How do I solve the differential equation xy′−y=3xy,y1=0?

Question

wyjadaczeqa8 · Accepted Answer

This is a separable equation. We can rewrite the equation as it follows:
x y' -y=3xy. Separating the x's and y's, we have that
xy'=y+3xy
xy'=(1+3x)y

\frac{y^{'}}{y} = \frac{1}{x} + 3

Now that variables are separated, we can integrate both sides with respect to x:

\int \frac{y^{'}}{y} d x = \int (\frac{1}{x} + 3) d x

Considering that y' dx=dy and that the integral of a sum is the sum of the integrals, we have

\int \frac{d y}{y} = \int \frac{1}{x} d x + \int 3 d x

Solving the ingrals, we have that

\log (y) = \log (x) + 3 x + c

Solving for y:

y (x) = e^{\log (x) + 3 x + c} = e^{\log (x)} \cdot e^{3 x} \cdot e^{c} = x e^{3 x} e^{c}

To determine the value of c, we should use the condition y(1)=0 (if this is what you meant with

y_{1} = 0

in your question).
I'm a little puzzled with that one, because evaluating y for x=1 we have

e^{3} \cdot e^{c}

,
and this is never zero for

c \in R

. I hope someone can correct me if I said something wrong, or that you can correct the question if the mistake was there.

How do I solve the differential equation