Find the domain for the particular solution to the differential equation dx/dy=1/2x, with initial condition y(1) = 1.
Boi just solve the question he had. He meant something, so solve it
srsly, Steve ur lackin
I’m sick and tired of this partially answer questions! We didn’t come here to be told the way to solve the question, we came here for the ANSWERS
dx/dy=1/2x
dx 2x= dy
x^2=y
The domain is all real numbers
I think he meant 1/(2x) as written....
To find the domain of the particular solution to the given differential equation dx/dy = 1/2x, we need to consider the initial condition y(1) = 1.
In order to solve the differential equation, we can separate the variables and integrate both sides:
dy = 2x dx
Integrating both sides gives:
∫1 dy = ∫2x dx
y = x^2 + C
where C is the constant of integration.
Now, using the initial condition y(1) = 1, we can substitute the values into the equation:
1 = 1^2 + C
1 = 1 + C
C = 0
Therefore, the particular solution to the differential equation is given by:
y = x^2
Now let's find the domain of this solution. The domain is the set of all x-values for which the solution is defined.
In this case, since y = x^2 does not have any restrictions on the x-values, the domain of the particular solution is all real numbers, or (-∞, +∞).
So, the domain for the particular solution to the differential equation dx/dy = 1/2x, with the initial condition y(1) = 1, is (-∞, +∞).
if you mean dx/dy = (1/2) x, then
dy/dx = 2/x
dy = 2/x dx
y = 2 lnx + c
domain: x>0
I'm sure you can find c.
I'm surprised that you give dx/dy when looking for y(x)...