Question

Let y=y(x) is the solution of the differential equation $\dfrac{dy}{dx}+y\tan x={{x}^{2}}+2x\tan x,x\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ such that y(0) = 1, then
[a] $y'\left( \dfrac{\pi }{4} \right)-y\left( \dfrac{\pi }{4} \right)=2$
[b] $y'\left( \dfrac{\pi }{4} \right)-y\left( \dfrac{\pi }{4} \right)=7$
[c] $y\left( \dfrac{\pi }{4} \right)-y\left( \dfrac{-\pi }{4} \right)=\dfrac{{{\pi }^{2}}}{2}+\sqrt{2}$
[d] $y\left( \dfrac{\pi }{4} \right)-y\left( \dfrac{-\pi }{4} \right)=\dfrac{{{\pi }^{2}}}{8}+\sqrt{2}$

Answer 1

Hint: Observe that the given differential equation is a linear differential equation. Use the fact that the if IF is the integrating factor of the linear differential equation $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ then the solution of the differential equation is given by $yIF=\int{Q\left( x \right)IFdx}+C$. Use the fact that the integrating factor of the linear differential equation $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ is given by $IF={{e}^{\int{P\left( x \right)dx}}}$. Hence find the solution of the given differential equation.

Complete step-by-step answer:
We have $\dfrac{dy}{dx}+y\tan x=2x+{{x}^{2}}\tan x$ which is of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$, where $P\left( x \right)=\tan x$ and $Q\left( x \right)=2x+{{x}^{2}}\tan x$
We know that the integrating factor of the linear differential equation $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ is given by $IF={{e}^{\int{P\left( x \right)dx}}}$.
Hence, we have
$IF={{e}^{\int{\tan xdx}}}={{e}^{\ln \left( \sec x \right)}}=\sec x$
We know that if IF is the integrating factor of the linear differential equation $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ then the solution of the differential equation is given by $yIF=\int{Q\left( x \right)IFdx}+C$.
Hence, we have
$y\sec x=\int{\sec x\left( 2x+{{x}^{2}}\tan x \right)}dx$
We know that $\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$
Hence, we have
$y\sec x=\int{\sec x2xdx}+\int{{{x}^{2}}\sec x\tan x}$
We know that if $\int{f\left( x \right)dx}=u\left( x \right)$ and $\dfrac{d}{dx}g\left( x \right)=v\left( x \right)$, then $\int{f\left( x \right)g\left( x \right)dx}=g\left( x \right)u\left( x \right)-\int{u\left( x \right)v\left( x \right)dx}$. This is known as integration by parts.
The function f(x) is called the second function and the function g(x) is called the first function.
The order of preference (in general) for choosing first function is given by ILATE rule
I = Inverse Trigonometric
L = Logarithmic
A = Algebraic
T = Trigonometric
E = Exponential
Hence according to ILATE rule, we choose $f\left( x \right)=2x$ and $g\left( x \right)=\sec x$, we have
$u\left( x \right)=\int{2xdx}={{x}^{2}}$ and $v\left( x \right)=\dfrac{d}{dx}\sec x=\sec x\tan x$
Hence, we have
$\int{\sec x2xdx}={{x}^{2}}\sec x-\int{{{x}^{2}}\sec x\tan xdx}$
Hence, we have
$\begin{align}
  & y\sec x={{x}^{2}}\sec x-\int{{{x}^{2}}}\sec x\tan x+\int{{{x}^{2}}}\sec x\tan x+C \\
 & y={{x}^{2}}+C\cos x \\
\end{align}$
Hence, we have
$\begin{align}
  & y\left( 0 \right)={{0}^{2}}+C=1 \\
 & \Rightarrow C=1 \\
\end{align}$
Hence, we have
$\begin{align}
  & y'\left( x \right)-y\left( x \right)=2x-\sin x-{{x}^{2}}-\cos x \\
 & \Rightarrow y'\left( \dfrac{\pi }{4} \right)-y\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{2}-\dfrac{1}{\sqrt{2}}-\left( \dfrac{{{\pi }^{2}}}{16} \right)-\dfrac{1}{\sqrt{2}} \\
 & =\dfrac{\pi }{2}-\dfrac{{{\pi }^{2}}}{16}-\sqrt{2} \\
\end{align}$
Also, we have
$\begin{align}
  & y\left( x \right)-y\left( -x \right)=2{{x}^{2}}+2\cos x \\
 & \Rightarrow y\left( \dfrac{\pi }{4} \right)-y\left( \dfrac{-\pi }{4} \right)=\dfrac{{{\pi }^{2}}}{8}+\sqrt{2} \\
\end{align}$

So, the correct answer is “Option d”.

Note: A common mistake done by the students in this type of question is that they apply integration by parts in both the integrals which leads to complicated results and does not simplify. It is always a good habit to apply integration by parts on one integrand and check whether the resulting integral might cancel the initial integral