Differential Equations

Differential Equations with Variables Separable

A separable differential equation is a common kind of differential calculus equation that is especially straightforward to solve. Separable equations have the form dy/dx = f(x) g(y), and are called separable because the variables x and y can be brought to opposite sides of the equation.

Then, integrating both sides gives y as a function of x, solving the differential equation. Separable differential equations are useful because they can be used to understand the rates of chemical reactions, the growth of populations, the movement of projectiles, and many other physical systems.

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Differential Calculus with Variables Separable

Many practically useful first-order differential equations can be reduced, by purely algebraic manipulations, to the form –

\( g(y)y’ = f(x)\)
\(g(y)dy = f(x)dx\)

where f and g are continuous functions This is known as the Variables Separable form of the differential equation. The name must be clear from the format of the equation since all of the terms with the independent variable x are on one side, while all of the terms with the dependent variable y are on the other side. Such an equation can be easily solved for its solution in the following way –

Integrate on both sides with respect to x

\(\int{g(y)y’dx} = \int{f(x)dx} + c\)

Now, y’dx is simply dy, so on the left-hand side, we can switch to y as the variable of integration.

\(\int{g(y)dy} = \int{f(x)dx} + c\)

This is the solution of the differential equation in variables separable form. This solution is only valid over the range of x, known as the domain of y(x). That’s all you need to know about this method. You must solve plenty of problems to make yourself proficient in applying it. The solved examples below will introduce you to some more new concepts, and should be enough to set you on track!

Solved Examples on Differential Calculus

Example 1: Solve \(y’ = e^{-x}y^2\)

Solution: Separating the variable, and integrating with respect to x –

\(y^{-2}y’ = e^{-x}\)
\(\int{y^{-2}dy} = \int{e^{-x}}\)
\(-\frac{1}{y} = -e^{-x} + c\)

where c is the constant of integration and must be introduced immediately after integration. The explicit i.e. of the form, y = f(x) solution can then be given as –

\(y = \frac{1}{e^{-x} + k}\)

Here, we have replaced by the previous arbitrary constant c, by a new constant k, where k = -c.

Radioactive Decay

You must have read in Physics that the phenomenon of radioactive decay is governed by the differential equation –

\(\frac{dN}{dt} = -kN\)

where N – the number of radioactive atoms present in the sample at time t, k – proportionality constant (the radioactive decay constant). This differential equation is clearly a variables-separable equation. Let us now solve it for the genral solution –

\(N'(t) = -kN\)
\(\frac{N'(t)}{N} = -k\)
\(\int{\frac{1}{N}dN} = -\int{kdt}\)
\(ln(N) = -kt + c\)
\(N = e^{-kt + c} \), where c is the arbitrary constant.

We can find the constant c by using initial conditions of the system. Put t = 0, and N = N0 (the initial number of radioactive atoms in the system) to get the value of c.

\(ln(N_0) = -k.0 + c\)
\(c = ln(N_0)\)

Using this value, one can get the particular solution as –

\(ln(N) = -kt + ln(N_0)\)
\(N = N_0e^{-kt}\)

This is just the famous law of Radioactive Decay!

Example 2: Solve the initial value problem

\(\frac{dy}{dx} = \frac{x}{y}\)
y(3) = 2

Solution: Using the variables-separable method –

\(y\frac{dy}{dx} = x\)
\(\int{ydy} = \int{xdx}\)
\(\frac{y^2}{2} = \frac{x^2}{2} + c\)

To find the arbitrary constant c, we use the value given in the problem.

\(\frac{2^2}{2} = \frac{3^2}{2} + c\)
\(c = -\frac{5}{2}\)

Using this value, we then get the particular solution as –

\(\frac{y^2}{2} = \frac{x^2}{2} – \frac{5}{2}\)
\(x^2 – y^2 = 5\)

This concludes our discussion on this topic of differential calculus.

Solved Questions for You

Question 1: Is Calculus 1 differential calculus?

Answer: Calculus 1 does cover differential calculus to an extent. Similarly, Calculus 2 covers integral calculus approximately. Further, Calculus 3 covers multivariable calculus roughly.

Question 2: What are the differential equations?

Answer: Separable differential equations are a very common type of differential calculus equation which is particularly quite simple to solve. Moreover, separable equations have the form dy/dx = f(x) g (y) and are referred to as separable because one can bring the variables x and y to the opposite sides of the equation.

Question 3: Why do we study differential calculus?

Answer: Differential calculus concerns with finding the instant rate at which one quantity changes with regards to the other, which we call the derivative of the first quantity with regards to the second. Further, integral calculus deals with the inverse of the derivative, that is to say, to find a function when we know the rate of change.

Question 4: How many types of calculus are there?       

Answer: There are two kinds of calculus. Differential calculus breaks up things in several small parts and tells how they change from one instant to the next. On the other hand, integral calculus connects the small pieces together and lets us know how much of something is made, on the whole, through a series of changes.

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