Question

How do you solve for `xy'-y=3xy` given `y(1)=0`?

Answer 1

I would write it as;
`xdy/dx=3xy+y`
`xdy/dx=y(3x+1)`
`dy/y=(3x+1)/xdx`
Integrating;
`intdy/y=int(3x+1)/xdx`
`intdy/y=int(3+1/x)dx`
`ln(y)=3x+ln(x)+c`
`y=e^(3x+ln(x)+c)`
And finally;
`y=e^(3x)×e^(ln(x))×e^c=Axe^(3x)`
where `A=e^c`

We can now find the value of the constant `A`;
`0=Ae^3`
`A=0`
The particular solution to your equation is then:
`y=0xe^(3x)=0`