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 $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$.

Concept: Verification of Solution — substitute the candidate function into the differential equation and check that it satisfies the equation identically.

Step 1: First derivative

$y = e^{ax}(c_1\cos bx + c_2\sin bx)$

Using the product rule:

$\frac{dy}{dx} = a e^{ax}(c_1\cos bx + c_2\sin bx) + e^{ax}(-b c_1\sin bx + b c_2\cos bx)$

$= e^{ax}\big[(a c_1 + b c_2)\cos bx + (a c_2 - b c_1)\sin bx\big]$

Step 2: Second derivative

Differentiate again (product rule on each term):

$\frac{d^2y}{dx^2} = a e^{ax}\big[(a c_1 + b c_2)\cos bx + (a c_2 - b c_1)\sin bx\big] + e^{ax}\big[-b(a c_1 + b c_2)\sin bx + b(a c_2 - b c_1)\cos bx\big]$

$= e^{ax}\big[(a^2 c_1 + a b c_2 + a b c_2 - b^2 c_1)\cos bx + (a^2 c_2 - a b c_1 - a b c_1 - b^2 c_2)\sin bx\big]$

$= e^{ax}\big[(a^2 - b^2)c_1 + 2ab c_2\big]\cos bx + \big[(a^2 - b^2)c_2 - 2ab c_1\big]\sin bx$

Step 3: Substitute into the DE

Compute $-2a\frac{dy}{dx} + (a^2+b^2)y$ and add to $\frac{d^2y}{dx^2}$.

After collecting coefficients of $\cos bx$ and $\sin bx$, each coefficient simplifies to zero. For example, the $\cos bx$ coefficient:

$(a^2 - b^2)c_1 + 2ab c_2 - 2a(a c_1 + b c_2) + (a^2+b^2)c_1 = 0$

Similarly for $\sin bx$. Hence the DE is satisfied for all $x$.

We need to check that $y = c_1 e^{ax}\cos bx + c_2 e^{ax}\sin bx$ satisfies

The key idea: this is a linear homogeneous ODE with constant coefficients. For such equations, if we can show that both $e^{ax}\cos bx$ and $e^{ax}\sin bx$ individually satisfy the ODE, then any linear combination (like the given $y$) will also satisfy it, by linearity. So we can either verify the combination directly, or verify each basis function separately. We'll do the direct substitution — it's cleaner and avoids repeating work.

1. First derivative Differentiate $y$ term by term. For $e^{ax}\cos bx$, use the product rule:

For $e^{ax}\sin bx$:

So

2. Second derivative Differentiate again. For the $c_1$ part:

Simplify:

For the $c_2$ part:

Simplify:

Hence

3. Form the combination We need to compute:

It's efficient to group terms by the two basis functions. Let’s collect coefficients of $e^{ax}\cos bx$ and $e^{ax}\sin bx$ separately.

Coefficient of $e^{ax}\cos bx$ (from all three pieces):

• From $\frac{d^2y}{dx^2}$: $c_1(a^2 - b^2) + c_2(2ab)$
• From $-2a\frac{dy}{dx}$: $-2a\big[c_1 a + c_2 b\big]$ (since $\frac{dy}{dx}$ has $c_1 a$ and $c_2 b$ multiplying $\cos bx$)
• From $(a^2+b^2)y$: $(a^2+b^2)c_1$

Sum these:

Simplify $c_1$ terms: $a^2 - b^2 - 2a^2 + a^2 + b^2 = 0$.

Simplify $c_2$ terms: $2ab - 2ab = 0$.

So the coefficient of $e^{ax}\cos bx$ is $0$.

Coefficient of $e^{ax}\sin bx$:

• From $\frac{d^2y}{dx^2}$: $c_1(-2ab) + c_2(a^2 - b^2)$
• From $-2a\frac{dy}{dx}$: $-2a\big[-c_1 b + c_2 a\big]$ (since $\frac{dy}{dx}$ has $-c_1 b$ and $c_2 a$ multiplying $\sin bx$)
• From $(a^2+b^2)y$: $(a^2+b^2)c_2$

Sum:

Simplify $c_1$ terms: $-2ab + 2ab = 0$.

Simplify $c_2$ terms: $a^2 - b^2 - 2a^2 + a^2 + b^2 = 0$.

Again zero.

4. Conclusion Both coefficients vanish, so the entire expression equals $0$ for all $x$, regardless of $c_1$ and $c_2$. Hence $y$ is indeed a solution.

Method: Verifying a two-constant candidate for a second-order equation

Use this when asked to verify that a function containing arbitrary constants (here $c_1, c_2$) solves a second-order linear equation. The technique is substitution, organised so the algebra stays manageable.

Steps

Step 1: Differentiate carefully with the product rule.

For a product like $e^{ax}\cos bx$, both factors depend on $x$, so $\dfrac{d}{dx}\big(e^{ax}\cos bx\big) = a e^{ax}\cos bx - b e^{ax}\sin bx$. Keep the constants $a,b,c_1,c_2$ fixed.

Step 2: Differentiate again for $\dfrac{d^2y}{dx^2}$, tracking cross terms.

Differentiating a $\cos$ term feeds the $\sin$ term and vice versa — do not drop these.

Step 3: Substitute into the equation and group by basis function.

Collect the coefficient of $e^{ax}\cos bx$ and, separately, the coefficient of $e^{ax}\sin bx$.

Step 4: Show each grouped coefficient is $0$.

If both coefficients vanish for all $c_1,c_2$, the left side is identically $0$ and the function is verified as a solution.

Common Mistakes

Mistake 1: Product-rule slips on $e^{ax}\cos bx$.

Why it's wrong: both factors depend on $x$, so $\dfrac{d}{dx}(e^{ax}\cos bx) = a e^{ax}\cos bx - b e^{ax}\sin bx$; forgetting either term breaks the verification. Correct approach: differentiate both factors every time.

Mistake 2: Dropping cross terms in the second derivative.

Why it's wrong: differentiating a $\cos$ part contributes to the $\sin$ part and vice versa; losing these makes the coefficients fail to cancel. Correct approach: keep every term and group by $e^{ax}\cos bx$ and $e^{ax}\sin bx$.

Mistake 3: Treating $a$, $b$, $c_1$, $c_2$ as variables.

Why it's wrong: they are constants, so they never get differentiated. Correct approach: hold them fixed and show each grouped coefficient reduces to $0$.