# How do I solve the differential equation xy'-y=3xy, y_1=0?

Jan 28, 2015

This is a separable equation. We can rewrite the equation as it follows:

$x \setminus y ' - y = 3 x y$. Separating the $x$'s and $y$'s, we have that

$x y ' = y + 3 x y$
$x y ' = \left(1 + 3 x\right) y$
$\frac{y '}{y} = \frac{1}{x} + 3$.

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

$\setminus \int \frac{y '}{y} \setminus \mathrm{dx} = \setminus \int \left(\frac{1}{x} + 3\right) \setminus \mathrm{dx}$

Considering that $y ' \setminus \mathrm{dx} = \mathrm{dy}$ and that the integral of a sum is the sum of the integrals, we have

$\setminus \int \frac{\mathrm{dy}}{y} = \setminus \int \frac{1}{x} \setminus \mathrm{dx} + \setminus \int 3 \setminus \mathrm{dx}$

Solving the ingrals, we have that

$\setminus \log \left(y\right) = \setminus \log \left(x\right) + 3 x + c$

Solving for $y$:

$y \left(x\right) = {e}^{\setminus \log \left(x\right) + 3 x + c} = {e}^{\setminus \log \left(x\right)} \setminus \cdot {e}^{3 x} \setminus \cdot {e}^{c} = x {e}^{3 x} {e}^{c}$.

To determine the value of $c$, we should use the condition $y \left(1\right) = 0$ (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 ${e}^{3} \setminus \cdot {e}^{c}$, and this is never zero for $c \setminus \in \setminus m a t h \boldsymbol{R}$. I hope someone can correct me if I said something wrong, or that you can correct the question if the mistake was there.