Get the answer to "Solve the differential equation:ddx(2yy′)=(y′)2"

parksinta8rkv

Answered question

2022-03-22

Solve the differential equation:

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

Avery Maxwell

Beginner2022-03-23Added 13 answers

Step 1
Problem is equivalent with

2 y^{' 2} + 2 y y^{″} = y^{' 2}

or with

y^{' 2} + 2 y y^{″} = 0

.
Use substitution

y^{'} = z

, where

z = z (y), z^{'} = y^{″} = z^{'} z

.
Your equation is then

z^{2} + 2 y z z^{'} = 0

, from where we get

z = 0

z + 2 y z^{'} = 0

.
From first equation we get

y^{'} = 0

and

y = C

and second equation is linear ODE

z^{'} + \frac{z}{2 y} = 0

, which is equivalent with

{(\sqrt{y} z)}^{'} = 0

z = \frac{C}{\sqrt{y}}

. By returning substitution, we get

\sqrt{y} d y = C d x

\frac{2}{3} y^{\frac{3}{2}} = C x + D

and

y = {(\frac{3}{2} (C x + D))}^{\frac{2}{3}}

Since

y (0) = y (1) = 0

, we get

D = 0

and

C + D = 0

, from where we get solution

y = 0

Dixie Reed

Beginner2022-03-24Added 15 answers

Step 1

\frac{y^{'}}{y}

= - \frac{2 y^{″}}{y^{'}} \to \ln y + 2 {\ln y}^{'}

= C_{0} \to \ln (y y^{' 2})

= C_{0} \to y y^{' 2}

= C_{1} \to y^{'} \sqrt{y} = \pm C_{2}

which is separable.

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