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NCERT Solutions Class 12 Maths Chapter-9 (Determinants)Exercise 9.5

NCERT Solutions Class 12 Maths Chapter-9 (Determinants)Exercise 9.5

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Exercise 9.5 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.
NCERT Solutions Class 12 Maths Chapter-9 (Determinants)Exercise 9.5

Exercise 9.5

Q1. 

Answer.  The given differential equation i.e., (x2+xy)dy=(x2+y2)dx can be written as: dydx=x2+y2x2+xy   ...(i) Let F(x,y)=x2+y2x2+xy  Now, F(λx,λy)=(λx)2+(λy)2(λx)2+(λx)(λy)=x2+y2x2+xy=λ0F(x,y) This shows that equation (i) is a homogeneous equation.  To solve it, we make the substitution as: y=vx Differentiating both sides with respect to x, we get:  dydx=v+xdvdx Substituting the values of v and dydx in equation (i), we get:  v+xdvdx=x2+(vx)2x2+x(vx)v+xdvdx=1+v21+vxdvdx=1+v21+vv=(1+v2)v(1+v)1+vxdvdx=1v1+v (1+v1v)=dv=dxx(21+v1v)dv=dxx(21v1)dv=dxx Integrating both sides, we get: v=2log(1v)logx+logkv=log[kx(1v)2] 

Q2. 

Answer. y=x+yxdydx=x+yx   ...(i)LetF(x,y)=x+yx Now, F(λx,λy)=λx+λyλx=x+yx=λ0F(x,y)  Thus, the given equation is a homogeneous equation.  To solve it, we make the substitution as: y=vx Differentiating both sides with respect to x, we get: dydx=v+xdvdx Substituting the values of y and dydx in equation (i), we get:  v+xdvdx=x+vxxv+xdvdx=1+vxdvdx=1dv=dxx Integrating both sides, we get:    v=logx+Cyx=logx+Cy=xlogx+Cx This is the required solution of the given differential equation.

Q3. 

Answer.  The given differential equation is: (xy)dy(x+y)dx=0dydx=x+yxy   ...(i) Let F(x,y)=x+yxy F(λx,λy)=λx+λyλxλy=x+yxy=λ0F(x,y) Thus, the given differential equation is a homogeneous equation.  To solve it, we make the substitution as: y=vx ddx(y)=ddx(vx)dydx=v+xdvdx Substituting the values of y and dydx in equation (i), we get: v+xdvdx=x+vxxvx=1+v1v xdvdx=1+v1vv=1+vv(1v)1vxdvdx=1+v21v1v(1+v2)dv=dxx(11+v2v1v2)dv=dxx  Integrating both sides, we get: tan1v12log(1+v2)=logx+Ctan1(yx)12log[1+(yx)2]=logx+Ctan1(yx)12log(x2+y2x2)=logx+C 

Q4. 

Answer.  The given differential equation is: (x2y2)dx+2xydy=0dydx=(x2y2)2xy   ...(i) Let F(x,y)=(x2y2)2xy F(λx,λy)=[(λx)2(λy)22(λx)(λy)]=(x2y2)2xy=λ0F(x,y) To solve it, we given differential equation is a homogeneous equation. y=vx ddx(y)=ddx(vx)dydx=v+xdvdx Substituting the values of y and dydx equation (i), we get: v+xdvdx=x2(vx)22x(vx)]v+xdvdx=v212v xdvdx=v212vv=v212v22vxdvdx=(1+v2)2v2v1+v2dv=dxx Integrating both sides, we get: log(1+v2)=logx+logC=logCx 

Q5. 

Answer.  The given differential equation is: x2dydx=x22y2+xy dydx=x22y2+xyx2   ...(i)LetF(x,y)=x22y2+xyx2F(λx,λy)=(λx)22(λy)2+(λx)(λy)(λx)2=x22y2+xyx2=λ0F(x,y)