rightarrow a_5=0$

For $n=4$: $6 \cdot 5 \cdot a_6+a_3=0 \Rightarrow 30 a_6-\frac{a_0}{6}=0 \Rightarrow a_6=\frac{a_0}{6 \cdot 30}=\frac{a_0}{180}$

For $n=5$: $7 \cdot 6 \cdot a_7+ a_4=0 \Rightarrow 7 \cdot 6 \cdot a_7-\frac{a_1}{12}=0 \Rightarrow a_7=\frac{a_1}{12 \cdot 42}$Will it be as follows?$$a_{3k+2}=0$$

$$a_{3k}=(-1)^k \frac{a_0}{(3k)!} \prod_{i=0}^{k-1} (3i+1)$$$$a_{3k+1}=(-1)^k \frac{a_1}{(3k+1)!} \prod_{i=0}^{k-1} (3i+2)$$

If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ? (Thinking)

Hi evinda, :)

Yes you do. So the final solution will be,

\[y(x)=\sum_{k=0}^{\infty}a_{3k}x^{3k}+\sum_{k=0}^{\infty}a_{3k+1}x^{3k+1}+\sum_{k=0}^{\infty}a_{3k+2}x^{3k+2}\]

And of course the first summation would be zero since all the coefficients are zero and the for the second and third summations you can substitute the $a_{3k+1}$ and $a_{3k+2}$ values that you have obtained.