Boundary Value Problems/Introduction to BVPs/IVP-student-1

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Problem: Differential equation: y=xy with initial condition: y(0)=3

Start Solution:

1ydy=xdx

ln(|y|)=x22+C

eln(|y|)=ex22+C

|y|=ex22eC

|y|=ex22C1

|y|=C1ex22

Use the initial condition y(0)=3 to solve for the constant C1.

|y(0)|=3=C1e0

C1=3

The particular solution to the initial value problem is:

y=3ex22 /Why did I drop the "absolute value" operation/?


End Solution:

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