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

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Algebra 2

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 $C$ $9$ .