Mathematics, 29.10.2020 16:30 christianhernan2

Find the solution of the differential equation that satisfies the given initial condition. xy' + y = y2, y(1) = −5

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Answer from: saabrrinnaaa

The solution to the differential equation

xy'+y=y^2

given the initial condition y(1)=-5 is $y=\frac{5}{5-6x}$

Given the differential equation

xy'+y=y^2

We can rearrange it as follows:

$x\frac{dy}{dx}+y=y^2\\\\x\frac{dy}{dx}=y^2-y\\\\\frac{1}{y^2-y}\frac{dy}{dx}=\frac{1}{x}\\\\\frac{1}{y^2-y}dy=\frac{1}{x}dx$

Factoring the denominators of the LHS, and decomposing into partial fractions, we get

$\frac{1}{y(y-1)}dy \implies \frac{1}{(y-1)}dy+\frac{1}{y}dy$

The final rearranged equation is

$\frac{1}{(y-1)}dy+\frac{1}{y}dy=\frac{1}{x}dx$

Integrating both sides;

$\int\frac{1}{y-1} dy +\int\frac{1}{y}dy=\int\frac{1}{x}dx\\\\ln(y-1)-ln(y)=ln(x)+c\\\\ln(\frac{y-1}{y})=ln(x)+c$

(We made of a law of logarithms on the last line to simplify the equation)

The initial condition $y(1)=-5\implies y=-5 \text{ when }x=1$

Substituting into the general solution we got earlier

$ln(\frac{y-1}{y})=ln(x)+c\\\\ln(\frac{-5-1}{-5})=ln(1)+c\\\\ln(\frac{-6}{-5})=ln(1)+c \\\\(\text{since }ln(1)=0)\\\\ln(\frac{-6}{-5})=c\\\\ln(\frac{6}{5})=c$

Substituting the value of back into the general solution

$ln(\frac{y-1}{y})=ln(x)+c\\\\ln(\frac{y-1}{y})=ln(x)+ln(\frac{6}{5})\\\\ln(\frac{y-1}{y})=ln(\frac{6x}{5})\\\\\frac{y-1}{y}=\frac{6x}{5}$

When is made the subject of the formula

$y=\frac{5}{5-6x}$

Therefore, the solution that satisfies the initial condition y(1)=-5 is $y=\frac{5}{5-6x}$

Learn more about solving differential equations here: link

Answer from: girlchamp654

$y=\dfrac{5}{5-6x}$

Step-by-step explanation:

The given differential equation: xy' + y = y^2

$\Rightarrow\ xy'=y^2-y$

$\Rightarrow\ \frac{1}{y^2-y}y'\:=\frac{1}{x}\\\\\Rightarrow\ \dfrac{1}{y(y-1)}\dfrac{dy}{dx}=\frac{1}{x}\\\\\Rightarrow\dfrac{y-(y-1)}{y(y-1)}dy=\dfrac{1}{x}dx\\\\\Rightarrow\dfrac{1}{(y-1)}dy+\dfrac{1}{y}dy=\dfrac{1}{x}dx$

Integrate both sides , we get

$\int\dfrac{1}{(y-1)}dy+\int\dfrac{1}{y}dy=\dfrac{1}{x}dx\\\\\Rightarrow\ \ln(y-1)-\ln y=\ln x+c\ \ \ \ (i)$

At x=1 , y=-5 (given)

$\ln(-5-1)-\ln -5=\ln 1+c\\\\\Rightarrow\ \ln (-6)-\ln(-5)=0+c\\\\\Rightarrow\ \ln(\dfrac{-6}{-5})=c\\\\\Rightarrow\ \ln(\dfrac{6}{5})=c$

$[\ \ln a+\ln b=\ln ab ,\ \ \ \ \ \ln a-\ln b=\ln\dfrac{a}{b}\ ]$

Put value of x in (i), we get

$\ln(y-1)-\ln y=\ln x+\ln (\dfrac65)\\\\\Rigtarrow\ \ln (\dfrac{y-1}{y})=\ln(\dfrac{6}{5}x)$

$\Rightarrow\ 1-\dfrac{1}{y}=\dfrac{6}{5}x\Rightarrow\ \dfrac{1}{y}=1-\dfrac{6}{5}x\\\\\Rightarrow\ \dfrac{1}{y}=\dfrac{5-6x}{5}\\\\\Rightarrow\ y=\dfrac{5}{5-6x}$

hence, the required solution: $y=\dfrac{5}{5-6x}$

Answer from: Quest

idk im srry

step-by-step explanation:

Answer from: Quest

see below

step-by-step explanation:

7. there are 10 data points in the set. the middle point is between 5 and 6. the middle point in the top half is 8 and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term point 8 = 100 point 3 = 88

iqr = 100-88 = 12

final point 8 = 93 point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6. b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical . there is an outlier for the males that skews the data

5. the iqr is a better measure of spread for the movies and basketball games. since the basketball game has an outlier, we do not want to use standard deviation for measurement.

3. the exam median is much higher than the class median. the median line is farther to the right on the exams

2. there is no way to tell what the means are from the box and whisker plots.

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