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dydx=2y2\frac{dy}{dx} = 2y^2 and y(1)=−1y(1)=-1. y(3)=?y(3)=?

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Full solution

Q. dydx=2y2\frac{dy}{dx} = 2y^2 and y(1)=−1y(1)=-1. y(3)=?y(3)=?
  1. Given differential equation: We are given the differential equation dydx=2y2\frac{dy}{dx} = 2y^2 and the initial condition y(1)=−1y(1) = -1. To find y(3)y(3), we first need to solve the differential equation.
  2. Separate and integrate variables: Separate the variables yy and xx to integrate them separately. We can write the equation as dyy2=2dx\frac{dy}{y^2} = 2dx.
  3. Use initial condition to find CC: Integrate both sides of the equation. The integral of 1y2\frac{1}{y^2} with respect to yy is −1y-\frac{1}{y}, and the integral of 22 with respect to xx is 2x2x.
    ∫(1y2)dy=∫2dx\int(\frac{1}{y^2}) dy = \int 2 dx
    −1y=2x+C-\frac{1}{y} = 2x + C, where CC is the constant of integration.
  4. Particular solution: Use the initial condition y(1)=−1y(1) = -1 to find the value of CC.−1−1=2(1)+C-\frac{1}{-1} = 2(1) + C1=2+C1 = 2 + CC=−1C = -1
  5. Solve for y in terms of x: Now we have the particular solution to the differential equation: \newline−1y=2x−1-\frac{1}{y} = 2x - 1
  6. Evaluate yy at x=3x=3: Solve for yy in terms of xx.y=−12x−1y = -\frac{1}{2x - 1}
  7. Evaluate yy at x=3x=3: Solve for yy in terms of xx.
    y=−12x−1y = -\frac{1}{2x - 1}Evaluate yy at x=3x = 3 using the particular solution.
    y(3)=−12(3)−1y(3) = -\frac{1}{2(3) - 1}
    y(3)=−16−1y(3) = -\frac{1}{6 - 1}
    y(3)=−15y(3) = -\frac{1}{5}

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