The discussion focuses on solving the differential equation dy/dx = 4yx^3 - y with the initial condition y(1) = -3. The integration process leads to the expression ln(y) = x^4 - x + ln(C), which simplifies to y = Ce^(x^4 - x). The key point raised is the use of ln(C) instead of just C during integration, which is clarified as a method to streamline the final expression for easier application of initial conditions.
PREREQUISITESStudents studying calculus, particularly those focusing on differential equations, as well as educators looking for clear explanations of integration techniques and initial conditions.
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...?