Solution Of A Differential Equation

Solution of a differential equation – General and particular:

Consider the following equation:

\(2x^2 – 5x – 7 = 0\)

The solution to this equation is a number i.e. -1 or \(\frac{7}{2}\)which satisfies the above equation. But in case of differential equation the solution is a function that satisfies the given differential equation.

Example: \(\frac{dy}{dx} = x^2 \)

\(\Rightarrow dy = x^2 dx \)

Integrating both sides, we get

\(\Rightarrow \int dy = \int x^2 dx \)

If we solve this equation to figure out the value of y we get

\( y = \frac{x^3}{3} + C \)

where C is any arbitrary constant.

In the above obtained solution, we see that \( y \) is a function of \( x \). On substituting this value of y in the given differential equation, both the sides of the differential equation becomes equal.

Solution of a Differential Equation – General 

The solution obtained above after integration consists of a function and an arbitrary constant. This represents general solution of the given equation.

Let the solution be represented as \( y = \phi(x) + C \) . It represents the solution curve or the integral curve of the given differential equation.

Thus, we can say that a general solution always involves a constant C.

Let us consider some more examples:

Example: Find the general solution of a differential equation \(\frac{dy}{dx} = e^x + cos2x  + 2x^3\).

Solution:\( \frac{dy}{dx} = e^x + cos 2x + 2x^3\).

\(\Rightarrow dy = (e^x + cos2x + 2x^3)dx \)

Integrating both the sides, we get

\( y = e^x + \frac{sin 2x}{2} + \frac{x^4}{2} + C\)

This represents the general solution of the given differential equation as it involves the constant C.

Solution of a Differential Equation – Particular :

When the arbitrary constant of the general solution takes some unique value, then the solution becomes the particular solution of the equation.

By using the boundary conditions (also known as the initial conditions) the particular solution of a differential equation is obtained.

So, to obtain the particular solution, first of all general solution is found out and then, by using the given conditions the particular solution is generated.

Suppose in the above mentioned example we are given to find the particular solution if

\(\frac{dy}{dx} = e^x + cos2x + 2x^3\), given that for

Then we know, the general solution is

\( y = e^x + \frac{sin2x}{2} + \frac{x^4}{2} + C\)

Now,\( x = 0, y = 5\) substituting this value in the general solution we get,

\(5 = e^{0} + \frac{sin(0)}{2} + \frac{(0)^4}{2} + C \)

\( \Rightarrow C = 4 \)

Hence substituting the value of C in the general solution we obtain,

\( y = e^x + \frac{sin2x}{2} + \frac{x^4}{2} + 4\)

This represents the particular solution of the given equation.

To get a better insight of the topic, let us have a look at the following example.

Example – Find out the particular solution of the differential equation \( ln \frac{dy}{dx} = 4y + ln x \), given that for \(x = 0, y = 0\).

Solution –\( \Rightarrow \frac{dy}{dx} = e^{4y + ln x} \)

\( \Rightarrow \frac{dy}{dx} = e^{4y} × e^{ln x} \)

\( \Rightarrow \frac{dy}{dx} = e^{4y} × x \)

\( \Rightarrow \frac{1}{e^{4y}}dy = x dx\)

\( \Rightarrow e^{-4y}dy = x dx \)

Integrating both the sides with respect to y and x respectively we get,

\( \frac{e^{-4y}}{-4} = \frac{x^2}{2} + C \)

This represents the general solution of the differential equation given.

Now, it is also given that \( y(0) = 0 \), substituting this value in the above general solution we get,

\(\frac{e^0}{-4} = \frac{0^2}{2} + C\)

\(\Rightarrow C = \frac{-1}{4}\)

Hence, the above equation can be rewritten as

\(\frac{e^{-4y}}{-4} = \frac{x^{2}}{2} – \frac{1}{4}\)

\(\Rightarrow e^{-4y} = -2x^{2} + 1\)

\(\Rightarrow \ln (e^{-4y}) = \ln (1 -2x^{2} )\)

\(\Rightarrow -4y = \ln (1 -2x^{2} )\)

\(\Rightarrow y = – \frac{\ln (1 -2x^{2} )}{4}\)

which is the particular solution of the differential equation given.

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Practise This Question

The degree of the differential equation satisfying 1x4+1y4=a(x2y2), is