y%26%2339%3B%3D1.jpeg "(y^2+xy^2)y'=1")
Solving the Differential Equation: (y^2 + xy^2)y' = 1
This article aims to provide a comprehensive solution to the given differential equation:
(y^2 + xy^2)y' = 1
We will utilize various techniques to arrive at the general solution.
1. Simplifying the Equation
First, let's simplify the equation by factoring out y² on the left-hand side:
y²(1 + x)y' = 1
2. Separating Variables
To solve this differential equation, we need to separate the variables 'x' and 'y'. Divide both sides of the equation by y²(1+x):
y' = 1/(y²(1+x))
Now, we can rewrite y' as dy/dx:
dy/dx = 1/(y²(1+x))
Multiply both sides by dx and y²:
y² dy = dx/(1+x)
3. Integrating Both Sides
Now we have successfully separated the variables. Let's integrate both sides of the equation:
∫ y² dy = ∫ dx/(1+x)
The left side integrates to (y³/3) and the right side integrates to ln|1+x| + C, where C is the constant of integration:
(y³/3) = ln|1+x| + C
4. Solving for y
To isolate y, we can perform the following steps:
- Multiply both sides by 3: y³ = 3ln|1+x| + 3C
- Take the cube root of both sides: y = (3ln|1+x| + 3C)^(1/3)
5. Simplifying the Constant
Since 3C is also a constant, we can rewrite the solution as:
y = (3ln|1+x| + C)^(1/3)
where C represents a general constant of integration.
Conclusion
The general solution to the differential equation (y² + xy²)y' = 1 is:
y = (3ln|1+x| + C)^(1/3)
This solution represents a family of curves. The specific curve that satisfies an initial condition can be obtained by plugging in the initial condition and solving for the constant C.