Does This Differential Equation Solution Satisfy Given Conditions?

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SUMMARY

The discussion focuses on verifying the general solution of the differential equation xy'' + y' = 0, with the proposed solution y = C1 + C2 ln(x). The user successfully derives the first and second derivatives, y' and y'', and substitutes them back into the equation, confirming that the equation holds true. The particular solution is then determined by applying the initial conditions y(2) = 0 and y'(2) = 1/2, leading to the conclusion that y'' = -1/4 when evaluated at x = 2.

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Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone seeking to understand the verification of solutions and initial conditions in mathematical modeling.

goaliejoe35
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Homework Statement
Verify that the general solution satisfies the differential equation. Then find the particular solution that satisfies the initial condition.

y=C1+C2 lnx

xy'' + y' = 0

y=0 when x=2
y'=(1/2) when x=2

The attempt at a solution

Here's what I did so far...

y=C1+C2 lnx
y'=C1+C2(1/x)
y''=C1+C2(-1/x^2)

Since xy''+y'=0 I then substituted y' and y'' into the equation.

x(C1+C2(-1/x^2))+(C1+C2(1/x))=0

After this step I am stuck. If you could help push me in the right direction that would be great. Also could you verify that what I already did is right?
 
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The derivative of a constant is zero, as I'm sure you're aware, so C1 drops out of y' and y''

then since [itex]y'+xy''=0[/itex] you have [tex]\frac{C2}{x}-\frac{C2x}{x^2}=\frac{C2}{x}-\frac{C2}{x}=0[/tex] as desired.

The second part is just plugging in some numbers, [itex](2)(y'')+\frac{1}{2}=0 \implies y''= - \frac{1}{4}[/itex]
 

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