A nonhomogeneous second-order linear equation and a complementary function&nbsp;yc&nbsp;are given. Find a particular solution of the equation.4x2y&nbsp;″−4xy′+3y=8x34

Consider the following differential equation:4x2y&nbsp;″−4xy′+3y=8x34To find the particular solutions first divide the equation by&nbsp;4x2&nbsp;then, to gety&nbsp;″−4x4x2y′+34x2y=8x434x2y&nbsp;″−1xy3′4x2y=2x−23Suppose that function isf(x)=2x−23Consider the complementary function&nbsp;yc(x):yc(x)=c1x12+c2x32Here,y1=x12,&nbsp;y2=x32And,y1′=12x−12,&nbsp;y2′=32x12Therefore:W=[(y1,y2),(y'2,y'2)]=[(x12,&nbsp;x32),(12x-12,32x12)]=32x12x12-12x-12x32=32x−12x=2x2=xyp(x)=-∫(y2(x)f(x)W(x)dx)y1(x)+∫y1(x)f(x)W(x))dxy2(x)=(-∫x32·2x-23xdx)x12+(∫x1/2·2x-23xdx)x32=(-2∫x-16dx)x12+(2∫x-76dx)x32=-2·x5656x12+2x-16-16x32Now simplify, we get:yp(x)=-125·x43-12x43=-12x43-60x435=-725·x43Therefore the particular solution is:yp=-725·x43

Expert answers to "A nonhomogeneous second-order linear equation and a complementary..."

College

Dexter Odom

Answered question

2022-03-29

A nonhomogeneous second-order linear equation and a complementary function $y_{c}$ are given. Find a particular solution of the equation.
$4 x^{2} y^{″} - 4 x y^{'} + 3 y = 8 x^{\frac{3}{4}}$

Answer & Explanation

pastuh7vka

Beginner2022-03-30Added 13 answers

Consider the following differential equation:
$4 x^{2} y^{″} - 4 x y^{'} + 3 y = 8 x^{\frac{3}{4}}$
To find the particular solutions first divide the equation by $4 x^{2}$ then, to get
$y^{″} - \frac{4 x}{4 x^{2}} y^{'} + \frac{3}{4 x^{2}} y = \frac{8 x^{\frac{4}{3}}}{4 x^{2}}$
$y^{″} - \frac{1}{x} \frac{y_{3}^{'}}{4 x^{2}} y = 2 x^{- \frac{2}{3}}$
Suppose that function is
$f (x) = 2 x^{- \frac{2}{3}}$
Consider the complementary function $y_{c} (x)$ :
$y_{c} (x) = c_{1} x^{\frac{1}{2}} + c_{2} x^{\frac{3}{2}}$
Here,
$y_{1} = x^{\frac{1}{2}}, y_{2} = x^{\frac{3}{2}}$
And,
$y_{1}^{'} = \frac{1}{2} x^{- \frac{1}{2}}, y_{2}^{'} = \frac{3}{2} x^{\frac{1}{2}}$
Therefore:
$W = [(y_{1}, y_{2}), (y'_{2}, y'_{2})] = [(x^{\frac{1}{2}}, x^{\frac{3}{2}}), (\frac{1}{2} x^{- \frac{1}{2}}, \frac{3}{2} x^{\frac{1}{2}})]$
$= \frac{3}{2} x^{\frac{1}{2}} x^{\frac{1}{2}} - \frac{1}{2} x^{- \frac{1}{2}} x^{\frac{3}{2}}$
$= \frac{32}{x} - \frac{12}{x} = \frac{2 x}{2}$
$= x$ $y_{p} (x) = - \int (\frac{y_{2} (x) f (x)}{W (x)} d x) y_{1} (x) + \int \frac{y_{1} (x) f (x)}{W (x)) d x} y_{2} (x)$
$= (- \int \frac{x^{\frac{3}{2}} \cdot 2 x^{- \frac{2}{3}}}{x} d x) x^{\frac{1}{2}} + (\int \frac{x^{1 / 2} \cdot 2 x^{- \frac{2}{3}}}{x} d x) x^{\frac{3}{2}}$
$= (- 2 \int x^{\frac{- 1}{6}} d x) x^{\frac{1}{2}} + (2 \int x^{\frac{- 7}{6}} d x) x^{\frac{3}{2}}$
$= - 2 \cdot \frac{x^{\frac{5}{6}}}{\frac{5}{6}} x^{\frac{1}{2}} + 2 \frac{x^{\frac{- 1}{6}}}{\frac{- 1}{6}} x^{\frac{3}{2}}$
Now simplify, we get:
$y_{p} (x) = - \frac{12}{5} \cdot x^{\frac{4}{3}} - 12 x^{\frac{4}{3}}$
$= \frac{- 12 x^{\frac{4}{3}} - 60 x^{\frac{4}{3}}}{5}$
$= - \frac{72}{5} \cdot x^{\frac{4}{3}}$
Therefore the particular solution is:
$y_{p} = - \frac{72}{5} \cdot x^{\frac{4}{3}}$

Jeffrey Jordon

Expert2022-08-23Added 2605 answers

Answer is given below (on video)

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