Question

Consider xưy" – 14xy' + 56y = 0. Find all values of r such that y=x" satisfies the differential equation for x > 0. Enter as a comma separated list: r= 7,8 help (numbers) Enter two linearly independent solutions of the form above: y1 = x7 help (formulas) y2 = 48 help (formulas) Now find a solution satisfying the initial values y(1) = 4, y'(1) = 3: y= 29x? – 25,28 help (formulas)

Answer 1

I bet the ODE is supposed to read

$x^2y''-14xy'+56y=0$

Then if $y=x^r$, we have $y'=rx^{r-1}$ and $y''=r(r-1)x^{r-2}$, and substituting these into the ODE gives

$r(r-1)x^r-14rx^r+56x^r=0\implies r(r-1)-14r+56=r^2-15r+56=0$

Solving for r, we find

$r^2-15r+56=(r-8)(r-7)=0\implies \boxed{r=8\text{ or }r=7}$

so that $y_1=x^8$ and $y_2=x^7$ are two fundamental solutions to the ODE. Thus the general solution is

$\boxed{y(x)=C_1x^8+C_2x^7}$

Given that $y(1)=4$ and $y'(1)=3$, we get

$\begin{cases}4=C_1+C_2\\3=8C_1+7C_2\end{cases}\implies C_1=-25\text{ and }C_2=29$

So the particular solution is

$\boxed{y(x)=29x^7-25x^8}$