#xy'-3y=x-1#? Solve DE

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Answer 1 · 2018-04-06T07:26:26

#y(x)=ax^3-1/2x+1/3#

We have

#xdy/dx-3y=x-1#

Divide all terms by #x#

#dy/dx-(3y)/x=(x-1)/x#

The ODE is now in a form where it can be solved using an integrating factor

Let #mu=expint-3/x=exp(-3lnx)=expln(1/x^3)=1/x^3#

Now multiply the whole equation by #mu#

#1/x^3 dy/dx-(3y)/x^4=(x-1)/x^4#

The LHS factors as a derivative and we can integrate the RHS

#intd/dx(y/x^3)dx=intx^-3-x^-4dx#

#y/x^3=-1/2x^-2+1/3x^-3+a#

Solve for #y#

#y=y(x)=ax^3-1/2x+1/3#