What is the particular solution to the differential equation (dy)/(dx)=((1-y)^(2))/(x+1) with the initial condition y(0)=5 ?

Separate variables: Separate the variables in the differential equation. We want to get all the y terms on one side and all the x terms on the other side. We can do this by multiplying both sides by dx and dividing both sides by (1-y)^2. (dy)/((1-y)^2) = (dx)/(x+1)Integrate sides: Integrate both sides of the equation. We need to integrate the left side with respect to y and the right side with respect to x. ∫(dy)/((1-y)^2) = ∫(dx)/(x+1)Perform integration: Perform the integration. The integral of 1/(1-y)^2 with respect to y is 1/(1-y), and the integral of 1/(x+1) with respect to x is ln|x+1|. Don't forget to add the constant of integration C. 1/(1-y) = ln|x+1| + CApply initial condition: Apply the initial condition to find the constant C. We are given that y(0) = 5, so we can plug in x = 0 and y = 5 into the equation to solve for C. 1/(1-5) = ln|0+1| + C 1/(-4) = ln(1) + C C = -1/4Substitute value of C: Substitute the value of C back into the equation. Now that we have found C, we can write the particular solution. 1/(1-y) = ln|x+1| - 1/4Solve for y: Solve for y in terms of x. To find y as a function of x, we need to solve the equation for y. 1-y = 1/(ln|x+1| - 1/4) y = 1 - 1/(ln|x+1| - 1/4)

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What is the particular solution to the differential equation
(dy)/(dx)=((1-y)^(2))/(x+1) with the initial condition
y(0)=5 ?

What is the particular solution to the differential equation $\frac{dy}{dx}=\frac{(1-y)^{2}}{x+1}$ with the initial condition $y(0)=5$ ?

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Q. What is the particular solution to the differential equation

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

with the initial condition

y(0)=5

Separate variables: Separate the variables in the differential equation. $\newline$ We want to get all the $y$ terms on one side and all the $x$ terms on the other side. We can do this by multiplying both sides by $dx$ and dividing both sides by $(1-y)^2$ . $\newline$ $\frac{dy}{(1-y)^2} = \frac{dx}{x+1}$
Integrate sides: Integrate both sides of the equation. $\newline$ We need to integrate the left side with respect to $y$ and the right side with respect to $x$ . $\newline$ $\int \frac{dy}{(1-y)^2} = \int \frac{dx}{x+1}$
Perform integration: Perform the integration. $\newline$ The integral of $\frac{1}{(1-y)^2}$ with respect to $y$ is $\frac{1}{1-y}$ , and the integral of $\frac{1}{x+1}$ with respect to $x$ is $\ln|x+1|$ . Don't forget to add the constant of integration $C$ . $\newline$ $\frac{1}{1-y} = \ln|x+1| + C$
Apply initial condition: Apply the initial condition to find the constant $C$ . We are given that $y(0) = 5$ , so we can plug in $x = 0$ and $y = 5$ into the equation to solve for $C$ . $\frac{1}{1-5} = \ln|0+1| + C$ $\frac{1}{-4} = \ln(1) + C$ $C = -\frac{1}{4}$
Substitute value of C: Substitute the value of C back into the equation. $\newline$ Now that we have found C, we can write the particular solution. $\newline$ $\frac{1}{1-y} = \ln|x+1| - \frac{1}{4}$
Solve for y: Solve for y in terms of x. $\newline$ To find y as a function of x, we need to solve the equation for y. $\newline$ $1-y = \frac{1}{\ln|x+1| - \frac{1}{4}}$ $\newline$ $y = 1 - \frac{1}{\ln|x+1| - \frac{1}{4}}$