Asked by Lucy
I am so confused on how to solve this problem. I have it set up with y's on one side and x's on the other but I don't know what to do from there.
Question: Solve the differential equation. The initial condition is y(0) = 1.
((x^2+1)^(1/2))(dy/dx) - (x/(2y)) = 0
y > 0
My work: 2y = (x/(x^2+1)^(1/2))(dx)
Thank You!!
Question: Solve the differential equation. The initial condition is y(0) = 1.
((x^2+1)^(1/2))(dy/dx) - (x/(2y)) = 0
y > 0
My work: 2y = (x/(x^2+1)^(1/2))(dx)
Thank You!!
Answers
Answered by
Steve
√(x^2+1) dy/dx - x/(2y) = 0
√(x^2+1) dy/dx = x/(2y)
2y dy = x/√(x^2+1) dx
y^2 = √(x^2+1) + c
y(0)=1, so
1 = 1+c
c = 0
y^2 = √(x^2+1)
y = ∜(x^2+1)
√(x^2+1) dy/dx = x/(2y)
2y dy = x/√(x^2+1) dx
y^2 = √(x^2+1) + c
y(0)=1, so
1 = 1+c
c = 0
y^2 = √(x^2+1)
y = ∜(x^2+1)
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