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

How do you solve for xy'−y=3xy given y(1)=0?
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Matteo Estes
$x\frac{dy}{dx}=3xy+y$
$x\frac{dy}{dx}=y\left(3x+1\right)$
$\frac{dy}{y}=\frac{3x+1}{x}dx$
Integrating;
$\int \frac{dy}{y}=\int \frac{3x+1}{x}dx$
$\int \frac{dy}{y}=\int \left(3+\frac{1}{x}\right)dx$
$\mathrm{ln}\left(y\right)=3x+\mathrm{ln}\left(x\right)+c$
$y={e}^{3x+\mathrm{ln}\left(x\right)+c}$
And finally;
$y={e}^{3x}×{e}^{\mathrm{ln}\left(x\right)}×{e}^{c}=Ax{e}^{3x}$
where $A={e}^{c}$
We can now find the value of the constant A;
$0=A{e}^{3}$
A=0
The particular solution to your equation is then:
$y=0x{e}^{3x}=0$