Verify that the function $y=c_{1} e^{a x} \cos b x+c_{2} e^{a x} \sin b x,$ where $c_{1}, c_{2}$ are arbitrary constants,is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0$.

The given function is $y=e^{a x}\left[c_{1} \cos b x+c_{2} \sin b x\right]$ .........$(1)$
Differentiating both sides of equation $(1)$ with respect to $x,$ we get
$\frac{d y}{d x}=e^{a x}\left[-b c_{1} \sin b x+b c_{2} \cos b x\right]+\left[c_{1} \cos b x+c_{2} \sin b x\right] e^{a x} \cdot a$
$\frac{d y}{d x}=e^{a x}\left[\left(b c_{2}+a c_{1}\right) \cos b x+\left(a c_{2}-b c_{1}\right) \sin b x\right]$ .........$(2)$
Differentiating both sides of equation $(2)$ with respect to $x,$ we get
$\frac{d^{2} y}{d x^{2}}=e^{a x}\left[\left(b c_{2}+a c_{1}\right)(-b \sin b x)+\left(a c_{2}-b c_{1}\right)(b \cos b x)\right]+\left[\left(b c_{2}+a c_{1}\right) \cos b x+\left(a c_{2}-b c_{1}\right) \sin b x\right] e^{a x} \cdot a$
$=e^{a x}\left[\left(a^{2} c_{2}-2 a b c_{1}-b^{2} c_{2}\right) \sin b x+\left(a^{2} c_{1}+2 a b c_{2}-b^{2} c_{1}\right) \cos b x\right]$
Substituting the values of $\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}$ and $y$ in the given differential equation,we get
$L.H.S. = e^{a x}\left[\left(a^{2} c_{2}-2 a b c_{1}-b^{2} c_{2}\right) \sin b x+\left(a^{2} c_{1}+2 a b c_{2}-b^{2} c_{1}\right) \cos b x\right] - 2 a e^{a x}\left[\left(b c_{2}+a c_{1}\right) \cos b x+\left(a c_{2}-b c_{1}\right) \sin b x\right] + \left(a^{2}+b^{2}\right) e^{a x}\left[c_{1} \cos b x+c_{2} \sin b x\right]$
$=e^{a x}\left[\left(a^{2} c_{2}-2 a b c_{1}-b^{2} c_{2}-2 a^{2} c_{2}+2 a b c_{1}+a^{2} c_{2}+b^{2} c_{2}\right) \sin b x + \left(a^{2} c_{1}+2 a b c_{2}-b^{2} c_{1}-2 a b c_{2}-2 a^{2} c_{1}+a^{2} c_{1}+b^{2} c_{1}\right) \cos b x\right]$
$=e^{a x}[0 \cdot \sin b x + 0 \cdot \cos b x] = 0 = R.H.S.$
Hence,the given function is a solution of the given differential equation.