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NCERT Solutions Class 12 Maths Chapter-9 (Determinants)Exercise 9.4

NCERT Solutions Class 12 Maths Chapter-9 (Determinants)Exercise 9.4

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Exercise 9.4 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 9.4

Exercise 9.4

Q1. 

Answer.  The given differential equation is: dydx=1cosx1+cosxdydx=2sin2x22cos2x2=tan2x2dydx=(sec2x21) Separating the variables, we get: dy=(sec2x21)dx 

Q2. 

Answer.  The given differential equation is: dydx=1cosx1+cosxdydx=2sin2x22cos2x2=tan2x2dydx=(sec2x21) Separating the variables, we get: dy=(sec2x21)dx 

Q3. 

Answer.  The given differential equation is: dydx+y=1dy+ydx=dxdy=(1y)dx Separating the variables, we get: dy1y=dx Now, integrating both sides, we get:  dy1y=dxlog(1y)=x+logClogClog(1y)=xlogC(1y)=xC(1y)=ex 

Q4. 

Answer.  The given differential equation is: sec2xtanydx+sec2ytanxdy=0sec2xtanydx+sec2ytanxdytanxtany=0sec2xtanxdx=sec2ytanydysec2xtanxdx=sec2ytanydy  Integrating both sides of this equation, we get: sec2xtanxdx=sec2ytanydy      ...(i) Let tanx=tddx(tanx)=dtdxsec2x=dtdxsec2xdx=dt  Now, sec2xtanxdx=1tdt=logt=log(tanx)  Similarly, tanxsec2xdy=log(tany) Substituting these values in equation (1), we get: log(tanx)=log(tany)+logC 

Q5. 

Answer.  The given differential equation is: (ex+ex)dy(exex)dx=0(ex+ex)dy=(exex)dxdy=[exexex+ex]dx Integrating both sides of this equation, we get:  dy=[exexex+ex]dx+Cy=[exexex+ex]dx+C   ...(i) Let (ex+ex)=t Differentiating both sides with respect to x, we get:  ddx(ex+ex)=dtdxexex=dtdt(exex)dx=dt Substituting this value in equation (i), we get:  

Q6. 

Answer.  The given differential equation is: dydx=(1+x2)(1+y2)dy1+y2=(1+x2)dx Integrating both sides of this equation, we get:  

Q7. 

Answer.  The given differential equation is: ylogydxxdy=0ylogydx=xdydyylogy=dxx Integrating both sides, we get: dyylogy=dxx   ...(i)  Let logy=tddy(logy)=dtdy1y=dtdy1ydy=dt Substituting this value in equation (i), we get:  

Q8. 

Answer.  The given differential equation is: x5dydx=y5dyy5=dxx5dxx5+dyy5=0  Integrating both sides, we get: dxx5+dyy5=k (where k is any constant) x5dx+y5dy=kx44+y44=kx4+y4=4k 

Q9. 

Answer.  The given differential equation is: dydx=sin1xdy=sin1xdx Integrating both sides, we get: dy=sin1x1)dxy=(sin1x1)dx