Find a solution of xdydx=y2−y that passes through the indicated points. a. (0,0) b. (14,14) c. (6,18)

By separing variables in xdydx=y2−y; we have dyy2−y=dxx dyy(y−1)=dxx [−1y]+1y−1]dy=dxx On integration, we obtain −ln⁡(y)+ln⁡(y−1)=ln⁡(x)+ln⁡c ln⁡(y−1y)=ln⁡(xc) y−1y=xc y−1=xyc y−xyc y(1−xc)=1 y=11−xc y=11−cx is a solution and y=0 is another constant solution a) At (0,0) y=11−xc 0=11−0 0=1 This is not possible Therefore in this case there is no solution.

c) At (6,18) Here, x=6,y=18 18=11−6c 1−6c=8 Substitute c=−76 in y=11−cx Thus, y=11+76x =66+7x The solution of the initial value problem is y=66+7x

By separing variables in xdydx=y2−y; we have dyy2−y=dxx dyy(y−1)=dxx [−1y]+1y−1]dy=dxx On integration, we obtain −ln⁡(y)+ln⁡(y−1)=ln⁡(x)+ln⁡c ln⁡(y−1y)=ln⁡(xc) y−1y=xc y−1=xyc y−xyc y(1−xc)=1 y=11−xc y=11−cx is a solution and y=0 is another constant solution a) At (0,0) y=11−xc 0=11−0 0=1 This is not possible Therefore in this case there is no solution.

c) At (6,18) Here, x=6,y=18 18=11−6c 1−6c=8 Substitute c=−76 in y=11−cx Thus, y=11+76x =66+7x The solution of the initial value problem is y=66+7x

Expert answers to "Find a solution of xdydx=y2−y that passes through..."

Adela Brown Answered2021-12-22

Find a solution of

x \frac{d y}{d x} = y^{2} - y

that passes through the indicated points.
a.

(0, 0)

(\frac{1}{4}, \frac{1}{4})

(6, \frac{1}{8})

Answer & Explanation

Melissa Moore

Expert

2021-12-23Added 32 answers

By separing variables in

x \frac{d y}{d x} = y^{2} - y

; we have

\frac{d y}{y^{2} - y} = \frac{d x}{x}

\frac{d y}{y (y - 1)} = \frac{d x}{x}

[\frac{- 1}{y}] + \frac{1}{y - 1}] d y = \frac{d x}{x}

On integration, we obtain

- \ln (y) + \ln (y - 1) = \ln (x) + \ln c

\ln (\frac{y - 1}{y}) = \ln (x c)

\frac{y - 1}{y} = x c

y - 1 = x y c

y - x y c

y (1 - x c) = 1

y = \frac{1}{1 - x c}

y = \frac{1}{1 - c x}

is a solution and

y = 0

is another constant solution
a) At

(0, 0)

y = \frac{1}{1 - x c}

0 = \frac{1}{1 - 0}

0 = 1

This is not possible
Therefore in this case there is no solution.

karton

Expert

2021-12-30Added 439 answers

c) At $(6, \frac{1}{8})$
Here, $x = 6, y = \frac{1}{8}$
$\frac{1}{8} = \frac{1}{1 - 6 c}$
$1 - 6 c = 8$
Substitute $c = \frac{- 7}{6}$ in $y = \frac{1}{1 - c x}$
Thus,
$y = \frac{1}{1 + \frac{7}{6} x}$
$= \frac{6}{6 + 7 x}$
The solution of the initial value problem is $y = \frac{6}{6 + 7 x}$

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