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

Answer: [tex]y=\dfrac{5}{5-6x}[/tex]

Step-by-step explanation:

The given differential equation: [tex]xy' + y = y^2[/tex]

[tex]\Rightarrow\ xy'=y^2-y[/tex]

[tex]\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[/tex]

Integrate both sides , we get

[tex]\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)[/tex]

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

[tex]\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[/tex]

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

Put value of x in (i), we get

[tex]\ln(y-1)-\ln y=\ln x+\ln (\dfrac65)\\\\\Rigtarrow\ \ln (\dfrac{y-1}{y})=\ln(\dfrac{6}{5}x)[/tex]

[tex]\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}[/tex]

hence, the required solution: [tex]y=\dfrac{5}{5-6x}[/tex]

The solution to the differential equation

[tex]xy'+y=y^2[/tex]

given the initial condition [tex]y(1)=-5[/tex]  is [tex]y=\frac{5}{5-6x}[/tex]

Given the differential equation

[tex]xy'+y=y^2[/tex]

We can rearrange it as follows:

[tex]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[/tex]

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

[tex]\frac{1}{y(y-1)}dy \implies \frac{1}{(y-1)}dy+\frac{1}{y}dy[/tex]

The final rearranged equation is

[tex]\frac{1}{(y-1)}dy+\frac{1}{y}dy=\frac{1}{x}dx[/tex]

Integrating both sides;

[tex]\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[/tex]

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

The initial condition [tex]y(1)=-5\implies y=-5 \text{ when }x=1[/tex]

Substituting into the general solution we got earlier

[tex]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[/tex]

Substituting the value of [tex]c[/tex] back into the general solution

[tex]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}[/tex]

When [tex]y[/tex] is made the subject of the formula

[tex]y=\frac{5}{5-6x}[/tex]

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

Learn more about solving differential equations here: https://brainly.com/question/4537000

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