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 = e^x/2

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

Recognize derivative and check:
D_x = [ve^x/2]=e^x/2x/2
Deriving that gets me back to my equation e^x/2dv/dx+e^x/2v/2=e^x/2x/2

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

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

Total integration:
e^x/2v=e^x/2(x-2)

Simplify for final answer:
y^3/2=Ce^-x/2+x-2


Does everything I did look correct?
Much appreciated.

 

[dextercioby]

You can always check if the y verifies the ODE.