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!!
Steve
answered
√(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)
Explain Bot
answered
((x^2+1)^(1/2))(dy/dx) - (x/(2y)) = 0, with the initial condition y(0) = 1.
The first step is to group the terms involving y and x on opposite sides of the equation. Rearranging the equation, we have:
((x^2+1)^(1/2))(dy/dx) = (x/(2y))
Now, we will multiply both sides of the equation by (2y)/(x^2+1)^(1/2):
(2y)(dy/dx) = (x/(x^2+1)^(1/2))
Next, let's integrate both sides with respect to x:
∫ (2y) dy = ∫ (x/(x^2+1)^(1/2)) dx
Integrating the left-hand side gives us:
y^2 = C + ∫ (x/(x^2+1)^(1/2)) dx
Where C is the constant of integration.
To solve the right-hand side integral, we can make a u-substitution, letting u = x^2 + 1:
The derivative of u is du/dx = 2x, which implies dx = du/(2x).
Substituting this into our integral:
∫ (x/(x^2+1)^(1/2)) dx = ∫ (x/u^(1/2)) (du/(2x))
Since the x terms cancel out, we are left with:
∫ du/(2 u^(1/2))
Using the power rule for integration, we get:
(1/2) ∫ u^(-1/2) du = (1/2) (2u^(1/2)) + C'
Simplifying further:
∫ (x/(x^2+1)^(1/2)) dx = u^(1/2) + C'
Substituting back u = x^2 + 1:
∫ (x/(x^2+1)^(1/2)) dx = (x^2 + 1)^(1/2) + C'
Now we can substitute this result back into our equation:
y^2 = C + (x^2 + 1)^(1/2) + C'
Since the initial condition is y(0) = 1, we can substitute x = 0 and y = 1 into the equation:
1 = C + (0^2 + 1)^(1/2) + C'
Simplifying further, we find C + C' = 0.
Therefore, our equation becomes:
y^2 = (x^2 + 1)^(1/2) - (x^2 + 1)^(1/2)
Simplifying this expression gives us:
y^2 = 0
Taking the square root of both sides, we find:
y = 0
Therefore, the solution to the differential equation with the initial condition y(0) = 1 is y = 0.
I hope this explanation helps you understand how to approach and solve the given problem. Let me know if you have any further questions!
