Question

Let $y=y\left(x\right)$ be the solution of the differential equation, $xy\text{'}-y=x^2(x\cos x+\sin x),x>0.Ify\left(\pi \right)=\pi ,$ then $y\text{'}\text{'}\left(\frac{\mathrm{\pi }}{2}\right)+y\left(\frac{\mathrm{\pi }}{2}\right)$ is equal to:

• A. $2+\left(\frac{\mathrm{\pi }}{2}\right)+\left(\frac{{\mathrm{\pi }}^2}{4}\right)$
• B. $2+\left(\frac{\mathrm{\pi }}{2}\right)$
• C. $1+\left(\frac{\mathrm{\pi }}{2}\right)$
• D. $1+\left(\frac{\mathrm{\pi }}{2}\right)+\left(\frac{{\mathrm{\pi }}^2}{4}\right)$

Answer 1

The correct option is B

$2+\left(\frac{\mathrm{\pi }}{2}\right)$

Explanation for the correct answer:

Step-1: Solution of linear differential equation:

Given, $xy\text{'}-y=x^2(x\cos x+\sin x)$

Expressing the differential equation in the general form:

$$\begin{gathered} xy\text{'}-y=x^2(x\cos x+\sin x) \\ \Rightarrow x\frac{dy}{dx}-y=x^2(x\cos x+\sin x) \\ \Rightarrow \frac{dy}{dx}-\frac{y}{x}=x(x\cos x+\sin x) \\ \Rightarrow \frac{dy}{dx}+Py=Q \end{gathered}$$

So,

$\begin{array}{rcl}I.F.& =& e^{\int -\frac{1}{x}dx}\\ & =& \frac{1}{\left|x\right|}\\ & =& \frac{1}{x}(x>0)\end{array}$

Calculating $y\left(\frac{\mathrm{\pi }}{2}\right)$:

$$\begin{gathered} \therefore \frac{y}{x}=\int \frac{1}{x}(x(x\cos x+\sin x\left)\right)dx \\ \Rightarrow \frac{y}{x}=x\sin x+c \\ \because y\left(\mathrm{\pi }\right)=\mathrm{\pi } \\ \Rightarrow \mathrm{c}=1 \\ \therefore y=x^2\sin x+x \end{gathered}$$

Step-2: Calculating the value of $y\text{'}\text{'}\left(\frac{\mathrm{\pi }}{2}\right)+y\left(\frac{\mathrm{\pi }}{2}\right)$:

$y=x^2\sin x+x$

$$\begin{gathered} \Rightarrow y\left(\frac{\mathrm{\pi }}{2}\right)=\frac{{\mathrm{\pi }}^2}{4}\sin \left(\frac{\mathrm{\pi }}{2}\right)+\frac{\mathrm{\pi }}{2} \\ =\frac{{\mathrm{\pi }}^2}{4}+\frac{\mathrm{\pi }}{2} \end{gathered}$$

And $y=x^2\sin x+x$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=x^2\cos x+2x\sin x+1 \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=-x^2\sin x+4x\cos x+2\sin x \\ \Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=\frac{\mathrm{\pi }}{2}}=-\frac{{\mathrm{\pi }}^2}{4}+2\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{c}\mathbf{o}\mathbf{s}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{]} \\ \therefore y\left(\frac{\mathrm{\pi }}{2}\right)+{\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=\frac{\mathrm{\pi }}{2}}=\frac{\mathrm{\pi }}{2}+2 \end{gathered}$$

Hence, the correct answer is option (B).