It's really the same thing, but it makes the final result look simpler. The two things are equivalent.PsychonautQQ said:Homework Statement
dy/dx =4yx^3-y y(1)=-3
dy/y = (4x^3-1)dx
ln(y) = x^4-x+C
y = e^(x^4-x+C)
But an answer source says that after the integration I get
ln(y) = x^4 - x + ln(C)
so then..
ln(y/c) = x^4 - x
y = Ce^(x^4 - x)
which makes it much easier to solve for the constant given the initial condition... My question is why when you take the integral is the constant ln(C) instead of just C...?
A separable differential equation is one in which the variables can be separated and solved for independently. This means that the equation can be written in the form of dy/dx = f(x)g(y), where f(x) is a function of x and g(y) is a function of y.
To solve a separable differential equation, you must first separate the variables on opposite sides of the equation. Then, you can integrate both sides and add a constant of integration. Finally, you can solve for y to get the general solution.
The initial condition in a separable differential equation is a known value for the dependent variable (usually denoted as y) at a specific value of the independent variable (usually denoted as x). This value is used to find the particular solution to the differential equation.
The initial condition is important because it helps to determine the particular solution to the differential equation. Without this known value, the general solution obtained may not accurately represent the behavior of the system being studied.
Yes, there are some limitations to using the separable method. This method can only be applied to first-order differential equations and is not always applicable to more complex equations. Additionally, the separable method may not always give the most general solution to a differential equation. Other methods, such as the integrating factor method, may be needed in some cases.