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Vidaara.orgClass 12 · Mathematics
CodeVID-M12-09-CT
Differential Equations — Full Chapter Test
Chapter: Differential Equations
Topic: Complete chapter — all topics
Maximum Marks: 35
Time: 80 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • This is a full-length, CBSE board-hard test covering the whole chapter. All questions are compulsory.
  • Section A carries 1 mark each (including an assertion-reason item), Section B 2 marks, Section C 3 marks and Section D 5 marks (long answer / case based).
  • Show all working for Sections B, C and D. Final answers are given in the Answer Key.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
The order and degree of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^3+\left(\dfrac{dy}{dx}\right)^2+\sin\left(\dfrac{dy}{dx}\right)+1=0$ are respectively:
  • A.$2$ and $3$
  • B.$2$ and $2$
  • C.$2$ and not defined
  • D.$1$ and not defined
2.
The integrating factor of the differential equation $x\dfrac{dy}{dx}+2y=x^2$ ($x\ne0$) is:
  • A.$\dfrac{1}{x^2}$
  • B.$x^2$
  • C.$e^{2x}$
  • D.$2x$
3.
Assertion (A): The order of the differential equation of the family of all circles of fixed radius $r$ with centres on the $x$-axis is $2$.
Reason (R): The order of the differential equation of a family of curves equals the number of independent arbitrary constants in the family.
  • A.Both A and R are true and R is the correct explanation of A
  • B.Both A and R are true but R is not the correct explanation of A
  • C.A is true but R is false
  • D.A is false but R is true
4.
The general solution of $\dfrac{dy}{dx}=e^{x+y}$ is:
  • A.$e^x+e^{-y}=C$
  • B.$e^x-e^{-y}=C$
  • C.$e^x+e^{y}=C$
  • D.$e^{-x}+e^{-y}=C$
5.
The general solution of $\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2xy}$ is:
  • A.$x^2+y^2=Cx$
  • B.$x^2-y^2=C$
  • C.$x^2-y^2=Cx$
  • D.$y^2-x^2=Cy$
Section B — Short Answer I 4 × 2 = 8 marks
6.
Form the differential equation representing the family of parabolas having vertex at the origin and axis along the positive $x$-axis.
7.
Find the particular solution of $(1+x^2)\dfrac{dy}{dx}=1+y^2$ given that $y=1$ when $x=0$.
8.
Find the general solution of $\dfrac{dy}{dx}+y=\cos x$.
9.
Verify that $y=Ax+\dfrac Bx$ is a solution of $x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}-y=0$.
Section C — Short Answer II 4 × 3 = 12 marks
10.
Solve: $x\dfrac{dy}{dx}-y+x\sin\dfrac yx=0$.
11.
Solve: $x\dfrac{dy}{dx}+2y=x^2\log x$, $x>0$.
12.
Find the particular solution of $(1+e^{2x})\,dy+(1+y^2)e^x\,dx=0$ given $y=1$ when $x=0$.
13.
Solve: $\dfrac{dy}{dx}=\dfrac{x+y+1}{x+y-1}$.
Section D — Long Answer / Case Based 2 × 5 = 10 marks
14.
Find the particular solution of $(x^2-1)\dfrac{dy}{dx}+2xy=\dfrac{2}{x^2-1}$ given that $y=0$ when $x=0$.
15.
Case study. The number $N$ of bacteria in a culture grows at a rate proportional to the number present. The culture starts with $1000$ bacteria and doubles in $3$ hours.
(a) Write the differential equation for $N(t)$. [1]
(b) Solve it using the given conditions to find $N(t)$. [2]
(c) Find the number of bacteria after $9$ hours. [1]
(d) After how many hours will the count reach $5000$? [1]

Answer Key

Section A — Multiple Choice Questions
  1. (C) $2$ and not defined
  2. (B) $x^2$
  3. (D) A is false but R is true
  4. (A) $e^x+e^{-y}=C$
  5. (C) $x^2-y^2=Cx$
Section B — Short Answer I
  1. The family is $y^2=4ax$. Differentiating, $2yy'=4a=\dfrac{y^2}{x}$. Hence the differential equation is $y^2-2xy\dfrac{dy}{dx}=0$.
  2. Separating, $\dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}\Rightarrow\tan^{-1}y=\tan^{-1}x+C$. At $(0,1)$: $C=\dfrac{\pi}{4}$. So $\tan^{-1}y=\tan^{-1}x+\dfrac{\pi}{4}$, i.e. $y=\dfrac{1+x}{1-x}$.
  3. I.F. $=e^{\int1\,dx}=e^x$. Then $ye^x=\displaystyle\int e^x\cos x\,dx=\dfrac{e^x(\sin x+\cos x)}{2}+C$. So $y=\dfrac{\sin x+\cos x}{2}+Ce^{-x}$.
  4. $y'=A-\dfrac{B}{x^2}$, $y''=\dfrac{2B}{x^3}$. Then $x^2y''+xy'-y=\dfrac{2B}{x}+Ax-\dfrac Bx-Ax-\dfrac Bx=0$. Hence verified.
Section C — Short Answer II
  1. Put $y=vx$: $v+x\dfrac{dv}{dx}-v+\sin v=0\Rightarrow\dfrac{dv}{\sin v}=-\dfrac{dx}{x}$. Integrating, $\log\left|\tan\dfrac v2\right|=-\log|x|+\log C$, i.e. $x\tan\dfrac{y}{2x}=C$.
  2. Write $\dfrac{dy}{dx}+\dfrac2xy=x\log x$. I.F. $=e^{2\log x}=x^2$. So $x^2y=\displaystyle\int x^3\log x\,dx=\dfrac{x^4}{4}\log x-\dfrac{x^4}{16}+C$, giving $y=\dfrac{x^2}{4}\log x-\dfrac{x^2}{16}+\dfrac{C}{x^2}$.
  3. $\dfrac{dy}{1+y^2}=-\dfrac{e^x\,dx}{1+e^{2x}}\Rightarrow\tan^{-1}y=-\tan^{-1}(e^x)+C$. At $(0,1)$: $\dfrac{\pi}{4}=-\dfrac{\pi}{4}+C\Rightarrow C=\dfrac{\pi}{2}$. So $\tan^{-1}y+\tan^{-1}e^x=\dfrac{\pi}{2}$, which means $ye^x=1$, i.e. $y=e^{-x}$.
  4. Put $v=x+y$: $\dfrac{dv}{dx}-1=\dfrac{v+1}{v-1}\Rightarrow\dfrac{dv}{dx}=\dfrac{2v}{v-1}\Rightarrow\dfrac{v-1}{v}\,dv=2\,dx$. Integrating, $v-\log|v|=2x+C$, i.e. $y-x-\log|x+y|=C$.
Section D — Long Answer / Case Based
  1. Write $\dfrac{dy}{dx}+\dfrac{2x}{x^2-1}y=\dfrac{2}{(x^2-1)^2}$. I.F. $=e^{\int\frac{2x}{x^2-1}dx}=x^2-1$. Then $y(x^2-1)=\displaystyle\int\dfrac{2}{x^2-1}dx=\log\left|\dfrac{x-1}{x+1}\right|+C$. At $(0,0)$: $0=\log1+C\Rightarrow C=0$. So the particular solution is $y(x^2-1)=\log\left|\dfrac{x-1}{x+1}\right|$.
  2. (a) $\dfrac{dN}{dt}=kN$. (b) $N=N_0e^{kt}$ with $N_0=1000$. At $t=3$, $2000=1000e^{3k}\Rightarrow k=\dfrac{\log2}{3}$. So $N(t)=1000\cdot2^{t/3}$. (c) $N(9)=1000\cdot2^3=8000$. (d) $5000=1000\cdot2^{t/3}\Rightarrow2^{t/3}=5\Rightarrow t=\dfrac{3\log5}{\log2}\approx6.97$ hours.
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