Q.Verify that the function $y = c_1 e^{ax}\cos bx + c_2 e^{ax}\sin bx$, where $c_1, c_2$ are arbitrary constants is a solution of the differential equation $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$.

Verifying a Solution of a Differential Equation

A function $y = \phi(x)$ is called a solution of a differential equation if, when you substitute it and its derivatives into the equation, the two sides become equal for every $x$ in the domain. Verification is the act of carrying out that substitution and checking that it holds as an identity.

The useful point: you do not have to solve the equation to verify a candidate. You are only checking a function that is already handed to you — which is exactly how many exam questions are phrased: "Show that $\dots$ is a solution of $\dots$."

The steps

1. From the given $y = \phi(x)$, compute exactly the derivatives that appear in the equation.
2. Substitute $y$ and those derivatives into the left-hand side.
3. Simplify and check that it equals the right-hand side for all $x$ (an identity, not just at one point).

Example 1

Verify that $y = e^{-3x}$ is a solution of $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$.

Here $y' = -3e^{-3x}$ and $y'' = 9e^{-3x}$. Substituting:

The left side is $0$ for every $x$, so $y = e^{-3x}$ is a solution.

Example 2 (a solution with constants)

Verify that $y = a\cos x + b\sin x$ satisfies $\dfrac{d^2y}{dx^2} + y = 0$ for any constants $a, b$.

Since $y'' = -a\cos x - b\sin x = -y$, we get $y'' + y = 0$. It holds for all $a, b$, so this two-constant family is a solution.

Verifying that a given function solves a differential equation is explicitly listed as an exercise type in the NCERT Class 12 Mathematics textbook's Differential Equations chapter, and "verify the solution of differential equation examples" is a common CBSE and JEE Main search. This is often the easiest full-mark question in the chapter once the substitution steps are practiced a few times.