# NCERT Solutions Class 12 Maths Chapter-9 (Determinants)Exercise 9.6

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Exercise 9.6 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### Exercise 9.6

Q1.

Answer. The given differential equation is $\frac{dy}{dx}+2y=\mathrm{sin}x$ This is in the form of  Now, I.F $={e}^{\int p\cdot dt}={e}^{\int 2dx}={e}^{2x}$ The solution of the given differential equation is given by the relation,   $\begin{array}{l}⇒I=\frac{{e}^{2x}\mathrm{sin}x}{2}-\frac{1}{2}\left[\mathrm{cos}x\cdot \int {e}^{2x}-\int \frac{d}{dx}\left(\mathrm{cos}x\right)\cdot \int {e}^{2x}dx\right)dx\right]\\ ⇒I=\frac{{e}^{2x}\mathrm{sin}x}{2}-\frac{1}{2}\left[\mathrm{cos}x\cdot \frac{{e}^{2x}}{2}-\int \left(-\mathrm{sin}x\right)\cdot \frac{{e}^{2x}}{2}\right]dx\right]\\ ⇒I=\frac{{e}^{2x}\mathrm{sin}x}{2}-\frac{{e}^{2x}\mathrm{cos}x}{4}-\frac{1}{4}\int \left(\mathrm{sin}x{e}^{2x}\right)dx\\ ⇒I=\frac{{e}^{2x}}{4}\left(2\mathrm{sin}x-\mathrm{cos}x\right)-\frac{1}{4}I\\ ⇒\frac{5}{4}I=\frac{{e}^{2x}}{4}\left(2\mathrm{sin}x-\mathrm{cos}x\right)\\ ⇒I=\frac{{e}^{2x}}{5}\left(2\mathrm{sin}x-\mathrm{cos}x\right)\end{array}$

Q2.

Answer.  Now, I.F $={e}^{\int p\mathrm{\Delta }x}={e}^{\int 3at}={e}^{3x}$ This is the required general solution of the given differential equation

Q3.

Answer.  Now, I.F $={e}^{\int pdx}={e}^{\int \frac{1}{x}dx}={e}^{\mathrm{log}x}=x$

Q4.

Answer.  Now I.F $={e}^{\int p\mathrm{\Delta }x}={e}^{\int \mathrm{sec}xdx}={e}^{\mathrm{log}\left(\mathrm{sec}x+\mathrm{tan}x\right)}=\mathrm{sec}x+\mathrm{tan}x$

Q5.

Answer.  Now I.F = ${e}^{\int pdx}={e}^{{\int }_{x}^{2}dx}={e}^{2\mathrm{log}x}={e}^{\mathrm{log}{x}^{2}}={x}^{2}$ The general solution of the given differential equation is given by the relation,  $\begin{array}{l}⇒{x}^{2}y=\mathrm{log}x\cdot \int {x}^{3}dx-\int \left[\frac{d}{dx}\left(\mathrm{log}x\right)\cdot \int {x}^{3}dx\right]dx+C\\ ⇒{x}^{2}y=\mathrm{log}x\cdot \frac{{x}^{4}}{4}-\int \frac{1}{x}\cdot \frac{{x}^{4}}{4}\right)dx+C\\ ⇒{x}^{2}y=\frac{{x}^{4}\mathrm{log}x}{4}-\frac{1}{4}\int {x}^{3}dx+C\end{array}$
Answer.  Now, I.F $={e}^{\int \rho dx}={e}^{\int \frac{1}{x\mathrm{log}}dx}={e}^{\mathrm{log}\left(\mathrm{log}x\right)}=\mathrm{log}x$ The general solution of the given differential equation is given by the relation,  $\begin{array}{rl}& =2\left[\mathrm{log}x\cdot \int \frac{1}{{x}^{2}}dx-\int \left\{\frac{d}{dx}\left(\mathrm{log}x\right)\cdot \int \frac{1}{{x}^{2}}dx\right\}dx\right]\\ & =2\left[\mathrm{log}x\left(-\frac{1}{x}\right)-\int \left(\frac{1}{x}\cdot \left(-\frac{1}{x}\right)\right)dx\right]\\ & =2\left[-2\left[-\frac{\mathrm{log}x}{x}+\int \frac{1}{{x}^{2}}dx\right]\\ & =2\left[-\frac{\mathrm{log}x}{x}-\frac{1}{x}\right]\\ & =2\left[-\frac{\mathrm{log}x}{x}-\frac{1}{x}\right]\\ & =-\frac{2}{x}\left(1+\mathrm{log}x\right)\end{array}$