solve the equation
y'=(sqrt(x))/2y
well,
dy/dx = √x/2y
2y dy = √x dx
...
Ah! I see. the y' threw me off. Thanks
To solve the given differential equation:
y' = sqrt(x)/(2y)
We will use the technique of separation of variables. The basic idea is to rewrite the equation in a way that separates the variables x and y, and then integrate both sides with respect to their respective variables.
First, let's rewrite the equation by multiplying both sides by 2y:
2yy' = sqrt(x)
Now, we can separate the variables:
2y dy = sqrt(x) dx
Next, we integrate both sides. On the left side, we integrate with respect to y, and on the right side, we integrate with respect to x:
∫2y dy = ∫sqrt(x) dx
Integrating the left side:
∫2y dy = y^2 + C1, where C1 is the constant of integration.
Integrating the right side:
∫sqrt(x) dx = (2/3)x^(3/2) + C2, where C2 is another constant of integration.
Now, we have:
y^2 + C1 = (2/3)x^(3/2) + C2
To solve for y, we isolate y on one side:
y^2 = (2/3)x^(3/2) + C2 - C1
y = ± sqrt((2/3)x^(3/2) + (C2 - C1))
So, the general solution to the given differential equation is:
y = ± sqrt((2/3)x^(3/2) + (C2 - C1))
where C1 and C2 are constants determined by any initial conditions or boundary conditions given in the problem.