I'm trying to find particular solution of the second-order linear equation but I can't find y1 and y2 according to y=c1y1+c2y2x2y″−2xy′+2y=0, y(1)=3, y′(1)=1If r is used, x2r2−2xr=−2 then xr(xr−2)=−2, I can't go on from there to find y1 and y2

y″(x)x2+2y′(x)x+2y(x)=0Assume a solution to this Euler-Cauchy equation will be proportional to eλ for some constant λSubstitute y=xλ into the differential equation:x2d2dx2(xλ)+2xddx(xλ)+2xλ=0Substitute d2dx2(xλ)=(λ−1)xλ−2 and ddx(xλ)=λxλ−1λ2xλ+λxλ+2xλ=0xλ(λ2+λ+2)=0Assuming x≠0, the zeros must come from the polynomial:λ2+λ+2=0λ=−12±i72The roots λ=−12±i22 give y1(x)=C1x−12+i72+C2x−12−i72 as solutions, where C1 and C2 are arbitrary constants. The general solution is the sum of the above solutions:y(x)=y1(x)+y2(x)=C1x−12+i72+C2x−12−i72Using xλ=eλln⁡(x), apply Euler's identity eα+bi=eαcos⁡(b)+ieαsin⁡(b)

"I'm trying to find particular solution of the..." solved and explained

protisvitfc

Answered question

2021-11-19

I'm trying to find particular solution of the second-order linear equation but I can't find

y_{1}

and

y_{2}

according to

y = c_{1} y_{1} + c_{2} y_{2}

x^{2} y^{″} - 2 x y^{'} + 2 y = 0, y (1) = 3, y^{'} (1) = 1

If r is used,

x^{2} r^{2} - 2 x r = - 2

then

x r (x r - 2) = - 2

, I can't go on from there to find

y_{1}

and

y_{2}

Answer & Explanation

Roger Noah

Beginner2021-11-20Added 17 answers

Assume

y = x^{r}

then if

y = x^{r}

then

y^{'} = r x^{r - 1}

And

y^{″} = r (r - 1) x^{r - 2}

Substitute in

x^{2} y^{″} - 2 x y^{'} + 2 y = 0

to get

r (r - 1) x^{r} - 2 r x^{r} + 2 x^{r} = 0

divide by y to get

r (r - 1) - 2 r + 2 = (r - 1) (r - 2) = 0

therefore

y = a x + b x^{2}

Solve

y (1) = a + b = 3

y^{'} (1) = a + 2 b = 1

to get

y = 5 x - 2 x^{2}

Charles Wee

Beginner2021-11-21Added 14 answers

$y^{″} (x) x^{2} + 2 y^{'} (x) x + 2 y (x) = 0$
Assume a solution to this Euler-Cauchy equation will be proportional to $e^{λ}$ for some constant $λ$
Substitute $y = x^{λ}$ into the differential equation:
$x^{2} \frac{d^{2}}{{d x}^{2}} (x^{λ}) + 2 x \frac{d}{d x} (x^{λ}) + 2 x^{λ} = 0$
Substitute $\frac{d^{2}}{{d x}^{2}} (x^{λ}) = (λ - 1) x^{λ - 2}$ and $\frac{d}{d x} (x^{λ}) = λ x^{λ - 1}$
$λ^{2} x^{λ} + λ x^{λ} + 2 x^{λ} = 0$
$x^{λ} (λ^{2} + λ + 2) = 0$
Assuming $x \neq 0$ , the zeros must come from the polynomial:
$λ^{2} + λ + 2 = 0$
$λ = - \frac{1}{2} \pm \frac{i \sqrt{7}}{2}$
The roots $λ = - \frac{1}{2} \pm \frac{i \sqrt{2}}{2}$ give $y_{1} (x) = C_{1} x^{- \frac{1}{2} + \frac{i \sqrt{7}}{2}} + C_{2} x^{- \frac{1}{2} - \frac{i \sqrt{7}}{2}}$ as solutions, where $C_{1}$ and $C_{2}$ are arbitrary constants. The general solution is the sum of the above solutions:
$y (x) = y_{1} (x) + y_{2} (x) = C_{1} x^{- \frac{1}{2} + \frac{i \sqrt{7}}{2} + C_{2} x^{- \frac{1}{2} - \frac{i \sqrt{7}}{2}}}$
Using $x^{λ} = e^{λ \ln (x)}$ , apply Euler's identity $e^{α + b i} = e^{α} \cos (b) + i e^{α} \sin (b)$

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