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

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

Exercise 9.6

Q1. 

Answer. The given differential equation is dydx+2y=sinx This is in the form of dydx+py=Q( where p=2 and Q=sinx) Now, I.F =epdt=e2dx=e2x The solution of the given differential equation is given by the relation, y(IF)=(Q×IF)dx+Cye2x=sinxe2xdx+C      ...(i)  Let I=sinxe2xI=sinxe2xdx(ddx(sinx)e2xdx)dxI=sinxe2x2 (cosxe2x2)dx I=e2xsinx212[cosxe2xddx(cosx)e2xdx)dx]I=e2xsinx212[cosxe2x2(sinx)e2x2]dx]I=e2xsinx2e2xcosx414(sinxe2x)dxI=e2x4(2sinxcosx)14I54I=e2x4(2sinxcosx)I=e2x5(2sinxcosx) 

Q2. 

Answer.   The given differential equation is  dydx+py=Q( where p=3 and Q=e2x) Now, I.F =epΔx=e3at=e3x The solution of the given differential equation is given by the relation, y( I.F. )=(Q× I.F. )dx+Cye3x=(e2x×e3x)+Cye3x=exdx+Cye3x=ex+Cy=e2x+Ce3x This is the required general solution of the given differential equation

Q3. 

Answer.  The given differential equation is: dydx+py=Q( where p=1x and Q=x2) Now, I.F =epdx=e1xdx=elogx=x 

Q4. 

Answer.  The given differential equation is: dydx+py=Q (where p=secx and Q=tanx) Now I.F =epΔx=esecxdx=elog(secx+tanx)=secx+tanx  The general solution of the given differential equation is given by the relation, y( I.F. )=(Q×1.F.)dx+Cy(secx+tanx)=tanx(secx+tanx)dx+Cy(secx+tanx)=secxtanxdx+tan2xdx+C 

Q5. 

Answer.  Let I=jπ2cos2xdxcos2xdx=(sin2x2)=F(x) By second fundamental theorem of calculus, we obtain  

Q6. 

Answer.  The given differential equation is: xdydx+2y=x2logxdydx+2xy=xlogx This equation is in the form of a linear differential equation as: dydx+py=Q( where p=2x and Q=xlogx) Now I.F = epdx=ex2dx=e2logx=elogx2=x2 The general solution of the given differential equation is given by the relation, y( I.F. )=(Q×I.F.)dx+Cyx2=(xlogxx2)dx+Cx2y=(x3logx)dx+C x2y=logxx3dx[ddx(logx)x3dx]dx+Cx2y=logxx441xx44)dx+Cx2y=x4logx414x3dx+C 

Q7. 

Answer.  The given differential equation is: xlogxdydx+y=2xlogxdydx+yxlogx=2x2 This equation is the form of a linear differential equation as: dydx+py=Q( where p=1xlogx and Q=2x2) Now, I.F =eρdx=e1xlogdx=elog(logx)=logx The general solution of the given differential equation is given by the relation, y(I.F.)=(Q×I.F)dx+Cylogx=(2x2logx)dx+C     ...(i) Now. (2x2logx)dx=2(logx1x2)dx =2[logx1x2dx{ddx(logx)1x2dx}dx]=2[logx(1x)(1x(1x))dx]=2[2[logxx+1x2dx]=2[logxx1x]=2[logxx1x]=2x(1+logx) 

Q8.