Question

Show that the substitution u = y’ leads to a Bernoulli equation. Solve this equation

xy''=y'+ (y' )³

Short answer

The solution is y(x)= -√ c₁-√ x²+c₂

Step-by-step solution

Solving for the given differential equation;

We have the differential equation

xy''=y'+ (y' )³

And we have to obtain the form bernoulli equation and solve it as following:

As given, let u=y’ then a differetial as

y''=u'

Subtitute with y’=u

xu'= u+u³

xu'-u=u³

u'- (1/x)u= (1/x)u³

Hence it is Bernoulli equation

Solving Bernoulli equation:

Now we have to solve this Bernoulli equation:

Let v=u^-2 [u=√ 1/√ v) the differential with respect to x as

dv/dx= -2u^-3 (du/dx)

du/dx= -(-1/2)u³ dv/dx

Since we have u= √ 1/√ v, then we can have

du/dx= -1/2 (1/v)^3/2 (dv/dx)

du/dx= -1/2 (1/v^3/2) dv/dx

After that substitiution with u= 1/v^1/2, we get

-1/2 1/v^3/2 dv/dx -1/x 1/v^1/2= 1/x 1/v^3/2

-1/2 dv/dx- (1/x)v= 1/x

dv/dx + (2/x)v = -2/x

integrating D.E;

Since the differential equation shows the first order D.E in the form dv/dx + p(x)v = q(x)

We have to obtain integrating factor for D.E as

I.F= e^{∫ p(x)dx}

⁼e^{∫ (2/x) dx}

=e^2lnx

=e^{lnx^2}

= x²

After multiplying both side and integrating we get;

d/dx(x²v) =-2x

∫ d (x²v) = -2∫ xdx

x²v= -x²+c₁

v= -1 + c₁/x²

Substituting with  v=u-2

Now we have to back substitute with v= u^-2, then we have

u^-2= c₁/x²-1

u^-2= c₁-x²/x²

u²= x²/c₁-x²

u= x/ ( √ c₁- √ x²)

Also substitute with u=dy/dx,

dy/dx= x/(√ c₁-√ x²)

Integrating both side

∫ dy=∫ x/(√ c₁-√ x²) dx

y= ∫x/(√ c₁-√ x²) dx

y (x)= -1/2∫ 1/√ s ds

=-1/2× 2√ s +c₂

= -√ s +c₂

= -(√ c₁-√ x²) +c₂

Hence the final answer is y (x)=-(√ c₁-√ x²) +c₂