X^2 y''+4xy'+2y=0 cauchy euler equation

1 answer:

7 0

Let $y=x^r$ , then

$x^2(r(r-1)x^{r-2})+4x(rx^{r-1})+2x^r=0$
$r(r-1)x^r+4rx^r+2x^r=0$
$r(r-1)+4r+2=0$
$r^2+3r+2=0$
$(r+2)(r+1)=0$
$r=-1,-2$

This admits a general solution of

$y=C_1x^{-1}+C_2x^{-2}$

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Let ????C be the positively oriented square with vertices (0,0)(0,0), (2,0)(2,0), (2,2)(2,2), (0,2)(0,2). Use Green's Theorem to

bonufazy [111]

Answer:

-48

Step-by-step explanation:

Lets call L(x,y) = 10y²x, M(x,y) = 4x²y. Green's Theorem stays that the line integral over C can be calculed by computing the double integral over the inner square of Mx - Ly. In other words

$\int\limits_C {L(x,y)} \, dx + M(x,y) \, dy = \int\limits_0^2\int\limits_0^2 (M_x - L_y ) \, dx \, dy$

Where Mx and Ly are the partial derivates of M and L with respect to the x variable and the y variable respectively. In other words, Mx is obtained from M by derivating over the variable x treating y as constant, and Ly is obtaining derivating L over y by treateing x as constant. Hence,

M(x,y) = 4x²y
Mx(x,y) = 8xy
L(x,y) = 10y²x
Ly(x,y) = 20xy
Mx - Ly = -12xy

Therefore, the line integral can be computed as follows

$\int\limits_C {10y^2x} \, dx + {4x^2y} \,dy = \int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy$

Using the linearity of the integral and Barrow's Theorem we have

$\int\limits_0^2\int\limits_0^2 -12xy \, dx \, dy = -12 \int\limits_0^2\int\limits_0^2 xy \, dx \, dy = -12 \int\limits_0^2\frac{x^2y}{2} |_{x = 0}^{x=2} \, dy = -12 \int\limits_0^22y \, dy \\= -24 ( \frac{y^2}{2} |_0^2) = -24*2 = -48$