Find out the answer to "The initial value problem is given by y‘=..."

High School
Math Word Problem

Janine Collamat

Answered question

2022-09-17

The initial value problem is given by y‘= (1 - 2x)y^2 ; y(0) = -1/6 solving its particular solution by variable separable, the value for the constant C is:

Answer & Explanation

madeleinejames20

Skilled2023-06-04Added 165 answers

To solve the initial value problem, let's use the method of variable separable.
The given initial value problem is:

\frac{d y}{d x} = (1 - 2 x) y^{2}

with the initial condition

y (0) = - \frac{1}{6}

.
Separate the variables.

\frac{d y}{y^{2}} = (1 - 2 x) d x

Integrate both sides of the equation.

\int \frac{d y}{y^{2}} = \int (1 - 2 x) d x

Using the power rule for integration, we have:

- \frac{1}{y} = x - x^{2} + C

where C is the constant of integration.
Solve for y.
To solve for y, we can rearrange the equation:

y = - \frac{1}{x - x^{2} + C}

Apply the initial condition.
We are given the initial condition

y (0) = - \frac{1}{6}

. Substituting this into the equation:

- \frac{1}{6} = - \frac{1}{0 - 0^{2} + C}

Simplifying further:

- \frac{1}{6} = - \frac{1}{C}

To find the value of C, we can cross-multiply:

- C = - 6

Dividing both sides by -1 gives:

C = 6

Therefore, the value of the constant C is 6.
The particular solution to the initial value problem is:

y = - \frac{1}{x - x^{2} + 6}

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