Final answer:
To solve the given differential equation with the initial condition y(3)=6, we find the integrating factor and integrate both sides. After applying the initial condition, we determine the constant of integration and arrive at the final solution for y.
Step-by-step explanation:
To find the solution of the given differential equation x² y' + x y = 1 that satisfies the initial condition y(3) = 6, we employ the method of separating variables or an integrating factor. However, since the provided information does not contain a clear method, we will proceed with the most likely applicable method given the first-order linear nature of the equation - using an integrating factor.
First, we rewrite the equation in the standard form for a linear first-order equation:
y' + \(\frac{1}{x}\)y = \(\frac{1}{x²}\)
The integrating factor \(\mu(x)\) is found by taking the exponential of the integral of \(\frac{1}{x}\):
\(\mu(x) = e^\int{\frac{1}{x} dx} = e^x = |x|\)
We multiply through by the integrating factor:
|x|y' + y = \(\frac{1}{x}\)
Now, we can notice that the left side of the equation is the derivative of |x|y. Thus, integrating both sides with respect to x:
\intx dx + \int{y dx} = \int{\frac{1}{x} dx}
|x|y = \ln|x| + C
Using the initial condition y(3) = 6, we can solve for the constant C:
3 \cdot 6 = \ln|3| + C
Therefore, C = 18 - \ln(3).
The solution is then given by:
y = \fracxx
Substituting back the value of C, we get:
y = \fracxx