Asked by Carrie
The general solution to the differential equation dy/dx=xy is y=±√(x^2+C). Let y=f(x) be the particular solution to the differential equation with the initial condition f(−5)=−4. What is an expression for f(x) and its domain?
Answers
Answered by
oobleck
The problem I see is that the solution to y'=xy is
y = c*e^(x^2/2)
and if y(-5) = -4 then
C = -4e^(25/2)
and so y = -4e^(1/2 (x^2-25))
with domain all real numbers
I think you meant y' = x/y
That gives y=±√(x^2+C)
so, if y(-5) =-4,
-√(25+C) = -4
25+C = 16
C = -9
y = -√(x^2-9)
so the domain is all |x| >= 3
y = c*e^(x^2/2)
and if y(-5) = -4 then
C = -4e^(25/2)
and so y = -4e^(1/2 (x^2-25))
with domain all real numbers
I think you meant y' = x/y
That gives y=±√(x^2+C)
so, if y(-5) =-4,
-√(25+C) = -4
25+C = 16
C = -9
y = -√(x^2-9)
so the domain is all |x| >= 3
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