# Power Series Solution to Differential Equation with Initial Conditions

I am trying to find the solution to $y''-2xy'+10y=$ given $y(0)=0, y'(0)=3$.

My work so far: $$\sum_{n=2}^\infty n(n-1)a_nx^{n-2} -2 \sum_{n=1}^\infty na_nx^{n-1} +10\sum_{n=0}^\infty a_nx^{n}=0$$ $$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n} -2\sum_{n=0}^\infty na_nx^{n} +10\sum_{n=0}^\infty a_nx^{n}=0$$ $$\sum_{n=0}^\infty [(n+2)(n+1)a_{n+2} -2 na_n+10 a_n]x^n=0$$ $$\sum_{n=0}^\infty [(n+2)(n+1)a_{n+2} +8 a_n]x^n=0$$ $$(n+2)(n+1)a_{n+2} +8a_n=0, n=0,1,2,...$$ $$a_{n+2} = \frac{-8a_n}{(n+1)(n+2)}, n=0,1,2,...$$ Next step is finding the recurrence relation and the generalized form for $a_n$, but I am having trouble with this. I know to plug in $n=0$, $n=1$, and so on until the relation can be reduced to $a_0$ and $a_1$ terms, but I cannot seem to successfully find the general form. Any hints or suggestions?

• I think you forgot to include the factor $x$ from $-2xy'$. – hamam_Abdallah May 21 '17 at 18:25
• I think the OP did it in the second line, but should not have switched the starting index (it does not matter though). It does affect the first term in the power series though $2a_2 + 10a_0 = 0$, the recurrence relation should be different though I think. – Gregory May 21 '17 at 18:30
• Not in the second line. he just shifted indices. – hamam_Abdallah May 21 '17 at 18:34

hint

After writing that the coefficient of $x^n$ is zero , you will get

$$a_{n+2}=\frac {2n-10}{(n+1)(n+2)} a_n$$

and treat the two cases : $n$ even to use $a_0$ and $n$ odd to use $a_1$.

Since $a_0=y (0)=0$, we conclude that $$a_{2p}=0$$

and since $a_1=y'(0)=3$, we find

$$a_3=\frac {-8}{2\times 3}3$$ $$a_5=\frac {4}{4\times 5}4$$ $$a_7=0=a_9=a_{11}=..$$

finally $$\boxed {y=3x-4x^3+\frac {4}{5}x^5}$$

• You forgot $a_n$ on the RHS – Gregory May 21 '17 at 18:30
• @Gregory I just added it before you... thanks. – hamam_Abdallah May 21 '17 at 18:31
• Thanks, I see where I forgot to include the $n$ term, but I am still having trouble finding a the equations for the general forms. @Salahamam_Fatima – Bryan Chen May 21 '17 at 19:29
• $$n=0: a_2=\frac{10}{2!}a_0$$ $$n=2: a_4=\frac{6\cdot10}{4!}a_0$$ $$n=4: a_6=\frac{2\cdot6\cdot10}{6!}a_0$$ $$n=6: a_8=\frac{-2\cdot2\cdot6\cdot10}{8!}a_0$$ and for the odd: $$n=1: a_3=\frac{8}{3!}a_1$$ $$n=3: a_5=\frac{4\cdot8}{5!}a_1$$ $$n=5: a_7=\frac{0\cdot4\cdot8}{7!}a_1 = 0$$ and the rest of the odd terms are 0, but what is the general form for this? Please help @Salahamam_ Fatima – Bryan Chen May 21 '17 at 20:17
• @BryanChen Is that helpful. – hamam_Abdallah May 21 '17 at 20:43