Check the solution to "Solve the following initial value problem for the..."

Jean Blumer

Answered question

2021-12-23

Solve the following initial value problem for the function y(x).

y^{'} = 2 x y^{2}; y (1) = \frac{1}{2}

Answer & Explanation

Bertha Jordan

Beginner2021-12-24Added 37 answers

y^{'} = 2 x y^{2}

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

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

= - y^{- 1} = x^{2} + c

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

given

y (1) = \frac{1}{2}

\Rightarrow - 2 = 1 + c

\Rightarrow c = - 3

∴ \frac{- 1}{y} = x^{2} - 3

\Rightarrow y = \frac{1}{3} - x^{2}

- is the solution

GaceCoect5v

Beginner2021-12-25Added 26 answers

Solve the separable equation $\frac{d y (x)}{d x} = 2 x y {(x)}^{2}$ , such that y(1) = 1/2:
Divide both sides by $y {(x)}^{2}$ :
$\frac{\frac{d y (x)}{d x}}{y} {(x)}^{2} = 2 x$
Integrate both sides with respect to $x$ :
$\int \frac{\frac{d y (x)}{d x}}{y} {(x)}^{2} d x = \int 2 x d x$
Evaluate the integrals:
$- \frac{1}{y} (x) = x^{2} + c_{1}$ , where $c_{1}$ is an arbitrary constant.
Solve for $y (x)$ :
$y (x) = - \frac{1}{x^{2} + c_{1}}$
Solve for $c_{1}$ using the initial conditions:
Substitute $y (1) = \frac{1}{2} \to y (x) = - \frac{1}{x^{2} + c_{1}}$ :
$- \frac{1}{c_{1} + 1} = \frac{1}{2}$
Solve the equation:
$c_{1} = - 3$
Substitute $c_{1} = - 3 \to y (x) = - \frac{1}{x^{2} + c_{1}}$ :
Answer:
$y (x) = \frac{1}{- x^{2} + 3}$

user_27qwe

Skilled2021-12-30Added 375 answers

Thank you, both answers are good!

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