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$\frac{dy}{dx} = 2y^2$ and $y(1)=-1$ . $y(3)=?$

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Find limits involving trigonometric functions

Full solution

Q.

\frac{dy}{dx} = 2y^2

and

y(1)=-1

.

y(3)=?

Given differential equation: We are given the differential equation $\frac{dy}{dx} = 2y^2$ and the initial condition $y(1) = -1$ . To find $y(3)$ , we first need to solve the differential equation.
Separate and integrate variables: Separate the variables $y$ and $x$ to integrate them separately. We can write the equation as $\frac{dy}{y^2} = 2dx$ .
Use initial condition to find $C$ : Integrate both sides of the equation. The integral of $\frac{1}{y^2}$ with respect to $y$ is $-\frac{1}{y}$ , and the integral of $2$ with respect to $x$ is $2x$ .
$\int(\frac{1}{y^2}) dy = \int 2 dx$
$-\frac{1}{y} = 2x + C$ , where $C$ is the constant of integration.
Particular solution: Use the initial condition $y(1) = -1$ to find the value of $C$ . $-\frac{1}{-1} = 2(1) + C$ $1 = 2 + C$ $C = -1$
Solve for y in terms of x: Now we have the particular solution to the differential equation: $\newline$ $-\frac{1}{y} = 2x - 1$
Evaluate $y$ at $x=3$ : Solve for $y$ in terms of $x$ . $y = -\frac{1}{2x - 1}$
Evaluate $y$ at $x=3$ : Solve for $y$ in terms of $x$ .
$y = -\frac{1}{2x - 1}$ Evaluate $y$ at $x = 3$ using the particular solution.
$y(3) = -\frac{1}{2(3) - 1}$
$y(3) = -\frac{1}{6 - 1}$
$y(3) = -\frac{1}{5}$

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