Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equation Miscellaneous Exercise

ffImage
banner

NCERT Solutions for Class 12 Maths Chapter 9 Miscellaneous Exercise - Free PDF Download

Class 12 Maths NCERT Solutions for Chapter 9 Differential Equation includes solutions to all Miscellaneous Exercise problems. Differential Equation Class 12 NCERT Solutions Miscellaneous Exercises are based on the ideas presented in Maths Chapter 9. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the latest CBSE Class 12 Maths Syllabus in PDF format and practice it offline.

Popular Vedantu Learning Centres Near You
centre-image
Sharjah, Sharjah
location-imgKing Abdul Aziz St - Al Mahatta - Al Qasimia - Sharjah - United Arab Emirates
Visit Centre
centre-image
Abu Dhabi, Abu-Dhabi
location-imgMohammed Al Otaiba Tower - 1401, 14th Floor - opposite to Nissan Showroom West Zone building - Al Danah - Zone 1 - Abu Dhabi - United Arab Emirates
Visit Centre
centre-image
22 No Phatak, Patiala
location-img#2, Guhman Road, Near Punjabi Bagh, 22 No Phatak-Patiala
Visit Centre
centre-image
Chhoti Baradari, Patiala
location-imgVedantu Learning Centre, SCO-144 -145, 1st & 2nd Floor, Chotti Baradari Scheme Improvement Trust, Patiala-147001
Visit Centre
centre-image
Janakpuri, Delhi
location-imgVedantu Learning Centre, A-1/173A, Najafgarh Road, Opposite Metro Pillar 613, Block A1, Janakpuri, New Delhi 110058
Visit Centre
centre-image
Tagore School, Gudha-Gorji
location-imgTagore Public School Todi, Gudha Gorji, Jhunjhunu, Rajasthan 333022
Visit Centre
View More
Courses
Competitive Exams after 12th Science

Access NCERT Class 12 Maths Chapter 9 Differential Equation

Miscellaneous Exercise

1. For each of the differential equations given below, indicate its order and degree (if defined).

(i)d2ydx2+5x(dydx)2-6y=logx 

Ans: The given differential equation is:

 d2ydx2+5x(dydx)26ylogx=0 

The highest order derivative in the equation is of the term d2ydx2, thus the order of the equation is 2 and its highest power is 1. Therefore its degree is 1.


(ii)(dydx)34(dydx)2+7y=sinx 

Ans: The given differential equation is:

 (dydx)34(dydx)2+7ysinx=0 

The highest order derivative in the equation is of the term (dydx)3, thus the order of the equation is 1 and its highest power is 3. Therefore its degree is 3.


(iii)d4ydx4-sin(d3ydx3)=0 

Ans: The given differential equation is:

 d4ydx4sin(d3ydx3)=0 

The highest order derivative in the equation is of the term d4ydx4, thus the order of the equation is 4.

As the differential equation is not polynomial in its derivative, therefore its degree is not defined.


2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

  1. xy=aex+be-x+x2:xd2ydx2+2dydx-xy+x2-2=0 

Ans: The given function is:
xy=aex+bex+x2 

Take derivative on both side:

y+xdydx=aexbex+2x 

Take derivative on both side:

dydx+xd2ydx2+dydx=aex+bex+2 

xd2ydx2+2dydx=aex+bex+2 ……(1)

The given differential equation is:
xd2ydx2+2dydxxy+x22=0 

Solving LHS:

Substitute xd2ydx2+2dydx from the result (1) and xy:

(xd2ydx2+2dydx)xy+x22 

(aex+bex+2)(aex+bex+x2)+x22 

2x2+x22 

0 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

  1. y=ex(acosx+bsinx):d2ydx2-2dydx+2y=0 

Ans: The given function is:
y=ex(acosx+bsinx) 

Take derivative on both side:

dydx=ex(acosx+bsinx)+ex(asinx+bcosx) 

dydx=ex((a+b)cosx+(ba)sinx) 

Take derivative on both side:

d2ydx2=ex((a+b)cosx+(ba)sinx)+ex((a+b)sinx+(ba)cosx)

d2ydx2=ex((a+b+ba)cosx+(baab)sinx) 

d2ydx2=ex(2bcosx2asinx)  

The given differential equation is:
d2ydx22dydx+2y=0 

Solving LHS:

ex(2bcosx2asinx)2ex((a+b)cosx+(ba)sinx)+2y 

ex((2b2a2b)cosx+(2a2b+2a)sinx)2y 

ex(2acosx2bsinx)2y 

2ex(acosx+bsinx)2y

0 

Thus LHS=RHS, the given function is the solution of the given differential equation.


  1. y=xsin3x:d2ydx2+9y-6cos3x=0 

Ans: The given function is:
y=xsin3x 

Take derivative on both side:

dydx=sin3x+3xcos3x 

Take derivative on both side:

d2ydx2=3cos3x+3(cos3x+x(3sin3x))

d2ydx2=3cos3x+3cos3x9xsin3x 

d2ydx2=6cos3x9xsin3x  

The given differential equation is:
d2ydx2+9y6cos3x=0 

Solving LHS:

d2ydx2+9y6cos3x 

(6cos3x9xsin3x)+9(xsin3x)6cos3x 

6cos3x9xsin3x+9xsin3x6cos3x 

0 

Thus LHS=RHS, the given function is the solution of the given differential equation. 

  1. x2=2y2log y:(x2+y2)dydx-xy=0 

Ans: The given function is:
x2=2y2logy 

Take derivative on both side:

2x=2(2ylogy+y2(1y))dydx 

dydx=x(2ylogy+y)

Multiply numerator and denominator by y:

dydx=xy(2y2logy+y2) 

dydx=xy(x2+y2)  

The given differential equation is:
(x2+y2)dydxxy=0 

Solving LHS:

(x2+y2)dydxxy 

(x2+y2)(xyx2+y2)xy 

xyxy 

0 

Thus LHS=RHS, the given function is the solution of the given differential equation.


3. Prove that x2-y2=c(x2+y2)2  is the general solution of differential equation (x3-3xy2)dx=(y3-3x2y)dy , where c is a parameter.

 Ans: Given differential equation:

(x33xy2)dx=(y33x2y)dy 

dydx=x33xy2y33x2y 

As it can be seen that this is an homogenous equation. Substitute y=vx:

d(vx)dx=x33x(vx)2(vx)33x2(vx) 

v+xdvdx=x3(13v2)x3(v33v) 

v+xdvdx=13v2v33v 

xdvdx=13v2v33vv 

xdvdx=13v2v4+3v2v33v 

xdvdx=1v4v33v 

Separate the differentials:

v33v1v4dv=dxx 

Integrate both side:

v33v1v4dv=dxx 

v33v1v4dv=logx+logC 

I=logx+logC(I=v33v1v4dv) ……(1)

Solving integral I:

I=v33v1v4dv 

v33v1v4=v33v(1v2)(1+v2) 

v33v1v4=v33v(1v)(1+v)(1+v2) 

Using partial fraction:

v33v1v4=A1v+B1+v+Cv+D1+v2 

Solving for A,B,CandD:

A=12 

B=12 

C=2 

D=0 

v33v1v4=121v+121+v+2v+01+v2 

I=1211vdv+1211+vdv2v1+v2dv

I=12(log(1v))+12(log(1+v))log(1+v2) 

I=12(log(1v2))22log(1+v2) 

I=12(log(1v2)(1+v2)2) 

I=12(log(1y2x2)(1+y2x2)2) 

I=12(logx2(x2y2)(x2+y2)2)

I=12(log(x2y2)(x2+y2)2)+12logx2 

I=12(log(x2y2)(x2+y2)2)+logx

Back substitute I in expression (1):

I=logx+logC 

12(log(x2y2)(x2+y2)2)+logx=logx+logC

log(x2y2)(x2+y2)2=2logC 

x2y2(x2+y2)2=C2 

x2y2=c(x2+y2)2(c=C2) 

Thus for given differential equation, its general solution is x2-y2=c(x2+y2)2.


4. Find the general solution of the differential equation dydx+1-y21-x2=0.

Ans: The given differential equation is:

 dydx+1y21x2=0 

dydx=1y21x2 

dy1y2=dx1x2 

Integrate both side:

dy1y2=dx1x2 

sin1y=sin1x+C

sin1y+sin1x=C 

Thus the general solution of given differential equation is sin-1y+sin-1x=C.


5. Show that the general solution of the differential equation dydx+y2+y+1x2+x+1=0 is given by (x+y+1)=A(1-x-y-2xy)  where A is a parameter.

Ans: The given differential equation is:

dydx+y2+y+1x2+x+1=0

dydx=y2+y+1x2+x+1

dydx=y2+2(12)y+1414+1x2+2(12)x+1414+1 

dydx=(y+12)2+34(x+12)2+34   

dy(y+12)2+34=dx(x+12)2+34 

Integrate both side:

dy(y+12)2+(32)2=dx(x+12)2+(32)2

1(32)tan1[y+1232]=1(32)tan1[x+1232]+C 

23(tan1[2y+13]+tan1[2x+13])=C 

tan1[2y+13]+tan1[2x+13]=32C 

Thus the general solution for given differential equation is tan-1[2y+13]+tan-1[2x+13]=32C.


6.Find the equation of the curve passing through the point (0,π4) whose differential equation is sinxcosydx+cosxsinydy=0.

Ans: Given differential equation is:

sinxcosydx+cosxsinydy=0

sinxcosydx+cosxsinydy=0

Divide both side by cosxcosy:

sinxcosydx+cosxsinydycosxcosy=0 

tanxdx+tanydy=0 

tanydy=tanxdx 

Integrate both side:

tanydy=tanxdx 

log(secy)=log(secx)+C 

log(secy)+log(secx)=C 

log(secxsecy)=C 

secxsecy=k(k=eC) 

As curve passes through (0, π 4):

sec0sec(π4)=k 

k=2 

secxsecy=2 

Thus the equation of required curve is secxsecy=2.


7. Find the particular solution of the differential equation (1+e2x)dy+(1+y2)exdx=0 given that y=1 when x=0.

Ans: The given differential equation is:

(1+e2x)dy+(1+y2)exdx=0 

Divide both side (1+e2x)(1+y2):

dy(1+y2)+ex(1+e2x)dx=0 

dy(1+y2)=ex(1+e2x)dx 

tan1y=ex(1+(ex)2)dx

Substitute t=ex:

dt=exdx 

tan1y=1(1+t2)dt 

tan1y=tan1t+C 

tan1y=tan1ex+C 

tan1y+tan1ex=C 

As y=1 when x=0:

tan1(1)+tan1(e0)=C 

π4+π4=C 

C=π2 

tan1y+tan1ex=π2 

Thus the required particular solution is tan-1y+tan-1ex= π 2.


8. Solve the differential equation yexydx=(xexy+y2)dy(y0).

Ans: The given differential equation is:

yexydx=(xexy+y2)dy 

yexydxdy=xexy+y2 

yexydxdyxexy=y2 

exy[ydxdyx]y2=1 

Substitute z=exy:

z=exy 

ddyz=ddyexy 

dzdy=ddy(exy) 

dzdy=exyddy(xy) 

dzdy=exy[(1y)dxdyxy2] 

dzdy=exy[ydxdyxy2] 

dzdy=1 

dz=dy 

dz=dy

z=y+C 

exy=y+C 

Thus the required general solution is exy=y+C.


9. Find a particular solution of the differential equation (x-y)(dx+dy)=dx-dy given that y=-1 when x=0. Hint (put x-y=t).

Ans: Given differential equation is:

(xy)(dx+dy)=dxdy 

(xy)dxdx=(yx)dydy

(xy+1)dy=(1x+y)dx 

dydx=1x+yxy+1 

Put x-y=t:

xy=t 

1dydx=dtdx 

1dtdx=dydx 

1dtdx=1t1+t 

dtdx=11t1+t

dtdx=1+t1+t1+t 

dtdx=2t1+t 

1+ttdt=2dx 

Integrate both side:

1+ttdt=2dx

1tdt+dt=2x+C 

log|t|+t=2x+C 

log|xy|+xy=2x+C 

log|xy|y=x+C 

As y=-1 when x=0:

log|0(1)|(1)=0+C 

log1+1=C 

C=1 

Thus the required particular solution is:

log|xy|y=x+1.


10. Solve the differential equation [e-2xx-yx]dxdy=1(x0).

Ans: Given differential equation is:
[e2xxyx]dxdy=1 

dydx=e2xxyx 

dydx+yx=e2xx 

It is linear differential equation of the form dydx+py=Q:

p=1x 

Q=e2xx 

Calculating integrating factor:

I.F=epdx

I.F=e1xdx 

I.F=e2x  

The general solution is given by:

y×I.F=(Q×I.F)dx+C 

y×(e2x)=(e2xx×e2x)dx+C 

ye2x=(e2x+2xx)dx+C 

ye2x=1xdx+C 

ye2x=2x+C 

Thus the general solution for the given differential equation is

ye2x=2x+C.


11. Find a particular solution of the differential equation dydx+ycotx=4xcosecx(x0) given that y=0 when x= π 2.

Ans:
The given differential equation is:

 dydx+ycotx=4xcosecx 

It is linear differential equation of the form dydx+py=Q:

p=cotx 

Q=4xcosecx 

Calculating integrating factor:

I.F=epdx

I.F=ecotxdx 

I.F=elog|sinx| 

I.F=sinx 

The general solution is given by:

y×I.F=(Q×I.F)dx+C

y×sinx=(4xcosecx)sinxdx+C 

ysinx=4xdx+C 

ysinx=4(x22)+C 

ysinx=2x2+C 

As y=0 when x= π 2:

0×sin(π2)=2(π2)2+C 

C=2(π24) 

C=π22 

Thus the required particular solution is:

ysinx=2x2π22


12. Find a particular solution of the differential equation (x+1)dydx=2e-y-1 given that y=0 when x=0.

Ans:
The given differential equation is:

 (x+1)dydx=2ey1

dy2ey1=dxx+1 

Integrate both side:

dy2ey1=dxx+1 

dy2ey1=log(x+1)+logC ……(1) 

Evaluating LHS integral:

dy2ey1=eydy2ey 

Put t=2-ey:

t=2ey

dt=eydy  

dy2ey1=dtt

dy2ey1=log(t) 

dy2ey1=log1t 

dy2ey1=log12ey 

Back substituting in expression (1):

dy2ey1=log(x+1)+logC 

log(12ey)=logC(x+1) 

2ey=1C(x+1) 

As y=0 when x=0:

2e0=1C(0+1) 

21=1C 

C=1 

Thus the required particular solution is:

2ey=1(x+1)

ey=21(x+1) 

ey=2x+21(x+1) 

ey=2x+1x+1 

y=log(2x+1x+1)  

Thus for given conditions the particular solution is y=log(2x+1x+1) .


13. The general solution of the differential equation ydx-xdyy=0.

A. xy=C 

B. x=Cy2 

C. y=Cx 

D. y=Cx2 

Ans: Given differential equation:

ydxxdyy=0

Divide both side by x :

ydxxdyxy=0

dxxdyy=0 

Integrate both side:

dxxdyy=0 

log|x|log|y|=logk 

log|xy|=logk 

xy=k 

y=Cx(C=1k)

Thus the correct option is ©


14. Find the general solution of a differential equation of the type dxdy+P1x=Q1.

A. yeP1dy=(Q1eP1dy)dy+C   

B. yeP1dx=(Q1eP1dx)dy+C 

C. xeP1dy=(Q1eP1dy)dy+C   

D. xeP1dy=(Q1eP1dx)dy+C   

Ans: The given differential equation is:

dxdy+P1x=Q1 

It is a linear differential equation and its general solution is:

xeP1dy=(Q1eP1dy)dy+C 

With integrating factor I.F=eP1dy.

Thus the correct option is (C).


15. Find the general solution of the differential equation exdy+(yex+2x)dx=0.

A. xey+x2=C 

B. xey+y2=C 

C. yex+x2=C 

D. yey+x2=C 

Ans: The given differential equation is:
exdy+(yex+2x)dx=0 

exdydx+yex=2x 

dydx+y=2xex   

The given differential equation is of the form:

dydx+Py=Q 

P=1 

Q=2xex 

Calculating integrating factor:
I.F=ePdx

I.F=edx 

I.F=ex  

It is a linear differential equation and its general solution is:

y(I.F)=(Q×I.F)dx+C

 y(ex)=(2xex×ex)dy+C 

yex=2xdx+C 

yex=2(x22)+C 

yex+x2=C 

Thus the correct answer is option (C).


Conclusion

NCERT solutions for class 12 maths Differential Equation miscellaneous exercise is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.


Class 12 Maths Chapter 9: Exercises Breakdown


CBSE Class 12 Maths Chapter 9 Other Study Materials


Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


Additional Study Materials for Class 12 Maths 

WhatsApp Banner

FAQs on NCERT Solutions for Class 12 Maths Chapter 9 Differential Equation Miscellaneous Exercise

1. What kind of problems are covered in the miscellaneous exercise?

The miscellaneous exercise in Chapter 9 deals with various aspects of solving differential equations, including:

  • Identifying the order and degree of a differential equation.

  • Finding general and particular solutions of differential equations using different methods (separation of variables,homogeneous equations, integrating factors, etc.).

  • Applying differential equations to solve real-world problems in various fields (e.g., population growth, motion,electrical circuits).

2. How do the NCERT solutions approach these problems?

The NCERT solutions typically:

  • Briefly remind you of the relevant concepts from differential equations, including order, degree, general and particular solutions, and different solution methods.

  • Guide you through the process of classifying the differential equation based on its order and degree.

  • Demonstrate how to apply the appropriate solution method for the given differential equation. This might involve separation of variables for separable equations, using integrating factors for specific types of equations, or other relevant methods.

  • Show you how to find the general solution and then apply initial conditions (if provided) to obtain the particular solution.

3. Where can I find additional resources for practising miscellaneous exercise problems?

  • The NCERT textbook itself might provide solutions to some problems within the miscellaneous exercise section.

  • Vedantu offers comprehensive solutions and explanations for these problems. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 9 Miscellaneous Exercise Differential Equations."

4. Differential Equation sample questions?

While I cannot provide specific solutions due to copyright, here are 2 sample questions to illustrate the types of problems you might encounter:

  1. Solve the differential equation dy/dx = x^2 + y (This question might involve using separation of variables)

  2. Find the general solution of the differential equation d^2y/dx^2 + 4y = 0 (This question requires identifying the order and degree and then solving the homogeneous equation)