Question collected form Telegram Group.
Solution 1:
Antiderivatives help us find a function whose derivative is
known. They are used to solve differential equations which relate an unknown
function and one or more of its derivatives.
For example:
If we have a differential equation dy/dx = 2x, we can find
its solution by integrating both sides with respect to x. The antiderivative of
2x is $x^2 + C$, where C is a constant of integration. Therefore, the general
solution of the differential equation is $y = x^2 + C$.
When we have a separable differential equation. We can separate the variables by writing it as dy/g(y) = f(x)dx and then integrate both sides. We get ln|g(y)| = F(x) + C3, where C3 = C2 - C1. We can then solve for y by taking the exponential of both sides.
Solution 2 (a bit lengthy):
The concept of anti-derivative is necessary for solving differential equations because the derivative of an anti-derivative is the original function. This means that if we know the derivative of a function, we can find the anti-derivative by taking the opposite of the derivative.
For example, if we know that the derivative of $f(x) = x^2$ is $2x$, then we can find the anti-derivative of $2x$ by taking the opposite of the derivative, which is $x^2 + C$, where $C$ is an arbitrary constant.
This can be used to solve differential equations by taking the derivative of both sides of the equation. This will give us an equation that contains the derivative of the unknown function. We can then use the concept of anti-derivatives to find the anti-derivative of the derivative of the unknown function, which will give us the unknown function itself.
For example, if we have the differential equation $y' = 2x$, we can take the derivative of both sides to get $y'' = 2$. We can then use the concept of anti-derivatives to find the anti-derivative of $2$, which is $x + C$, where $C$ is an arbitrary constant. This means that the solution to the differential equation is $y = x^2 + C$.
The concept of anti-derivative is also necessary for solving initial value problems. An initial value problem is a differential equation along with a set of initial conditions that specify the value of the unknown function at a particular point.
For example, the initial value problem $y' = 2x$, $y(0) = 1$ can be solved by using the concept of anti-derivatives. We know that the solution to the differential equation is $y = x^2 + C$. We can then use the initial condition $y(0) = 1$ to solve for $C$. This gives us $C = 1$. Therefore, the solution to the initial value problem is $y = x^2 + 1$.
In conclusion, the concept of anti-derivative is necessary for solving differential equations and initial value problems.