Solution of Differential Equation

Here you will learn how to find solution of differential equation i.e. general solution and particular solution with examples.

Let’s begin –

Solution of Differential Equation

The solution of differential equation is a relation between the variables involved which satisfies the differential equation.

for example, y = \(e^x\) is a solution of the differential equations \(dy\over dx\) = y.

General Solution

The solution which contains as many as arbitrary constants as the order of the differential equations is called the general solution.

for example, y = A cos x + B sin x is the general solution of the equation \(d^2y\over dx^2\) + y = 0.

Particular Solution

The solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equations is called a particular solution.

for example, y = 3 cos x + 2 sin x is the particular solution of the equation \(d^2y\over dx^2\) + y = 0.

Example : Show that y = Ax  + \(B\over x\), x \(\ne\) 0 is a solution of the differential equation \(x^2\)\(d^2y\over dx^2\) + x\(dy\over dx\) – y = 0

Solution : We have,

y = Ax  + \(B\over x\) = 0, x \(\ne\) 0                ……..(i)

Differentiating both sides with respect to x, we get

\(dy\over dx\) = A – \(B\over x^2\)                   ……….(ii)

Again differentiating with respect to x, we get

\(d^2y\over dx^2\) = \(2B\over x^3\)            …………..(iii)

Substituting the values of y, \(dy\over dx\) and \(d^2y\over dx^2\) in \(x^2\)\(d^2y\over dx^2\) + x\(dy\over dx\) – y , we get

\(x^2\)\(d^2y\over dx^2\) + x\(dy\over dx\) – y = \(x^2\)(\(2B\over x^3\)) + x(A – \(B\over x^2\)) – (Ax + \(B\over x\)) 

= \(2B\over x\) + Ax – \(B\over x\) – Ax – \(B\over x\) = 0

Thus, the function y = Ax  + \(B\over x\) satisfies the given equation.

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