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

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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 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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