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.
