- #1

- 86

- 1

## Homework Statement

Solve the differential equation:

√(y)*(3*y'+y)=x

## Homework Equations

Bernoulli equations:

dy/dx+P(x)y=Q(x)y

^{n}

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/2}dv/dx+e

^{x/2}v/2=e

^{x/2}x/2

Recognize derivative and check:

D

_{x}= [ve

^{x/2}]=e

^{x/2}x/2

Deriving that gets me back to my equation e

^{x/2}dv/dx+e

^{x/2}v/2=e

^{x/2}x/2

Integrate equation:

e

^{x/2}v=∫(e

^{x/2}x/2)dx+C

Integration of ∫(e

^{x/2}x/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/2}dx

(x/2)*2*e

^{x/2}- ∫2*(1/2)*e

^{x/2}dx

x*e

^{x/2}-∫e

^{x/2}dx

x*e

^{x/2}-2*e

^{x/2}

e

^{x/2}(x-2)

Total integration:

e

^{x/2}v=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.

Last edited: