# Differential Equation (Bernoulli?)

gmmstr827

## Homework Statement

Solve the differential equation:
√(y)*(3*y'+y)=x

## Homework Equations

Bernoulli equations:
dy/dx+P(x)y=Q(x)yn
dv/dx+(1-n)P(x)v=(1-n)Q(x)

Integration by Parts:
∫u*dv = u*v - ∫v*du

## The Attempt at a Solution

Since it's not separable, failed the homogeneous test, and failed the exact method test, I tried putting it into Bernoulli's format, and it seems to fit.

I rewrite the equation in Bernoulli's equation form:
dy/dx+(1/3)y=(x/3)y-1/2 where n=-1/2, P(x)=1/3, Q(x)=x/3

Therefore, using Bernoulli's equation dv/dx+(1-n)P(x)v=(1-n)Q(x)
I substitute the values in and simplify to get:
dv/dx+v/2=x/2; now P(x)=1/2 and Q(x)=x/2

Find rho:
ρ=e∫P(x)dx = e∫1/2dx = ex/2

Multiply equation by rho:
ex/2dv/dx+ex/2v/2=ex/2x/2

Recognize derivative and check:
Dx = [vex/2]=ex/2x/2
Deriving that gets me back to my equation ex/2dv/dx+ex/2v/2=ex/2x/2

Integrate equation:
ex/2v=∫(ex/2x/2)dx+C

Integration of ∫(ex/2x/2)dx using integration by parts:
∫u*dv = u*v - ∫v*du where u = x/2 and dv = ex/2
u = x/2, du = 1/2dx; v = 2*ex/2, dv = ex/2dx
(x/2)*2*ex/2 - ∫2*(1/2)*ex/2dx
x*ex/2-∫ex/2dx
x*ex/2-2*ex/2
ex/2(x-2)

Total integration:
ex/2v=ex/2(x-2)

y3/2=Ce-x/2+x-2

Does everything I did look correct?
Much appreciated.

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