Solved: "Nonlinear Ordinary differential equation.d2ydx2−2y2=0with y(0)=a and y′(0)=0Where a..."

Mabel Breault

Answered question

2022-01-18

Nonlinear Ordinary differential equation.

\frac{d^{2} y}{{d x}^{2}} - \frac{2}{y^{2}} = 0

with

y (0) = a and y^{'} (0) = 0

Where a is a known constant.

Jonathan Burroughs

Beginner2022-01-19Added 37 answers

Change variables so that the independent variable is y and the dependent on

p = y^{'}

. Then

p p = y^{″}

(dots stand for derivatives w.r.t. y, 's w.r.t. x) and the equation becomes

p p - \frac{2}{y^{2}} = 0

\frac{d}{d y} (\frac{p^{2}}{2}) = \frac{2}{y^{2}}

This can be solved by integrating into a first order equation for y as a function of x.

Hector Roberts

Beginner2022-01-20Added 31 answers

Maybe this is the same technique, but what I have done before for

y^{″} = f (y)

is treated as

\frac{{d y}^{'}}{d x} = f (y)

\frac{{d y}^{'}}{d y} \frac{d y}{d x} = f (y)

y^{'} \frac{{d y}^{'}}{d y} = f (y)

\int y^{'} {d y}^{'} = \int f (y) d y + K_{0}

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

with

g (y) = \int f (y) d y + K_{0}

y^{'} = \sqrt{2 g (y)}

\frac{d y}{\sqrt{2 g (y)}} = d x

x = \int \frac{1}{\sqrt{2 g (y)}} d y + K_{1}

With your example

f (y) = \frac{2}{y^{2}}

g (y) = \int \frac{2}{y^{2}} d y + K_{0} = - \frac{2}{y} + K_{0}

x = \int \frac{1}{\sqrt{2 (- \frac{2}{y} + K_{0})}} d y + K_{1}

and with the IC given

x = \frac{\sqrt{a}}{4} (2 a \ln (\sqrt{y - a} + \sqrt{y}) - a \ln a + 2 \sqrt{y (y - a)})

alenahelenash

Expert2022-01-24Added 556 answers

\frac{d^{2} y}{d x^{2}} - \frac{2}{y^{2}} = 0

Multiply by 2y

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