dy%2Fdt%2Bt%2F2y%5E4%3D0.jpeg "(3y^2-t^2/y^5)dy/dt+t/2y^4=0")
Solving the Differential Equation: (3y^2 - t^2/y^5) dy/dt + t/2y^4 = 0
This article will explore the solution to the given differential equation:
(3y^2 - t^2/y^5) dy/dt + t/2y^4 = 0
This equation is a first-order, non-linear, ordinary differential equation. To solve it, we will utilize techniques like separation of variables and integration.
1. Rearranging the equation
First, we need to rearrange the equation to separate the variables 'y' and 't':
-
Move the term with 't' to the right side: (3y^2 - t^2/y^5) dy/dt = -t/2y^4
-
Divide both sides by (3y^2 - t^2/y^5): dy/dt = -t/(2y^4 * (3y^2 - t^2/y^5))
-
Simplify the right side: dy/dt = -t/(6y^6 - 2t^2)
2. Separation of variables
Now, we can separate the variables 'y' and 't' by multiplying both sides by dt and dividing by the expression containing 'y':
(6y^6 - 2t^2) dy = -t dt
3. Integration
Integrate both sides of the equation with respect to their respective variables:
∫(6y^6 - 2t^2) dy = ∫-t dt
This gives us:
y^7 - (2/3)t^3 = - (1/2)t^2 + C
Where 'C' is the constant of integration.
4. Solving for y
To get an explicit solution for 'y', we need to isolate it:
y^7 = (2/3)t^3 - (1/2)t^2 + C
y = ( (2/3)t^3 - (1/2)t^2 + C )^(1/7)
Conclusion
The solution to the differential equation (3y^2 - t^2/y^5) dy/dt + t/2y^4 = 0 is given by:
y = ( (2/3)t^3 - (1/2)t^2 + C )^(1/7)
where 'C' is the constant of integration, determined by initial conditions or boundary values.
This solution represents a family of curves, each corresponding to a different value of 'C'. Understanding the behavior of these curves and their relationship to the initial conditions can provide valuable insights into the system modeled by the differential equation.