(dy)/(dx)=2y^(2) and y(1)=-1. y(3)=

Separate Variables: First, we need to solve the differential equation (dy)/(dx)=2y^(2). This is a separable differential equation, so we can separate the variables y and x.Integrate Both Sides: Separate the variables by dividing both sides by y^(2) and multiplying both sides by dx to get (1/y^(2))dy = 2dx.Find Constant of Integration: Now, integrate both sides. The integral of (1/y^(2))dy is -1/y, and the integral of 2dx is 2x + C, where C is the constant of integration.Determine Particular Solution: After integrating, we get -1/y = 2x + C.Find Final Solution: Use the initial condition y(1)=-1 to find the value of C. Plug in x=1 and y=-1 into the equation -1/y = 2x + C to get -1/(-1) = 2(1) + C.Simplify to get 1 = 2 + C.Solve for C by subtracting 2 from both sides to get C = -1.Now we have the particular solution -1/y = 2x - 1.To find y(3), plug in x=3 into the equation -1/y = 2(3) - 1 to get -1/y = 6 - 1.Simplify to get -1/y = 5.Solve for y by dividing both sides by -1 to get 1/y = -5.Take the reciprocal of both sides to get y = -1/5.

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$\frac{d y}{d x}=2 y^{2}$ and $y(1)=-1$ . $\newline$ $y(3)=$

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Simplify variable expressions using properties

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

and

y(1)=-1

\newline

y(3)=

Separate Variables: First, we need to solve the differential equation $(\frac{dy}{dx}=2y^{2})$ . This is a separable differential equation, so we can separate the variables $y$ and $x$ .
Integrate Both Sides: Separate the variables by dividing both sides by $y^{2}$ and multiplying both sides by $dx$ to get $(1/y^{2})dy = 2dx$ .
Find Constant of Integration: Now, integrate both sides. The integral of $\frac{1}{y^{2}}$ dy is $-\frac{1}{y}$ , and the integral of $2dx$ is $2x + C$ , where $C$ is the constant of integration.
Determine Particular Solution: After integrating, we get $-\frac{1}{y} = 2x + C$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{(-1)} = 2(1) + C$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{(-1)} = 2(1) + C$ . Simplify to get $1 = 2 + C$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{(-1)} = 2(1) + C$ .Simplify to get $1 = 2 + C$ .Solve for $C$ by subtracting $2$ from both sides to get $C = -1$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{(-1)} = 2(1) + C$ . Simplify to get $1 = 2 + C$ . Solve for $C$ by subtracting $2$ from both sides to get $C = -1$ . Now we have the particular solution $C$ $0$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{-1} = 2(1) + C$ . Simplify to get $1 = 2 + C$ . Solve for $C$ by subtracting $2$ from both sides to get $C = -1$ . Now we have the particular solution $C$ $0$ . To find $C$ $1$ , plug in $C$ $2$ into the equation $C$ $3$ to get $C$ $4$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{(-1)} = 2(1) + C$ . Simplify to get $1 = 2 + C$ . Solve for $C$ by subtracting $2$ from both sides to get $C = -1$ . Now we have the particular solution $C$ $0$ . To find $C$ $1$ , plug in $C$ $2$ into the equation $C$ $3$ to get $C$ $4$ . Simplify to get $C$ $5$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{-1} = 2(1) + C$ . Simplify to get $1 = 2 + C$ . Solve for $C$ by subtracting $2$ from both sides to get $C = -1$ . Now we have the particular solution $C$ $0$ . To find $C$ $1$ , plug in $C$ $2$ into the equation $C$ $3$ to get $C$ $4$ . Simplify to get $C$ $5$ . Solve for $C$ $6$ by dividing both sides by $C$ $7$ to get $C$ $8$ .
Find Final Solution: Use the initial condition $y(1)=-1$ to find the value of $C$ . Plug in $x=1$ and $y=-1$ into the equation $-\frac{1}{y} = 2x + C$ to get $-\frac{1}{-1} = 2(1) + C$ . Simplify to get $1 = 2 + C$ . Solve for $C$ by subtracting $2$ from both sides to get $C = -1$ . Now we have the particular solution $C$ $0$ . To find $C$ $1$ , plug in $C$ $2$ into the equation $C$ $3$ to get $C$ $4$ . Simplify to get $C$ $5$ . Solve for $C$ $6$ by dividing both sides by $C$ $7$ to get $C$ $8$ . Take the reciprocal of both sides to get $C$ $9$ .

$\frac{d y}{d x}=2 y^{2}$ and $y(1)=-1$ . $\newline$ $y(3)=$

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dydx=2y2 \frac{d y}{d x}=2 y^{2} dxdy​=2y2 and y(1)=−1 y(1)=-1 y(1)=−1.\newliney(3)= y(3)= y(3)=

$\frac{d y}{d x}=2 y^{2}$ and $y(1)=-1$ . $\newline$ $y(3)=$