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

Question

Charles Arocha · Accepted Answer

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#

Emma Aldridge · Answer

undefined This is a first-order linear ordinary differential equation. We can solve it using an integrating factor method. First, rearrange the equation to get it in standard form:

xy' - y = 3xy

Next, identify $ P(x) $ and $ Q(x) $:

$ P(x) = x $
$ Q(x) = 3x $

Now, find the integrating factor $ \mu(x) $:

$ \mu(x) = e^{\int P(x) \, dx} $
$ \mu(x) = e^{\int x \, dx} $
$ \mu(x) = e^{\frac{x^2}{2}} $

Multiply both sides of the equation by $ \mu(x) $:

$ e^{\frac{x^2}{2}}xy' - e^{\frac{x^2}{2}}y = 3xe^{\frac{x^2}{2}}y $

Now, the left side can be rewritten as the derivative of $ y $ multiplied by $ \mu(x) $:

$ \frac{d}{dx}(e^{\frac{x^2}{2}}y) = 3xe^{\frac{x^2}{2}}y $

Integrate both sides with respect to $ x $:

$ \int \frac{d}{dx}(e^{\frac{x^2}{2}}y) \, dx = \int 3xe^{\frac{x^2}{2}}y \, dx $

$ e^{\frac{x^2}{2}}y = \int 3xe^{\frac{x^2}{2}}y \, dx + C $

Solve the integral on the right side:

$ \int 3xe^{\frac{x^2}{2}}y \, dx = \frac{3}{2}e^{\frac{x^2}{2}}y + K $

Now, solve for $ y $:

$ y = e^{-\frac{x^2}{2}}(\frac{3}{2}e^{\frac{x^2}{2}}y + K) $

$ y = \frac{3}{2}y + Ke^{-\frac{x^2}{2}} $

$ y - \frac{3}{2}y = Ke^{-\frac{x^2}{2}} $

$ \frac{-1}{2}y = Ke^{-\frac{x^2}{2}} $

$ y = -2Ke^{-\frac{x^2}{2}} $

Now, use the initial condition $ y(1) = 0 $ to find $ K $:

$ 0 = -2Ke^{-\frac{1}{2}} $

$ K = 0 $

So, the particular solution to the differential equation is $ y = 0 $.