UPSC Maths 2013 Paper 1 Q6b — Solution

10 marks · Section B

Question

Using the method of variation of parameters, solve the differential equation

\frac{d^{2}y}{dx^{2}}+a^{2}y=\sec ax.

Standard variation-of-parameters formula; the key trig identities $\sin\,\sec=\tan$ and $\cos\,\sec=1$ simplify both integrals.

Strategy. Find homogeneous solutions $y_1,y_2$ ; compute Wronskian $W$ ; apply the formula

y_p=-y_1\int\frac{y_2\,g(x)}{W}\,dx+y_2\int\frac{y_1\,g(x)}{W}\,dx

where $g(x)$ is the inhomogeneous term.

Characteristic equation of $y''+a^{2}y=0$ is $r^{2}+a^{2}=0$ , giving $r=\pm ai$ . So

y_1=\cos ax,\qquad y_2=\sin ax.

W=\begin{vmatrix}y_1 & y_2\\ y_1' & y_2'\end{vmatrix}=\begin{vmatrix}\cos ax & \sin ax\\ -a\sin ax & a\cos ax\end{vmatrix}=a\cos^{2}ax+a\sin^{2}ax=a.

With $g(x)=\sec ax$ :

First integral: $-y_1\int\dfrac{y_2\,g}{W}\,dx=-\cos ax\cdot\dfrac{1}{a}\int\sin ax\,\sec ax\,dx=-\dfrac{\cos ax}{a}\int\tan ax\,dx$ .

$\int\tan ax\,dx=-\dfrac{1}{a}\ln|\cos ax|$ , so

-\dfrac{\cos ax}{a}\cdot\!\left(-\dfrac{1}{a}\ln|\cos ax|\right)=\dfrac{\cos ax}{a^{2}}\ln|\cos ax|.

Second integral: $y_2\int\dfrac{y_1\,g}{W}\,dx=\dfrac{\sin ax}{a}\int\cos ax\,\sec ax\,dx=\dfrac{\sin ax}{a}\int 1\,dx=\dfrac{x\sin ax}{a}$ .

y_p=\frac{\cos ax}{a^{2}}\ln|\cos ax|+\frac{x\sin ax}{a}.

\boxed{\;y=C_1\cos ax+C_2\sin ax+\frac{\cos ax}{a^{2}}\ln|\cos ax|+\frac{x\sin ax}{a}.\;}