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MATHS M.C.Q (1 Marks)

DIFFERENTIAL EQUATIONS · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 176 questions.

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M.C.Q (1 Marks)

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Question 11 Mark
The number of arbitrary constants in the particular solution of a differential equation of second order is (are):
  1. 0
  2. 1
  3. 2
  4. 3
Answer
  1. 0
Solution:
In the particular solution of a differential equation of third order, there is no arbitrary constant because in the particular solution of any differential equation, we remove all the arbitrary constant by substituting some particular values.
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MCQ 21 Mark
What is the general solution of the differential equation $x \ dy - y \ dx \ y^2$ ?
  • A
    $x = cy$
  • B
    $y^2 = cx$
  • C
    $x + xy - cy = 0$
  • ✓
    None of the above
Answer
Correct option: D.
None of the above
$xdy - y^3dx = 0$ rearranging
$\frac{\text{dx}}{\text{x}}=\frac{\text{dy}}{\text{y}^3}$
integrating both sides ln
$\text{x}=\frac{\text{y}^{-2}}{-2}+\text{c}$
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Question 31 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^{2}+\text{xy}^{2}, \text{y}=(0)$ is: 
  1. $\text{y}^{2}=\text{exp}\big(\text{x}+\frac{\text{x}^{2}}{2}-1\big)$
  2. $\text{y}^{2}=1+\text{C}\ \text{exp}\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$
  3. $\text{y}=\tan (\text{C}+\text{x}+\text{x}^{2})$
  4. $\text{y}=\tan\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$
Answer
  1. $\text{y}=\tan\big(\text{x}+\frac{\text{x}^{2}}{2}\big)$
Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^{2}+\text{xy}^{2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=(\text{x}+1)\text{y}^{2}(\text{x}+1)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=(\text{x}+1)(1+\text{y})$
$\Rightarrow \frac{\text{dy}}{(1+\text{y}^{2})}=(\text{x}+1)\text{dx}$
Integrating both sides, we get
$\int \frac{\text{dy}}{(1+\text{y}^{2})}=\int(\text{x}+1)\text{dx}$
$\Rightarrow \tan^{-1}=\frac{\text{x}^{2}}{2}+\text{x}+\text{C}\ ...(\text{i})$
Now, y(0) = 0
$\therefore\ \tan^{-1}(0)=\frac{\text{0}}{2}+\text{0}+\text{C}$
$\Rightarrow \text{C}=0$
Putting the value of C in (i),
$\Rightarrow \tan^{-1}\text{y}=\frac{\text{x}^{2}}{2}+\text{x}$
$\Rightarrow \text{y}=\tan\big(\frac{\text{x}^{2}}{2}+\text{x}\big)$
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Question 41 Mark
If $\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\text{x}\sin\text{x},$ then $(\text{y}-1)\sin\text{x}=$
  1. $\text{c}-\text{x}\sin\text{x}$
  2. $\text{c}+\text{x}\cos\text{x}$
  3. $\text{c}-\text{x}\cos\text{x}$
  4. $\text{c}+\text{x}\sin\text{x}$
Answer
  1. $\text{c}-\text{x}\cos\text{x}$
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Question 51 Mark
Solution of differential equation x.dy – y.dx = Q represents:
  1. A rectangular hyperbola
  2. Parabola whose vertex is at the origin
  3. Straight line passing through the origin
  4. A circle whose centre is at the origin
Answer
  1. Straight line passing through the origin
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Question 61 Mark
The degree and the order of the differential equation
$\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2$  are respectively:
  1. 1, 1
  2. 2, 1
  3. 4, 1
  4. 1, 4
Answer
  1. 4, 1
Solution:
Given differential equation is
$\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2$
$\Rightarrow\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+1$
Here, degree is 44 and order is 1.
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Question 71 Mark
The differential equation  $\frac{\text{dx}}{\text{dx}}=\frac{1}{\text{ax}+\text{by}+\text{c}},$  where a, b, c are all non zero real numbers, is:
  1. Linear in y
  2. Linear in x
  3. Linear in both x and y
  4. Homogeneous equation
Answer
  1. Linear in x
Solution:
$=\frac{\text{dx}}{\text{dx}}={\text{ax}+\text{by}+\text{c}}$
$=\frac{\text{dx}}{\text{dx}}-{\text{ax}=\text{by}+\text{c}}$
$=\text{Linear in x}$
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Question 81 Mark
A homogeneous dofferential equation of the from $\frac{\text{dx}}{\text{dy}}=\text{h}(\frac{\text{x}}{\text{y}})$ can be solved by making the substitution:
  1. y = vx
  2. v = yx
  3. x = vy
  4. x = v
Answer
  1. x = vy
Solution:
A homogeneous differential of the from $\frac{\text{dx}}{\text{dy}}=\text{h}(\frac{\text{x}}{\text{y}})$ can be solved by sunstituting x = vy.
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Question 91 Mark
Choose the correct answer from the given four option.
Solution of differential equation xdy - ydx = 0 represents:
  1. A rectangular hyperbola.
  2. Parabola whose vertex is at origin.
  3. Straight line passing through origin.
  4. A circle whose centre is at origin.
Answer
  1. Straight line passing through origin.
Solution:
Given that, $\text{xdy}-\text{ydx}=0$
$\Rightarrow\text{xdy}=\text{ydx}$
$\Rightarrow\frac{\text{dy}}{\text{y}}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\Rightarrow\int\frac{\text{dy}}{\text{y}}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\log\text{y}=\log\text{x}+\log\text{C}$
$\Rightarrow\log\text{y}=\log\text{Cx}$
$\Rightarrow\text{y}=\text{Cx}$
Which is a straight line passing through origin.
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Question 101 Mark
The order of the differential equation $\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5\Bigg]^\frac{2}{3}=\frac{\text{d}^3\text{y}}{\text{dx}}$ is:
  1. $2$
  2. $1$
  3. $3$
  4. $\frac{2}{3}$
Answer
  1. $3$
Solution:
$=\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5\Bigg]^\frac{2}{3}=\frac{\text{d}^3\text{y}}{\text{dx}}$
$\Rightarrow\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5\Bigg]^2=\Big(\frac{\text{d}^3\text{y}}{\text{dx}}\Big)^3$
Hence order of above differential equation is 3
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Question 111 Mark
Choose the correct answer from the given four option.
The solution of the differential equation $\cos\text{x}\ \sin\text{y}\ \text{dx}+\sin\text{x}\ \cos\text{y}\ \text{dy}=0$ is:
  1. $\frac{\sin\text{x}}{\sin\text{y}}=\text{C}$
  2. $\sin\text{x}\ \sin\text{y}=\text{C}$
  3. $\sin\text{x}+\sin\text{y}=\text{C}$
  4. $\cos\text{x}\ \cos\text{y}=\text{C}$
Answer
  1. $\sin\text{x}\ \sin\text{y}=\text{C}$
Solution:
Given differential equation is
$\cos\text{x}\ \sin\text{y}\ \text{dx}+\sin\text{x}\ \cos\text{y}\ \text{dy}=0$
$\Rightarrow\cos\text{x}\ \sin\text{y}\ \text{dx}=-\sin\text{x}\ \cos\text{y}\ \text{dy}=0$
$\Rightarrow\frac{\cos\text{x}}{\sin\text{x}}\text{dx}=-\frac{\cos\text{y}}{\sin\text{y}}\text{dy}$
$\Rightarrow\cot\text{x}\ {\text{dx}}=-\cot\text{y}\ {\text{dy}}$
On integrating both sides, we get
$\log\ \sin\text{x}=-\log\ \sin\text{y}+\log\ \text{C}$
$\Rightarrow\log\ \sin\text{x}\ \sin\text{y}=\log\ \text{C}$
Carrying the exponent on both sides, we get
$\Rightarrow\sin\text{x}\ \sin\text{y}=\text{C}$
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Question 121 Mark
Which of the following differentials equation has $\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$ as the general solution?
  1. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
  2. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
  3. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{1}=0$
  4. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{1}=0$ 
Answer
  1. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-1=0$
Solution:
We have,
$\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{i})$
Differentiating both sides of (i) with we get,
$\frac{\text{dy}}{\text{dx}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{ii})$
Differentiating both sides of (ii) with we get,
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{y}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
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MCQ 131 Mark
$\text{x}\frac{\text{b}-\text{c}}{\text{b}\text{c}}\ \text{x}\frac{\text{c}-\text{a}}{\text{c}\text{a}}\text{x}\frac{\text{a}-\text{b}}{\text{a}\text{c}}=$
  • A
    $a^{a+b+c}$
  • B
    $x^{abc}$
  • ✓
    $1$
  • D
    $0$
Answer
Correct option: C.
$1$
$\text{x}\frac{\text{b}-\text{c}}{\text{b}\text{c}}\text{x}\frac{\text{c}-\text{a}}{\text{ac}}\text{x}\frac{\text{a}-\text{b}}{\text{ac}}\text{x}\frac{\text{b}-\text{c}}{\text{bc}}+\frac{\text{c}-\text{a}}{\text{ca}}+\frac{\text{a}-\text{b}}{\text{ac}}$
$\text{x}^0=1$
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Question 141 Mark
The solution of differential equation $\frac{\text{dy}}{\text{dx}}-3\text{y}=\sin2\text{x}$ is:
  1. $\text{y}=\text{e}^{-3\text{x}}\Big[\frac{\cos2\text{x}+3\sin2\text{x}}{13}\Big]+\text{c}$
  2. $\text{y}=\text{e}^{-3\text{x}}\Big[\frac{\cos2\text{x}-3\sin2\text{x}}{13}\Big]+\text{c}$
  3. $\text{y}^{-3\text{x}}=-\text{e}^{-3\text{x}}\Big[\frac{2\cos2\text{x}+3\sin2\text{x}}{13}\Big]+\text{c}$
  4. $\text{None of these}$
Answer
  1. $\text{y}^{-3\text{x}}=-\text{e}^{-3\text{x}}\Big[\frac{2\cos2\text{x}+3\sin2\text{x}}{13}\Big]+\text{c}$
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Question 151 Mark
Choose the correct answer from the given four option.
The differential equation $\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{x}=\text{C}$ represents:
  1. Family of hype.
  2. Family of parabolas.
  3. Family of ellipses.
  4. Family of circles.
Answer
  1. Family of circles.
Solution:
Given that, $\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{x}=\text{C}$
$\Rightarrow\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}=\text{C}-\text{x}$
$\Rightarrow\text{ydy}=(\text{C}-\text{x})\text{dx}$
On integrating both sides, we get
$\int\text{ydy}=\int(\text{C}-\text{x})\text{dx}$
$\Rightarrow\frac{\text{y}^2}{2}=\text{Cx}-\frac{\text{x}^2}{2}+\text{k}$
$\Rightarrow\frac{\text{x}^2}{2}+\frac{\text{y}^2}{2}=\text{Cx}+\text{k}$
$\Rightarrow\frac{\text{x}^2}{2}+\frac{\text{y}^2}{2}-\text{Cx}=\text{k}$
which represent family of circles.
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Question 161 Mark
The number of arbitrary constants in the particular solution of a differential equation of third order is:
  1. 3
  2. 2
  3. 1
  4. 0
Answer
  1. 0
Solution:
The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.
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Question 171 Mark
The degree of the differential equation $\big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\big)^{3}+\big(\frac{\text{dy}}{\text{dx}}\big)^{2}+\sin\big(\frac{\text{dy}}{\text{dx}}\big)+1=0$ is:
  1. 3
  2. 2
  3. 1
  4. Not defined.
Answer
  1. Not defined
Solution:
We have,
$\big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\big)^{3}+\big(\frac{\text{dy}}{\text{dx}}\big)^{2}+\sin\big(\frac{\text{dy}}{\text{dx}}\big)+1=0$
The highest order derivative in this equation is $\frac{\text{d}^{2}\text{y}}{\text{d}^{2}\text{x}}.$
But the equation cannot be as a polynomial in differential coeffcient.
Hence, the degree is not defined.
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Question 181 Mark
What is integrating factor of $\frac{\text{dy}}{\text{dx}}+\text{y}\sec\text{x}=\tan\text{x}?$
  1. $\sec\text{x}+\tan\text{x}$
  2. $\log(\sec\text{x}+\tan\text{x})$
  3. $\text{e}^{\sec\text{x}}$
  4. $\sec{\text{x}}$ 
Answer
  1. $\sec{\text{x}}+\tan\text{x}$
Solution:
We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\sec\text{x}=\tan\text{x}$
Comparing with We get,
$\text{P}=\sec{\text{x}}, \text{Q}=\tan{\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\sec\text{x}\text{dx}}$
$=\text{e}^{\log(\sec\text{x}+\tan\text{x})}$
$=\sec\text{x}+\tan\text{x}$
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Question 191 Mark
 The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=0$ with y(1) = 1 is given by.
  1. $\text{y}=\frac{1}{\text{x}^{2}}$
  2. $\text{x}=\frac{1}{\text{y}^{2}}$
  3. $\text{x}=\frac{1}{\text{y}}$
  4. $\text{y}=\frac{1}{\text{x}}$ 
Answer
  1. $\text{y}=\frac{1}{\text{x}^{2}}$
Solution:
We have,
$\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{-2\text{y}}{\text{x}}$
$\Rightarrow\frac{1}{2}\times\frac{1}{\text{y}}\text{dy}=\frac{-1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\Rightarrow\frac{1}{2}\int\frac{1}{\text{y}}\text{dy}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\frac{1}{2}\log\text{y}=-\log{\text{x}}+\log\text{C}$
$\Rightarrow\log\text{y}^{\frac{1}{2}}+\log{\text{x}}=\log\text{C}$
$\Rightarrow \log(\sqrt{\text{yx}})=\log\text{C}$
$\Rightarrow\sqrt{\text{yx}}=\log\text{C}\ ...(\text{i})$
As (i) y(1) = 1, we get
$1=\text{C}$
Putting the valur of C in (i)
$\Rightarrow\sqrt{\text{yx}}=1$
$\Rightarrow\text{y}=\frac{1}{\text{x}^{2}}$
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Question 201 Mark
Choose the correct answer from the given four option.
The degree of the differential equation $\Big[1+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^2\Big]^{\frac{3}{2}}=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$ is:
  1. 4
  2. $\frac{3}{2}$
  3. Not defined
  4. 2
Answer
  1. 2
Solution:
Given is, $\Big[1+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^2\Big]^{\frac{3}{2}}=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$
On squaring both sides, we get
$\Big[1+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^2\Big]^{3}=\Big(\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}\Big)^2$
So, the degree of differential equation is 2.
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Question 211 Mark
The number of arbitrary constants in the particular solution of differential equation of fourth order is:
  1. 3
  2. 2
  3. 1
  4. 0
Answer
  1. 0
Solution:
The number of arbitray constant in the particular solution of a differential equation is always zero.
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Question 221 Mark
Choose the correct answer from the given four option.
Integrating factor of $\frac{\text{xd}\text{y}}{\text{d}\text{x}}-\text{y}=\text{x}^4-3\text{x}$ is:
  1. $\text{x}$
  2. $\log\text{x}$
  3. $\frac{1}{\text{x}}$
  4. $-\text{x}$
Answer
  1. $\frac{1}{\text{x}}$
Solution:
Given that $\frac{\text{xd}\text{y}}{\text{d}\text{x}}-\text{y}=\text{x}^4-3\text{x}$
Dividing both sides by x, we get
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}-\frac{\text{y}}{\text{x}}=\text{x}^3-3$
Here, $\text{P}=-\frac{1}{\text{x}},\text{Q}=\text{x}^3-3$
$\therefore\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{-\log\text{x}}=\frac{1}{\text{x}}$
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Question 231 Mark
Choose the correct answer from the given four option.
The differential equation for $\text{y}=\text{A}\cos\alpha\text{x}+\text{B}\sin\alpha\text{x},$ where A and B are arbitrary constants is:
  1. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha^2\text{y}=0$
  2. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
  3. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha\text{y}=0$
  4. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha\text{y}=0$
Answer
  1. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
Solution:
Given, $\text{y}=\text{A}\cos\alpha\text{x}+\text{B}\sin\alpha\text{x}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=-\alpha\text{A}\sin\alpha\text{x}+\alpha\text{B}\cos\alpha\text{x}$
Again, differentiating both sides w.r.t. x, we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\text{A}\alpha^2\cos\alpha\text{x}-\alpha^2\text{B}\sin\alpha\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\alpha^2(\text{A}\cos\alpha\text{x}-\text{B}\sin\alpha\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\alpha^2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
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Question 241 Mark
The solution of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}+\text{x}\ \tan\frac{\text{y}}{\text{x}}$ is:
  1. $\sin\frac{\text{x}}{\text{y}}=\text{x}+\text{C}$
  2. $\sin\frac{\text{y}}{\text{x}}=\text{Cx}$
  3. $\sin\frac{\text{x}}{\text{y}}=\text{Cy}$
  4. $\sin\frac{\text{y}}{\text{x}}=\text{Cy}$ 
Answer
  1. $\sin\frac{\text{y}}{\text{x}}=\text{Cx}$
Solution:
We have,
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}+\text{x}\ \tan\frac{\text{y}}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\tan\frac{\text{y}}{\text{x}}\ ...(\text{i})$
Let $\text{y}=\upsilon\text{x}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}$
Putting both value in (i)
$\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}=\upsilon+\tan\upsilon$
$\Rightarrow \frac{\text{d}\upsilon}{\tan\upsilon}=\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get
$\log\sin\upsilon=\log\text{x}+\log\text{C}$
$\Rightarrow \log\frac{\sin\upsilon}{\text{x}}=\log\text{C}$
$\Rightarrow \frac{\sin\upsilon}{\text{x}}=\text{C}$
$\Rightarrow\sin\upsilon=\text{Cx}$
$\Rightarrow\sin(\frac{\text{y}}{\text{x}})=\text{Cx}$
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Question 251 Mark
The solution of the differential equation $2\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=3$ resresents:
  1. circles
  2. straight lines
  3. ellipses
  4. parabolas
Answer
  1. parabolas
Solution:
We have,
$2\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=3$
$\Rightarrow 2\text{x}\frac{\text{dy}}{\text{dx}}=3+\text{y}$
$\Rightarrow \frac{1}{3+\text{y}}\text{dy}=\frac{1}{2\text{x}}\text{dx}$
Interating both sides, we get
$\Rightarrow \int\frac{1}{3+\text{y}}\text{dy}=\frac{1}{2}\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \log|3+\text{y}|=\frac{1}{2}\log|\text{x}|+\log|\text{C}|$
$\Rightarrow\log|\frac{3+\text{y}}{\sqrt{\text{x}}}|=\log\text{C}$
$\Rightarrow\frac{3+\text{y}}{\sqrt{\text{x}}}=\text{C}$
$\Rightarrow 3+\text{y}=\text{C}\sqrt{\text{x}}$
Squaring both sides, we get
$(3+\text{y})^{2}=\text{C}{\text{x}}\ ...(\text{i})$
Thus, (i) the equation of parabolas.
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Question 261 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}-\text{Ky}=0, \text{y}(0)=1$ approaches to zero when $\text{x}\rightarrow\propto$ if,
  1. K = 0
  2. K > 0
  3. K < 0
  4. None of these.
Answer
  1. K < 0
Solution:
We have,
$\Rightarrow \frac{\text{dy}}{\text{dx}}-\text{Ky}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{Ky}$
$\Rightarrow \frac{1}{\text{y}}\text{dy}=\text{K}\ \text{dx}$
Integrating both sides, we get
$ \int\frac{1}{\text{y}}\text{dy}=\text{K}\int\text{dx}$
$\Rightarrow \log|\text{y}|=\text{Kx}+\text{C}\ ...(\text{i})$
Now,
$\text{y}(0)=1$
$\text{C}=0$
Putting C = 0 in (i),
$\log|\text{y}|=\text{Kx}$
$\Rightarrow \text{e}^{\text{Kx}}=\text{y}$
According to the quation,
$\text{e}^{\text{K}\propto}=0$
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Question 271 Mark
Choose the correct answer from the given four options.The general solution of the differential equation $(\text{e}^{\text{x}}+1)\text{ydy}=(\text{y}+1)\text{e}^{\text{x}}$ is:
  1. $(\text{y}+1)=\text{k}(\text{e}^{\text{x}}+1)$
  2. $\text{y}+1=\text{e}^{\text{x}}+1+\text{k}$
  3. $\text{y}=\log\left\{\text{k}(\text{y}+1)(\text{e}^{\text{x}}+1)\right\}$
  4. $\text{y}=\log\left\{\frac{\text{e}^{\text{x}}+1}{\text{y}+1}\right\}+\text{k}$
Answer
  1. $\text{y}=\log\left\{\text{k}(\text{y}+1)(\text{e}^{\text{x}}+1)\right\}$
Solution:
Given differential equation
$(\text{e}^{\text{x}}+1)\text{ydy}=(\text{y}+1)\text{e}^{\text{x}}\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}(1+\text{y})}{(\text{e}^{\text{x}}+1)\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{e}^{\text{x}}+1)\text{y}}{\text{e}^{\text{x}}(1+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}\text{y}}{\text{e}^{\text{x}}(1+\text{y})}\frac{\text{y}}{\text{e}^{\text{x}}(1+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}+\frac{\text{y}}{(1+\text{y})\text{e}^{\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(1+\frac{1}{\text{e}^{\text{x}}}\Big)\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(1+\frac{1}{\text{e}^{\text{x}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(\frac{\text{e}^{\text{x}}+1}{\text{e}^{\text{x}}}\Big)$
$\Rightarrow\Big(\frac{\text{y}}{1+\text{y}}\Big)\text{dy}=\Big(\frac{\text{e}^{\text{x}}}{\text{e}^{\text{x}}+1}\Big)\text{dx}$
On integrating both sides, we get
$\int\frac{\text{y}}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\int\frac{1+\text{y}-1}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\int1\text{dy}-\int\frac{\text{y}}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\text{y}-\log|(1+\text{y})+\log(1+\text{e}^{\text{x}})|+\log(\text{k})$
$\Rightarrow\text{y}=\log\left\{\text{k}(1+\text{y})(1+\text{e}^{\text{x}})\right\}$
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Question 281 Mark
The differential equation satisfied by $\text{ax}^{2}+\text{by}^{2}=1$ is:
  1. $\text{xyy}_{2}+\text{y}_{1}^{2}+\text{yy}_{1}=0$
  2. $\text{xyy}_{2}+\text{xy}_{1}^{2}-\text{yy}_{1}=0$
  3. $\text{xyy}_{2}+\text{xy}_{1}^{2}+\text{yy}_{1}=0$
  4. None of these. 
Answer
  1. $\text{xyy}_{2}+\text{xy}_{1}^{2}-\text{yy}_{1}=0$
Solution:
We have,
$\text{ax}^{2}+\text{by}^{2}=1\ ...(\text{i})$
Differential both sides of (i) with x, we get
$2\text{ax}+2\text{by}\frac{\text{dy}}{\text{dy}}=0\ ...(\text{ii})$
Differential both sides of (ii) with x, we get
$2\text{ax}+2\text{b}\Big(\frac{\text{dy}}{\text{dy}}\Big)^{2}+2\text{}by\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$
$\Big[\text{y}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big(\frac{\text{dy}}{\text{dx}^{2}}\Big)\Big]=-\frac{2\text{a}}{2\text{b}}$
$\text{x}\Big[\text{y}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big(\frac{\text{dy}}{\text{dx}^{2}}\Big)\Big]=-\Big(-\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}\Big)$
$\text{xy}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\big)^{2}-\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\text{xyy}_{2}+\text{x}(\text{y}_{1}^{2})-\text{yy}_{1}=0$
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Question 291 Mark
Choose the correct answer from the given four options.The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}-\text{y}}+\text{x}^2\text{e}^{-\text{y}}$ is:
  1. $\text{y}=\text{e}^{\text{x}-\text{y}}-\text{x}^2\text{e}^{-\text{y}}+\text{c}$
  2. $\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{c}$
  3. $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$
  4. $\text{e}^{\text{x}}-\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$
Answer
  1. $\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{c}$
Solution:
We have, $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}-\text{y}}+\text{x}^2\text{e}^{-\text{y}}$
$\Rightarrow\text{e}^{\text{y}}\text{dy}=(\text{e}^{\text{x}}+\text{x}^2)\text{dx}$
$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int(\text{e}^{\text{x}}+\text{x}^2)\text{dx}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}}+\frac{\text{x}^3}{3}+\text{C}$
$\Rightarrow\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{C}$
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Question 301 Mark
Choose the correct answer from the given four options.The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is:
  1. An ellipse.
  2. parabola.
  3. circle.
  4. rectangular hyperbola.
Answer
  1. rectangular hyperbola.
Solution:
Slope of langent to the curve $=\frac{\text{dy}}{\text{dx}}$
According to the question, $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\text{ydy}=\text{xdx}$
On integrating both sides, we get
$\frac{\text{y}^2}{2}=\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow\text{y}^2-\text{x}^2=2\text{C}$ which is an equation of rectangular hyperbola.
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Question 311 Mark
A homogeneous differential equation of the from $\frac{\text{dx}}{\text{dy}}=\text{h}\Big(\frac{\text{x}}{\text{y}}\Big)$ can be solved by making the substitution.
  1. y = vx
  2. v = yx
  3. x = vy
  4. x = v
Answer
  1. x = vy
We know that a homogeneous differential equation of the form $\frac{\text{dx}}{\text{dy}}=\text{h}\Big(\frac{\text{x}}{\text{y}}\Big)$  can be solved by the substitution $\frac{\text{x}}{\text{y}}=\text{v i.e., x}=\text{vy}.$
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Question 321 Mark
Choose the correct answer from the given four options.The solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}},$ y(0) = 0 is:
  1. $\text{y}=\text{e}^{-\text{x}}(\text{x}-1)$
  2. $\text{y}=\text{x}\text{e}^{\text{x}}$
  3. $\text{y}=\text{x}\text{e}^{-\text{x}}+1$
  4. $\text{y}=\text{x}\text{e}^{-\text{x}}$
Answer
  1. $\text{y}=\text{x}\text{e}^{-\text{x}}$
Solution:
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}}$
Which is a linear differential equation.
Here, $\text{P}=1$ and $\text{Q}=\text{e}^{-\text{x}}$
$\text{I.F.}=\text{e}^{\int\text{dx}}=\text{e}^{\text{x}}$
The general solution is
$\text{y}.\text{e}^{-\text{x}}=\int\text{e}^{-\text{x}}.\text{e}^{\text{x}}\ \text{dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{\text{x}}=\int\text{dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{-\text{x}}=\text{x}+\text{C}\ ....(\text{i})$
Given, when x = 0 and y = 0
$\Rightarrow0=0+\text{C}$
$\Rightarrow\text{C}=0$
Eq, (i) resuces to $\text{y}.\text{e}^{-\text{x}}=\text{x}$ or $\text{y}=\text{x}\text{e}^{-\text{x}}.$
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Question 331 Mark
Choose the correct answer from the given four options.General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$ is:
  1. $\text{y}\sec\text{x}=\tan\text{x}+\text{c}$
  2. $\text{y}\tan\text{x}=\sec\text{x}+\text{c}$
  3. $\tan\text{x}=\sec\text{x}+\text{c}$
  4. $\text{x}\sec\text{x}=\tan\text{y}+\text{c}$
Answer
  1. $\text{y}\sec\text{x}=\tan\text{x}+\text{c}$
Solution:
Given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$
This is a linear differential equation
Here, $\text{P}=\tan\text{x},\text{Q}=\sec\text{x},$
$\therefore\text{I.F.}=\text{e}^{\int\tan\text{xdx}}$
$=\text{e}^{\log|\sec\text{x}|}=\sec\text{x}$
Thus, the general solution is
$\text{y}.\sec\text{x}=\int\sec\text{x}.\sec\text{x}+\text{C}$
$\Rightarrow\text{y}.\sec\text{x}=\int\sec^2\text{x}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\sec\text{x}=\tan\text{x}+\text{C}$
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Question 341 Mark
The Integrating Factor of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\text{x}^2\ \text{is}$
  1. $\text{e}^{-\text{x}}$
  2. $\text{e}^{-\text{y}}$
  3. $\frac{1}{\text{x}}$
  4. $\text{x}$
Answer
  1. $\frac{1}{\text{x}}$
The given differential equation is:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=2\text{x}$
This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}​​\ (\text{where p}=-\frac{1}{\text{x}}\ \text{and}\ \text{Q}=2\text{x})$
The integrating factor (I.F) is given by the relation,
$​​\text{e}^{\int​​\text{pdx}}$
$\therefore\text{I.F}=\text{e}^{\int-\frac{1}{\text{x}}\text{dx}}=\text{e}^{-\log\text{x}}=\text{e}^{\log(\text{x}^-1)}=\text{x}^{-1}=\frac{1}{\text{x}}$
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Question 351 Mark
The number of arbitrary constants in the particular solution of a differential equation of third order are:
  1. 3
  2. 2
  3. 1
  4. 0
Answer
  1. 0
Solution:
The number of arbitrary constants in a particular solution of a differential equation of order n is always 0.
So here it will be 0
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Question 361 Mark
Choose the correct answer from the given four options.Which of the following is the general solution of $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=0?$
  1. $\text{y}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
  2. $\text{y}=(\text{Ax}+\text{B})\text{e}^{-\text{x}}$
  3. $\text{y}=\text{Ax}\text{e}^{\text{x}}+\text{B}\text{e}^{\text{x}}$
  4. $\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}$
Answer
  1. $\text{y}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
Solution:
Given that, $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=0$
$\text{D}^2\text{y}-2\text{D}\text{y}+\text{y}=0,$
Where, $\text{D}=\frac{\text{d}}{\text{dx}}$
$\text{D}^2\text{y}-2\text{D}\text{y}+\text{y}=0$
The auxiliary equation is $\text{m}^2-2\text{m}+1=0$
$(\text{m}-1)^2=0$
$\Rightarrow\text{m}=1,1$
Since, the roots are real and equal.
$\therefore\text{CF}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
$\Rightarrow\text{y}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
$\big[$ Since, if roots of Auxiliary equation are real and equal say (m), then $\text{CF}=(\text{C}_1\text{x}+\text{C}_2)\text{e}^{\text{mx}}\big]$
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Question 371 Mark
Choose the correct answer from the given four options.The order and degree of the differential equation $\Big(\frac{\text{d}^3\text{y}}{\text{d}\text{x}^3}\Big)^2-3\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^4=\text{y}^4$ are:
  1. 1, 4.
  2. 3, 4.
  3. 2, 4.
  4. 3, 2.
Answer
  1. 3, 2.
Solution:
Given that $\Big(\frac{\text{d}^3\text{y}}{\text{d}\text{x}^3}\Big)^2-3\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^4=\text{y}^4$
$\therefore\text{Order}=3\text{ and degree}=2$
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MCQ 381 Mark
The order of the differential whose general solution is given by ${\text{y}}=\text{C}_1\cos(2\text{x}+\text{C}_{2})+(\text{C}_{3}+\text{C}_{4})\text{a}^{\text{x}+\text{C}_{5}}+\text{C}_{6}\sin(\text{x}-\text{C}_{7}).$ 
  • A
    $3$
  • B
    $4$
  • ✓
    $5$
  • D
    $2$
Answer
Correct option: C.
$5$

The given equation can be reduced to :
${\text{y}}=\text{C}_1\cos(2\text{x}+\text{C}_{2})+(\text{C}_{3}+\text{C}_{4})\text{a}^{\text{x}+\text{C}_{5}}+\text{C}_{6}\sin(\text{x}-\text{C}_{7})$
Where $C = C_3 + C_4$ be a constant
There are $5$ constant $(C_1, C_2, C_3, C_6, C_7)$ in the given differential equation.
Hence, the order of the dfifferential equation is 5.

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Question 391 Mark
Choose the correct answer from the given four options.The order and degree of the differential equation $\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$ are:
  1. $2,\frac{3}{2}$
  2. 2, 3
  3. 2, 1
  4. 3, 4
Answer
  1. 2, 1
Solution:
Given that, $\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$
$\therefore\text{Order}=2\text{ and degree}=1$
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Question 401 Mark
If P and q are the order and degree of the differention $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^{3}\frac{\text{d}^{2}\text{y}}{\text{dx}^{3}}+\text{xy}=\cos\text{x}$ then: 
  1. $\text{p}<\text{q}$ 
  2. $\text{p}=\text{q}$
  3. $\text{p}>\text{q}$
  4. None of these. 
Answer
  1. $\text{p}>\text{q}$
Solution:
We have,
$\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^{3}\frac{\text{d}^{2}\text{y}}{\text{dx}^{3}}+\text{xy}=\cos\text{x}$
The highest order is $\frac{\text{d}^{2}\text{y}}{\text{dz}^{2}}$ and it's degree is 1.
So, the order is 2 and the degree is 1.
$\text{p}=2, \text{q}=1$
Clearly, $\text{p}>\text{q}$ 
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Question 411 Mark
Choose the correct answer from the given four options.The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ is:
  1. $\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
  2. $\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$
  3. $\text{y}\log(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
  4. $\text{y}(1+\text{x}^2)=\text{C}+\sin^{-1}\text{x}$
Answer
  1. $\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
Solution:
Given is, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$
Here, $\text{P}=\frac{2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\frac{1}{(1+\text{x}^2)^2}$
This ia a linerar differential equation.
$\therefore\text{I.F.}=\text{e}^{\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
Put $1+\text{x}^2=\text{t}\Rightarrow2\text{x}\text{ dx}=\text{dt}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{\text{dt}}{\text{t}}}=\text{e}^{\log\text{t}}$
$=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2$
Thus, the general solution is
$\text{y}.(1+\text{x}^2)=\int(1+\text{x}^2)\frac{1}{(1+\text{x}^2)}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\int\frac{1}{(1+\text{x}^2)}\text{dx}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}+\text{C}$
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Question 421 Mark
Choose the correct answer from the given four options.The differential equation for which $\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}$ is a solution, is:
  1. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{y}=0$
  2. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{y}=0$
  3. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+(\text{a}+\text{b})\text{y}=0$
  4. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+(\text{a}-\text{b})\text{y}=0$
Answer
  1. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{y}=0$
Solutuion:
Given equation is, $\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}$
On differentiating both sides w.r.t.x. we get
$\frac{\text{dy}}{\text{dx}}=-\text{a}\sin\text{x}+\text{b}\cos\text{dx}$
Again, differentiating w.r.t.x. we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\text{a}\sin\text{x}+\text{b}\cos\text{dx}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{y}=0$
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MCQ 431 Mark
If m and n are the order and degree of the differential equation $(\text{y}_{2})^{5}+\frac{4(\text{y}_{2})^{3}}{\text{y}^{3}}+\text{y}^{3}=\text{y}_{3}=\text{x}^{2}-1$, then
  • A
    $\text{m}=3, \text{n}=3$
  • ✓
    $\text{m}=3, \text{n}=2$
  • C
    $\text{m}=3, \text{n}=5$
  • D
    $\text{m}=3,\text{n}=1$ 
Answer
Correct option: B.
$\text{m}=3, \text{n}=2$

We have,
$(\text{y}_{2})^{5}+\frac{4(\text{y}_{2})^{3}}{\text{y}^{3}}+\text{y}_{3}=\text{x}^{2}-1$
$\text{y}_{3}(\text{y}_{2})^{5}+{4(\text{y}_{2})^{3}}+(\text{y}_{3})^{2}=\text{y}_{3}(\text{x}^{2}-1)$
The highest order is $y_3$ and its highest in this equation is $2.$
Hence, $m = 3, n = 2.$

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Question 441 Mark
Choose the correct answer from the given four options.Solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$ is:
  1. $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
  2. $\text{x}(\text{y}-\cos\text{x})=\sin\text{x}+\text{c}$
  3. $\text{x}\text{y}\cos\text{x}=\sin\text{x}+\text{c}$
  4. $\text{x}(\text{y}+\cos\text{x})=\cos\text{x}+\text{c}$
Answer
  1. $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
Solution:
Given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}\frac{1}{\text{x}}=\sin\text{x}$
Which is liner differential equations.
Here, $\text{P}=\frac{1}{\text{x}}$ and $\text{Q}=\sin\text{x}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
 $=\text{e}^{\log\text{x}}=\text{x}$
The general solution is
$\text{yx}=\int\text{x}\sin\text{xdx}+\text{c}\ .....(\text{i})$
Take $\text{I}=\int\text{x}\sin\text{xdx}$
$-\text{x}\cos\text{x}-\int-\cos\text{xdx}$
$=-\text{x}\cos\text{x}+\sin\text{x}$
Put the value of 1 in Eq. (i), we get
$\text{xy}=-\text{x}\cos\text{x}+\sin\text{x}+\text{c}$
$\Rightarrow\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
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Question 451 Mark
The solution of the differential equation $(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}+(\text{y}^{2}+1)=0$ is:  
  1. $\text{y}=2+\text{x}^{2}$
  2. $\text{y}=\frac{1+\text{x}}{1-\text{x}}$ 
  3. $\text{y}=\text{x}(\text{x}-1)$
  4. $\text{y}=\frac{1+\text{y}}{1-\text{y}}$ 
Answer
  1. $\text{y}=\frac{1-\text{x}}{1+\text{x}}$
Solution:
We have,
$(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}=-(\text{y}^{2}+1)=0$
$\Rightarrow (\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}=-(\text{y}^{2}+1)$
$\Rightarrow \frac{1}{(\text{y}^{2}+1)}\text{dy}=-\frac{1}{(\text{x}^{2}+1)}\text{dx}$
Intergrating both sides, we get
$\Rightarrow \int\frac{1}{(\text{y}^{2}+1)}\text{dy}=-\int\frac{1}{(\text{x}^{2}+1)}\text{dx}$
$\Rightarrow \tan^{-1}\text{y}=-\tan^{-1}\text{x}+\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big) =\tan^{-1}\text{C}$
$\Rightarrow \frac{\text{x}+\text{y}}{1+\text{xy}}=\text{C}$
Disclaimer : The initial value given, So the find will be C = 1, So
 $\Rightarrow \text{x}+\text{y}=1-\text{xy}$
$\Rightarrow \text{y}+\text{xy}=1-\text{x}$
$\Rightarrow \text{y}(1+\text{x})=1-\text{x}$
$\Rightarrow \text{y}=\frac{1-\text{x}}{1+\text{x}}$
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MCQ 461 Mark
The solution of the differential equation $xdy + ydy = x^2y dy - y^2x dx$, is:
  • ✓
    $x^2 - 1 = C(1 + y^2)$
  • B
    $x^2+ 1 = C(1 + y^2)$
  • C
    $x^3 - 1 = C(1 + y^3)$
  • D
    $x^3 + 1 = C(1 - y^3)$
Answer
Correct option: A.
$x^2 - 1 = C(1 + y^2)$

We have,
$\text{x}\ \text{dx}+\text{y}\ \text{dy}=\text{x}^{2}\ \text{dy}-\text{y}^{2}\text{x}\ \text{dx}$
$\Rightarrow (\text{x}+\text{xy}^{2})\text{dx}=(\text{x}^{2}\text{y}-\text{y})\text{dy}$
$\Rightarrow \frac{\text{x}}{(\text{x}^{2}-1)}\text{dx}=\frac{\text{y}}{(1+\text{x})^{2}}\text{dy}$
$\Rightarrow \frac{2\text{x}}{2(\text{x}^{2}-1)}\text{dx}=\frac{2\text{y}}{2(1+\text{y})^{2}}\text{dy}$
Integrating both sides, we get
$\frac{1}{2}\int\frac{2\text{y}}{(1+\text{y})^{2}}\text{dy}=\frac{1}{2}\int \frac{2\text{x}}{(1+\text{x})^{2}}\text{dx}$
$\Rightarrow \log|(1+\text{y}^{2})|=\frac{1}{2}\log|(\text{x}^{2}-1)|-\frac{1}{2}\log|\text{C}|$
$\Rightarrow \log|(1+\text{y}^{2})|=\log|(\text{x}^{2}-1)|-\log|\text{C}|$
$\Rightarrow \log|(1+\text{y}^{2})|=\log|\frac{\text{x}^{2}-1}{\text{C}}|$
$\Rightarrow 1+\text{y}^{2}=\frac{\text{x}^{2}-1}{\text{C}}$
$\Rightarrow \text{C}(1+\text{y}^{2})=\text{x}^{2}-1$

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MCQ 471 Mark
The differential equation $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}^{2}$ has the general solution:
 
  • A
    $y - x^3 = 2Cx$
  • ✓
    $2y - x^3 = Cx$
  • C
    $2y + x^2 = 2Cx$
  • D
    $y + x^2 = 2Cx $
Answer
Correct option: B.
$2y - x^3 = Cx$

We have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}^{2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}=\text{x}^{2}$
Comparing with we get, 
$\text{P}=-\frac{1}{\text{x}}$
$\text{Q}=\text{x}^{2}$
Now,
$\text{I.F}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log|\text{x}|}$
$=\text{e}^{\log|\frac{1}{\text{x}}|}$
$=\frac{1}{\text{x}}$
$\text{y}\times\text{I.F}=\int\text{x}^{2}\times\text{I.F}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\int\text{x}^{2}\times\frac{1}{\text{x}}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\int\text{x}^{2}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\frac{\text{x}^{2}}{2}+\text{C}$
$\Rightarrow 2\text{y}-\text{x}^{3}=\text{Cx}$

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Question 481 Mark
Choose the correct answer from the given four options.The differential equation of the family of curves $\text{y}^2=4\text{a}(\text{x}+\text{a})$ is:
  1. $\text{y}^2=4\frac{\text{dy}}{\text{dx}}\Big(\text{x}+\frac{\text{dy}}{\text{dx}}\Big)$
  2. $2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
  3. $\text{y}\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$
  4. $2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0$
Answer
  1. $2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0$
Solution:
We have equation of the curve as
$\text{y}^2=4\text{a}(\text{x}+\text{a})\ .....(\text{i})$
On differentiating both sides w.r.t.x, we gat
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\frac{1}{2}\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}$
On putting the value of a in Eq. (i), we get
$\text{y}^2=2\text{y}\frac{\text{dy}}{\text{dx}}\Big(\text{x}+\frac{1}{2}\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\text{y}^2=2\text{xy}\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0$
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Question 491 Mark
Choose the correct answer from the given four options.The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\ \text{e}^{\frac{\text{x}^2}{2}}+\text{xy}$ is:
  1. $\text{y}=\text{c}\text{e}^{\frac{-\text{x}^2}{2}}$
  2. $\text{y}=\text{c}\text{e}^{\frac{\text{x}^2}{2}}$
  3. $\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
  4. $\text{y}=(\text{c}-\text{x})\text{e}^{\frac{\text{x}^2}{2}}$
Answer
  1. $\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
Solution:
Given that, $\frac{\text{dy}}{\text{dx}}\ \text{e}^{\frac{\text{x}^2}{2}}+\text{xy}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\text{xy}=\text{e}^{\frac{\text{x}^2}{2}}$
It is a linear differential equation.
Here, $\text{P}=-\text{x},\text{ Q}=\text{e}^{\frac{\text{x}^2}{2}}$
$\therefore\text{I.F.}=\text{e}^{\int-\text{xdx}}=\text{e}^{\frac{-\text{x}^2}{2}}$
The general solution is
$\text{y}\text{e}^{\frac{-\text{x}^2}{2}}=\int\text{e}^{\frac{-\text{x}^2}{2}}.\text{e}^{\frac{-\text{x}^2}{2}}\text{dx}+\text{c}$
$\Rightarrow\text{y}{\text{e}}^{\frac{-\text{x}^2}{2}}=\int1\text{dx}+\text{c}$
$\Rightarrow\text{y}\text{e}^{\frac{-\text{x}^2}{2}}=\text{x}+\text{c}$
$\Rightarrow\text{y}=\text{x}\text{e}^{\frac{\text{x}^2}{2}}+\text{c}\text{e}^{\frac{\text{x}^2}{2}}$
$\Rightarrow\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
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Question 501 Mark
The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\text{is}$
  1. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
  2. $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{C}$
  3. $\text{e}^{-\text{x}}+\text{e}^{\text{y}}=\text{C}$
  4. $\text{e}^{-\text{x}}+\text{e}^{-\text{y}}=\text{C}$
Answer
  1. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
The given differential equation is
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}\ \ \text{or}\ \ \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}.\text{e}^{\text{y}}$
Separating the variables, we get,
$\frac{1}{\text{e}^{\text{y}}}\text{dy}=\text{e}^\text{x}\ \text{dx}$
$\text{Integrating},\ \int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^\text{x}\ \text{dx}$
$\therefore\ \frac{\text{e}^{-\text{y}}}{-1}=\text{e}^\text{x}+\text{c}'$
$\therefore\ -\text{e}^{-\text{y}}=\text{e}^\text{x}+\text{c}'\ \ \text{or}\ \ \text{e}^\text{x}+\text{e}^{-\text{y}}=-\text{c}'$
$\therefore\ \text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C},$ which is required solution.
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip