M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip

Questions · Page 1 of 2

M.C.Q (1 Marks)

🎯

Test yourself on this topic

50 questions · timed · auto-graded

Question 11 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:

$\sin\frac{\text{x}}{\text{y}}=\text{x}+\text{C}$
$\sin\frac{\text{y}}{\text{x}}=\text{Cx}$
$\sin\frac{\text{x}}{\text{y}}=\text{Cy}$
$\sin\frac{\text{y}}{\text{x}}=\text{Cy}$

Answer

$\sin\frac{\text{y}}{\text{x}}=\text{Cx}$

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
$\log\sin\upsilon=\log\text{x}+\log\text{C}$
$\Rightarrow \log\frac{\sin\upsilon}{\text{x}}=\log\text{C}$
$\Rightarrow \frac{\sin\upsilon}{\text{x}}=\text{C}$
$\Rightarrow\sin\upsilon=\text{Cx}$
$\Rightarrow\sin(\frac{\text{y}}{\text{x}})=\text{Cx}$

View full question & answer→

Question 21 Mark

The solution of the differential equation $2\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=3$ resresents:

circles
straight lines
ellipses
parabolas

Answer

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.

View full question & answer→

Question 31 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,

K = 0
K > 0
K < 0
None of these.

Answer

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,
$\text{e}^{\text{K}\propto}=0$

View full question & answer→

Question 41 Mark

The differential equation satisfied by $\text{ax}^{2}+\text{by}^{2}=1$ is:

$\text{xyy}_{2}+\text{y}_{1}^{2}+\text{yy}_{1}=0$
$\text{xyy}_{2}+\text{xy}_{1}^{2}-\text{yy}_{1}=0$
$\text{xyy}_{2}+\text{xy}_{1}^{2}+\text{yy}_{1}=0$
None of these.

Answer

$\text{xyy}_{2}+\text{xy}_{1}^{2}-\text{yy}_{1}=0$

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
$2\text{ax}+2\text{b}\Big(\frac{\text{dy}}{\text{dy}}\Big)^{2}+2\text{}by\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$
$\Big[\text{y}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big(\frac{\text{dy}}{\text{dx}^{2}}\Big)\Big]=-\frac{2\text{a}}{2\text{b}}$
$\text{x}\Big[\text{y}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big(\frac{\text{dy}}{\text{dx}^{2}}\Big)\Big]=-\Big(-\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}\Big)$
$\text{xy}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\big)^{2}-\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\text{xyy}_{2}+\text{x}(\text{y}_{1}^{2})-\text{yy}_{1}=0$

View full question & answer→

MCQ 51 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 $5.$

View full question & answer→

Question 61 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:

$\text{p}<\text{q}$
$\text{p}=\text{q}$
$\text{p}>\text{q}$
None of these.

Answer

$\text{p}>\text{q}$

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, $\text{p}>\text{q}$

View full question & answer→

MCQ 71 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, $m = 3, n = 2.$

View full question & answer→

Question 81 Mark

The solution of the differential equation $(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}+(\text{y}^{2}+1)=0$ is:

$\text{y}=2+\text{x}^{2}$
$\text{y}=\frac{1+\text{x}}{1-\text{x}}$
$\text{y}=\text{x}(\text{x}-1)$
$\text{y}=\frac{1+\text{y}}{1-\text{y}}$

Answer

$\text{y}=\frac{1-\text{x}}{1+\text{x}}$

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
$\Rightarrow \text{x}+\text{y}=1-\text{xy}$
$\Rightarrow \text{y}+\text{xy}=1-\text{x}$
$\Rightarrow \text{y}(1+\text{x})=1-\text{x}$
$\Rightarrow \text{y}=\frac{1-\text{x}}{1+\text{x}}$

View full question & answer→

MCQ 91 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
$\frac{1}{2}\int\frac{2\text{y}}{(1+\text{y})^{2}}\text{dy}=\frac{1}{2}\int \frac{2\text{x}}{(1+\text{x})^{2}}\text{dx}$
$\Rightarrow \log|(1+\text{y}^{2})|=\frac{1}{2}\log|(\text{x}^{2}-1)|-\frac{1}{2}\log|\text{C}|$
$\Rightarrow \log|(1+\text{y}^{2})|=\log|(\text{x}^{2}-1)|-\log|\text{C}|$
$\Rightarrow \log|(1+\text{y}^{2})|=\log|\frac{\text{x}^{2}-1}{\text{C}}|$
$\Rightarrow 1+\text{y}^{2}=\frac{\text{x}^{2}-1}{\text{C}}$
$\Rightarrow \text{C}(1+\text{y}^{2})=\text{x}^{2}-1$

View full question & answer→

MCQ 101 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,
$\text{I.F}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log|\text{x}|}$
$=\text{e}^{\log|\frac{1}{\text{x}}|}$
$=\frac{1}{\text{x}}$
$\text{y}\times\text{I.F}=\int\text{x}^{2}\times\text{I.F}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\int\text{x}^{2}\times\frac{1}{\text{x}}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\int\text{x}^{2}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\frac{\text{x}^{2}}{2}+\text{C}$
$\Rightarrow 2\text{y}-\text{x}^{3}=\text{Cx}$

View full question & answer→

Question 111 Mark

The degree of the differntial equation $\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{y}^{3}$ is:

$\frac{1}{2}$
$2$
$3$
$4$

Answer

$2$

Solution:
We have,
$\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{y}^{3}$
The highest order derivative is $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$ and its power is 2.
Hence, the degree is 2.

View full question & answer→

Question 121 Mark

Which of the following differentials equation has y = x as one of its particular solution?

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$

Answer

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=0$

Solution:
We have,
$\text{y}=\text{x}\ ...(\text{i})$
Differentiating both sides of (i) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=1\ ...(\text{ii})$
Differentiating both sides of (ii) with respect to x, we get
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}=\text{x}^{2}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}\times\text{x}=\text{x}^{2}\times1$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{xy}=\text{x}^{2}\times1$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{xy}=\text{x}^{2}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=0$

View full question & answer→

Question 131 Mark

The solution of the differential equartion $\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$ is given by:

$\text{y}=\text{xe}^{\text{x}+\text{C}}$
$\text{x}=\text{ye}^{\text{x}}$
$\text{y}=\text{x}+\text{c}$
$\text{xy}=\text{e}^{\text{x}}+\text{C}$

Answer

$\text{y}=\text{xe}^{\text{x}+\text{C}}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+1)}{\text{x}}\text{dx}$
Integrating both sides, we get
$ \int\frac{\text{dy}}{\text{y}}=\int\frac{(\text{x}+1)}{\text{x}}\text{dx}$
$ \Rightarrow \int\frac{\text{dy}}{\text{y}}=\int\text{dx}+\int\frac{1}{\text{x}}\text{dx}$
$ \Rightarrow \log{\text{y}}=\text{x}+\log\text{x}+\text{C}$
$\Rightarrow \log{\text{y}}-\log\text{x}=\text{x}+\text{C}$
$ \Rightarrow \log\frac{{\text{y}}}{\text{x}}=\text{x}+\text{C}$
$\Rightarrow \frac{\text{y}}{\text{x}}=\text{e}^{\text{x}+\text{C}}$
$\Rightarrow{\text{y}}=\text{xe}^{\text{x}+\text{C}}$

View full question & answer→

Question 141 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:

y = vx
v = yx
x = vy
x = v

Answer

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.

View full question & answer→

MCQ 151 Mark

The general solution of differention eqution of the $e^x dy + (ye^x + 2x)dx = 0$ is :

✓
$xe^y + x^2= C$
B
$xe^y+ y^2= C$
C
$ye^x + y^2= C$
D
$ye^y+ x^2 = C$

Answer

Correct option: A.

$xe^y + x^2= C$

We have,
$e^x dy + (ye^x + 2x) dx = 0$
Diving both sides by we get,
$\frac{\text{dy}}{\text{dx}}+(\text{y}+\frac{2\text{x}}{\text{e}^{x}})=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\text{y}=-\frac{2\text{x}}{\text{e}^{\text{x}}}$
Comping with $\frac{\text{dy}}{\text{dx}}=\text{Q}$ we get,
$\text{P}=1, \text{Q}=-\frac{2\text{x}}{\text{e}^{\text{x}}}$
Now,
$\text{I.F}=\text{e}^{\int\text{dx}}$
$=\text{e}^{\text{x}}$
Solution is given by,
$\text{y}\times\text{I.F}=\int(\text{Q}\times\text{I.F}) \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-\int\text{e}^{\text{x}}\times \frac{2\text{x}}{\text{e}^{\text{x}}}\text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-2\int\text{x}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-\text{x}^{2}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}+\text{x}^{2}=\text{C}$

View full question & answer→

Question 161 Mark

The integrating factor of the differential equation $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\ \log\text{x}$ is given by:

$\log(\log\text{x})$
$\text{e}^{\text{x}}$
$\log\text{x}$
$\text{x}$

Answer

$\log\text{x}$

Solution:
We have,
$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\ \log\text{x}$
Dividing both sides by,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}+\log\text{x}}=\frac{2}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{1}}{\text{x}+\log\text{x}}\Big)\text{y}=\frac{2}{\text{x}}$
Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{1}{\text{x}\log\text{x}}$
$\text{Q}=\frac{2}{\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\text{P}\text{dx}}$
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}}\text{dx}$
$=\text{e}^{\log(\log\text{x})}$
$=\log\text{x}$

View full question & answer→

Question 171 Mark

The integrating factor of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\text{x}^{2}.$

$\text{e}^{-\text{x}}$
$\text{e}^{-\text{y}}$
$\frac{1}{\text{x}}$
$\text{x}$

Answer

$\frac{1}{\text{x}}$

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

View full question & answer→

Question 181 Mark

The solution of the differention $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$ is:

$(\text{x}+\text{y})\text{e}^{\text{x}+\text{y}}=0$
$(\text{x}+\text{C})\text{e}^{\text{x}+\text{y}}=0$
$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}=1$
$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$

Answer

$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$
Let $\text{x}+\text{y}=\text{u}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+1=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{u}}$
$\Rightarrow \text{e}^{-\text{u}}\text{du}=\text{dx}$
Intergrating both sides, we get
$\Rightarrow \text{e}^{-\text{u}}=\text{x}-\text{C}$
$\Rightarrow -1=\text{e}^{-\text{u}}(\text{x}-\text{C})$
$\Rightarrow (\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$

View full question & answer→

Question 191 Mark

Integrating factor of the differntial equation $\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$is:

$\sin\text{x}$
$\sec\text{x}$
$\tan\text{x}$
$\cos\text{x}$

Answer

$\sec\text{x}$

Solution:
We have,
$\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$
Dividing both sides by we get
$\frac{\text{dy}}{\text{dx}}+\frac{\sin\text{x}}{\cos\text{x}}\text{y}=\frac{1}{\cos\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+(\tan\text{x})\text{y}=\frac{1}{\cos\text{x}}$
Comparing with $ \frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\tan\text{x}$
$\text{Q}=\frac{1}{\cos\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\tan\text{x}}\text{dx}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec\text{x}$

View full question & answer→

Question 201 Mark

The differential equation which respresents the famliy of curves $\text{y}=\text{e}^{\text{Cx}}$ is:

$\text{y}_{1}=\text{C}^{2}\text{y}$
$\text{xy}_{1}-\log\text{y}=0$
$\text{x}\log\text{y}=\text{yy}_{1}$
$\text{y}\log\text{y}=\text{xy}_{1}$

Answer

$\text{y}\log\text{y}=\text{xy}_{1}$

Solution:
We have,
$\text{y}=\text{e}^{\text{Cx}}$
Taking in both sides, we get
$\Rightarrow \log\text{y}=\text{Cx}\ ...(\text{1})$
Differentiating both sides of (i) with respect to x, we get
$\frac{1}{\text{y}_{1}}=\text{C}$
Substituting the value of C in in (i). we get
$\log\text{y}=\frac{\text{y}_{1}}{\text{y}}\text{x}$
$\Rightarrow \text{y}\ \log\text{y}=\text{y}_{1}\text{x}$

View full question & answer→

Question 211 Mark

The equation of the curve whose slope is given by $\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}};\text{x}>0,\text{y}>0$ and which passes through the point (1, 1) is:

$\text{x}^{2}=\text{y}$
$\text{y}^{2}=\text{x}$
$\text{x}^{2}=2\text{y}$
$\text{y}^{2}=2\text{x}$

Answer

$\text{x}^{2}=\text{y}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\Rightarrow\frac{1}{2}\times\frac{1}{\text{y}}\text{dy}=\frac{1}{\text{x}}\text{dx}$
Interating both sides, we get
$\Rightarrow\frac{1}{2}\int\frac{1}{\text{y}}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\frac{1}{2}\ \log{\text{y}}=\log{\text{x}}+\log\text{C}$
$\Rightarrow\log{\text{y}}^{\frac{1}{2}}-\log{\text{x}}=\log\text{C}$
$\Rightarrow\log\big(\frac{\sqrt{\text{y}}}{2}\big)=\log\text{C}$
$\Rightarrow\frac{\sqrt{\text{y}}}{2}=\text{C}$
$\Rightarrow\sqrt{\text{y}}=\text{Cx}\ ...(\text{i})$
As (i) passes through (1, 1), we get
$1=\text{C}$
Putting the value of C in (1), we get
$\Rightarrow\sqrt{\text{y}}=\text{x}$
$\Rightarrow{\text{y}}=\text{x}^{2}$

View full question & answer→

Question 221 Mark

The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\ \text{g}(\text{x})=\text{g}(\text{x})\ \text{g}'(\text{x})$ is a given function of x, is:

$\text{g}(\text{x})+\log(1+\text{y}+\text{g}(\text{x}))=\text{C}$
$\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
$\text{g}(\text{x})-\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
None of these.

Answer

$\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\ \text{g}'(\text{x})=\text{g}(\text{x})\ \text{g}'(\text{x})\ ...(\text{i})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\text{g}'(\text{x}), \text{Q}=\text{g}(\text{x})\ \text{g}'(\text{x})$
$\text{I.F}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\text{g}'(\text{x})\text{dx}}$
$=\text{e}^{\text{g}(\text{x})}$
Multiplying both sides, we get
$\text{e}^{\text{g}(\text{x})}\Big(\frac{\text{dy}}{\text{dx}}+\text{yg}'({\text{x}})\Big)=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})$
$\text{e}^{\text{g}(\text{x})}\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{g}(\text{x})}\text{yg}'({\text{x}})=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})$
Integrating both sides with respect to x, we get
$\text{y}\ \text{e}^{\text{g}(\text{x})}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}+\text{K}$
$\text{y}\ \text{e}^{\text{g}(\text{x})}=\text{I}+\text{K}$
Where, $\text{I}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}$
Now,
$\text{I}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}$
Putting $\text{g}'(\text{dx})=\text{dt}$
$\text{I}=\int\text{t}\ \text{e}^{\text{t}}\ \text{dt}$
$=\text{t}\int\ \text{e}^{\text{t}}\ \text{dt}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{t})\int\text{e}^{\text{t}}\ \text{dt}\Big]\text{dt}$
$=\text{t}\text{e}^{\text{t}}-\text{e}^{\text{t}}$
$=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}$
$\Rightarrow \text{y}\ \text{e}^{\text{g}(\text{x})}=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}+\text{K}$
$\Rightarrow \text{y}\ \text{e}^{\text{g}(\text{x})}=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}=\text{K}$
Taking log on both sides, we get
$\log\Big[\text{y+1}-\text{g}(\text{x})\Big]=-\text{g}(\text{x})+\log\text{K}$

View full question & answer→

Question 231 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:

$\text{y}^{2}=\text{exp}\big(\text{x}+\frac{\text{x}^{2}}{2}-1\big)$
$\text{y}^{2}=1+\text{C}\ \text{exp}\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$
$\text{y}=\tan (\text{C}+\text{x}+\text{x}^{2})$
$\text{y}=\tan\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$

Answer

$\text{y}=\tan\big(\text{x}+\frac{\text{x}^{2}}{2}\big)$

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),
$\Rightarrow \tan^{-1}\text{y}=\frac{\text{x}^{2}}{2}+\text{x}$
$\Rightarrow \text{y}=\tan\big(\frac{\text{x}^{2}}{2}+\text{x}\big)$

View full question & answer→

Question 241 Mark

Which of the following is the integrating factor of $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\log\text{x}?$

$\text{x}$
$\text{e}^{\text{x}}$
$\log\text{x}$
$\log(\log\text{x})$

Answer

$\log{\text{x}}$

Solution:
We have,
$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\log\text{x}$
Dividing both sides by, we get
$ \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=2\frac{\log\text{x}}{\text{x}\log\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{2}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\big(\frac{\text{1}}{\text{x}\log\text{x}}\big)\text{y}=\frac{2}{\text{x}}$
Comparing with we get,
$\text{P}=\frac{1}{\text{x}\log\text{x}}, \text{Q}=\frac{2}{\text{x}}$
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}\text{dx}}$
$=\text{e}^{\log({\log\text{x})}}$
$=\log\text{x}$

View full question & answer→

Question 251 Mark

The solution of the differention equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}$ is:

$\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)-\log\text{y}+\text{C}$
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
$\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)=\log\text{x}+\text{C}$
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$

Answer

$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}\ ...(\text{i})$
This is homogenous differential equation.
Let $\text{y}=\text{ux}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$
Now, putting equation (i),
$\text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\frac{\text{x}^{2}+\text{x}^{2}\text{u}+\text{x}^{2}\text{u}^{2}}{\text{x}^{2}}$
$\Rightarrow \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}+\text{u}^{2}$
$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}^{2}$
$\Rightarrow \big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\frac{1}{\text{x}}\text{dx}$
Intergreting both sides, we get
$\int\big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \tan^{-1}\text{u}=\log\text{x}+\text{C}$
$\Rightarrow \tan^{-1}\big(\frac{\text{y}}{2}\big)=\log\text{x}+\text{C}$

View full question & answer→

MCQ 261 Mark

The solution of the differential equartion $y_1y_3 = y_2$ is:

A
$\text{x}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{y}}+\text{C}_{3}$
B
$\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
C
$2\text{x}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{y}}+\text{C}_{3}$
✓
None of these.

Answer

Correct option: D.

None of these.

$\text{y}_{1}\text{y}_{3}=\text{y}^{2}_{2}$$\frac{\text{y}_{3}}{\text{y}_{2}}=\frac{\text{y}_{2}}{\text{y}_{1}}$
$\Rightarrow \frac{\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\frac{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$
$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$
$\Rightarrow\log\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)=\log\Big(\frac{\text{dy}}{\text{dx}}\Big)+\log\text{C}_{4}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}}^{2}=\text{C}_{4}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\text{C}_{4}\ \text{dx}$
$\Rightarrow \log\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{C}_{4}\text{x}+\text{C}_{5}$
$\Rightarrow \int\text{dy}=\int\text{e}^{\text{C}_{4}\text{x+}\text{C}_{5}}\ \text{dx}$
$\Rightarrow\text{y}=\frac{\text{e}^{\text{C}_{4}\text{x}+\text{C}_{3}}}{\text{C}_{4}}+\text{C}_{6}$
$\Rightarrow\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
Where,
$\text{C}_{1}=\frac{\text{e}^{\text{C}_{5}}}{\text{C}_{4}}$
$\text{C}_{4}=\text{C}_{2}$
$\text{C}_{6}=\text{C}_{3}$

View full question & answer→

Question 271 Mark

Integrating factor of the differential equation $\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$ is:

$\cos\text{x}$
$\tan\text{x}$
$\sec\text{x}$
$\sin\text{x}$

Answer

$\sec\text{x}$

Solution:
We have,
$\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$
Dividing both sides by, we get
$\frac{\text{dy}}{\text{dx}}+\frac{\sin\text{x}}{\cos\text{x}}\text{y}=\frac{1}{\cos\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+(\tan\text{x})\text{y}=\frac{1}{\cos\text{x}}$
Comparing with we get,
$\text{P}=\tan\text{x}, \text{Q}=\frac{2}{\cos\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\tan\text{x}\text{dx}}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec{\text{x}}$

View full question & answer→

Question 281 Mark

The differential equation of the ellipse $\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=\text{C}$ is:

$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}+\frac{1}{\text{x}}=0$
$\frac{\text{y}''}{\text{y}'}-\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
None of these.

Answer

$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$

Solution:
We have,
$\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=\text{C}\ ...(\text{i})$
Differentiating with respect to x, we get
$\frac{2\text{x}^{2}}{\text{a}^{2}}+\frac{2\text{y}^{2}}{\text{b}^{2}}\text{y}'=0$
$\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}\text{y}'=0\ ...(\text{ii})$
Again differentiating with respect to x, we get
$\Rightarrow \frac{1}{\text{a}^{2}}+\frac{1}{\text{b}^{2}}(\text{y'}^{2})+\frac{\text{xy}}{\text{b}^{2}}\text{y}''=0\ ...(\text{iii})$
Multiplying throughout by x, we get
$\Rightarrow \frac{\text{x}}{\text{a}^{2}}+\frac{\text{x}}{\text{b}^{2}}(\text{y'}^{2})+\frac{\text{xy}}{\text{b}^{2}}\text{y}''=0\ ...(\text{iv})$
Subtracting (ii) from (iv),
$\frac{1}{\text{b}^{2}}\big[\text{x}(\text{y}')^{2}+\text{xyy}''-\text{yy}''\big]=0$
$\Rightarrow \text{x}(\text{y}')^{2}+\text{xyy}''-\text{yy}''=0$
Diving both sides by,
$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
$\Rightarrow \frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$

View full question & answer→

Question 291 Mark

The differention equation $\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}},\text{n}>2$ can be reduced to linear from by substituting:

$\text{z}=\text{y}^{\text{n}-1}$
$\text{z}=\text{y}^{\text{n}}$
$\text{z}=\text{y}^{\text{n}+1}$
$\text{z}=\text{y}^{1-\text{n}}$

Answer

$\text{z}=\text{y}^{1-\text{n}}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}}$
$\Rightarrow \text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}+\text{Py}^{1-\text{n}}=\text{Q}\ ...(\text{i})$
Put $\text{z}=\text{y}^{1-\text{n}}$
Integrating both sides with respect to x, we get
$\frac{\text{dz}}{\text{dx}}=(1-\text{n})\text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}=\frac{1}{(1-\text{n})}\frac{\text{dz}}{\text{dx}}$
Now, (i),
$\frac{1}{(1-\text{n})}\frac{\text{dz}}{\text{dx}}+\text{Pz}=\text{Q}$
$\Rightarrow \frac{\text{dz}}{\text{dx}}+\text{P}(1-\text{n})=\text{Q}(1-\text{n})$
Which is linear from of differential equation.
Therefore the given differential equation can be to linear by the $\text{z}=\text{y}^{1-\text{n}}.$

View full question & answer→

Question 301 Mark

The degree of the differntial equation $\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{\frac{5}{3}}=\text{x}^{5}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)$ is:

4
2
5
10

Answer

Solution:
We have,
$\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{\frac{5}{3}}=\text{x}^{5}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)$
Taking cube power on both sides, we get
$\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{5}=\text{x}^{15}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{3}$
The highest order derivative is $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$ and its power is 3.
Hence, the degree is 3.
Disclaimer: The correct potion is not given in the quation.

View full question & answer→

Question 311 Mark

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$ is:

$\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
$\text{x}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
$\phi(\frac{\text{y}}{\text{x}})=\text{Ky}$
$\text{y}\phi(\frac{\text{y}}{\text{x}})=\text{K}$

Answer

$\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$
Let $\text{y}=\text{ux}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$
$\therefore \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\text{u}+\frac{\phi(\text{u})}{\phi'(\text{u})}$
$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=\frac{\phi(\text{u})}{\phi'(\text{u})}$
$\Rightarrow \frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$ \int\frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \log|\phi(\text{v})|=\log|\text{x}|+\log|\text{K}|$
$\Rightarrow \log|\phi(\frac{\text{y}}{2})|-\log|\text{x}|=\log\text{K}$
$\Rightarrow \log|\phi(\frac{\text{y}}{2})|=\log\text{K}$
$\Rightarrow\phi(\frac{\text{y}}{2})|=\text{Kx}$

View full question & answer→

MCQ 321 Mark

Which of the following is a homogeneous differnetial equation?

A
$(4x + 6y + 5)dy - (3y + 2x + 4)dx = 0$
B
$xy dx - (x^3 + y^3)dy = 0$
C
$(x^3 + 2y^2)dx + 2xy dy = 0$
✓
$y^2dx + (x^2 - xy - y^2) = 0$

Answer

Correct option: D.

$y^2dx + (x^2 - xy - y^2) = 0$

A differential equation is said to be homogenous if all the in the terms in the equation have equal degree and it can be written in the from $\frac{\text{dy}}{\text{dx}}=\frac{\text{f}(\text{x,}\text{y})}{\text{g}(\text{x,}\text{y})}.$In $(a), (b)$ and $(c),$ the degree of all the terms is not equal.
But in the equation $y^2 dx + (x^2- xy - y^2)dy = 0,$
the degree of all the terms is $2.$
Thus$, (d)$ constant a homogeneous differential equation.

View full question & answer→

Question 331 Mark

The number of arbitrary constants in the general solution of differential equation of fourth order is:

0
2
3
4

Answer

4

Solution:
The number of arbitray constant in the general solution of a differential equation of order n is n.
Thus, the number of arbitrary conatant tn the general solution of differential equation of fourth order is 4.

View full question & answer→

Question 341 Mark

The differential equation obtained on eliminating A and B from $\text{y}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}$ is:

$\text{y}''+\text{y}'=0$
$\text{y}''-\omega^{2}\text{y}=0$
$\text{y}''=-\omega^{2}\text{y}=0$
$\text{y}''+\text{y}=0$

Answer

$\text{y}''=-\omega^{2}\text{y}$

Solution:
We have,
$\text{y}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}\ ...(\text{i})$
Differentiating both sides of (i) with respect to x, we get
$\frac{\text{dy}}{\text{dt}}=-\text{A}\omega\sin\omega\text{t}+\text{B}\omega\cos\omega\text{t}\ ...(\text{ii})$
Differentiating both sides of (ii)
$\frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\text{A}\omega^{2}\cos\omega\text{t}+\text{B}\omega^{2}\sin\omega\text{t}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}(\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t})$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}\text{y}$
$\text{y}''=-\omega^{2}\text{y}$

View full question & answer→

Question 351 Mark

The order of the differential equartion $\sqrt{1-\text{x}^{4}}+\sqrt{1-\text{y}^{4}}=\text{a}(\text{x}^{2}-\text{y}^{2})$ is:

1
2
3
4

Answer

1

Solution:
The order of a differention depends on the number of constent in it.
Since $\sqrt{1-\text{x}^{4}}+\sqrt{1-\text{y}^{4}}=\text{a}(\text{x}^{2}-\text{y}^{2})$ constant only 1 constant, the order of the differential equation is 1.

View full question & answer→

Question 361 Mark

The general solution of differention eqution of the type $\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$ is:

$\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$
$\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
$\text{xe}^{\int\text{P}_{1}\text{dx}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$

Answer

$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$

Solution:
We have,
$\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$
Comparing with the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ we get,
$\text{P}=\text{P}_{1}, \text{Q}=\text{Q}_{1}$
The solution of the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ is given by
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}\ ...(\text{i})$
Putting the value of P and Q in (i),
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$

View full question & answer→

Question 371 Mark

The general solution of the dofferential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$ is:

$\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{C}$
$\text{e}^{-\text{x}}+\text{e}^{\text{y}}=\text{C}$
$\text{e}^{-\text{x}}+\text{e}^{-\text{y}}=\text{C}$

Answer

$\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\times\text{e}^{\text{y}}$
$\Rightarrow \text{e}^{-\text{y}}\text{dy}=\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^{\text{x}}\text{dx}$
$\Rightarrow \text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{C}$

View full question & answer→

MCQ 381 Mark

The famliy of curve in which the sub tangent at any point of a curve is double is the abscissae, is

✓
$x = Cy^2$
B
$y = Cx^2$
C
$x^2 = Cy^2$
D
$Y = Cx$

Answer

Correct option: A.

$x = Cy^2$

It is given that subtangent at any point of a curve is doble of the abscissa.
$\therefore \frac{\text{y}}{\frac{\text{dy}}{\text{dx}}}=2\text{x}$
$\text{y}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\int\frac{\text{dx}}{\text{x}}=2\int\frac{\text{dy}}{\text{y}}$
$\log\text{x}=2\log\text{y}+\text{a}$
$\log\text{x}=\log\text{y}^{2}+\log\text{C}$
$\log\text{x}=\log\text{Cy}^{2}$
$\text{x}=\text{Cy}^{2}$

View full question & answer→

Question 391 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?

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{1}=0$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{1}=0$

Answer

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-1=0$

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,
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{y}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$

View full question & answer→

Question 401 Mark

The solution of the differention equation $(1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}+1+\text{y}^{2}=0$ is:

$\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\text{C}$
$\tan^{-1}\text{y}-\tan^{-1}\text{x}=\tan^{-1}\text{C}$
$\tan^{-1}\text{y}\pm\tan^{-1}\text{x}=\tan^{-1}\text{C}$
$\tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$

Answer

$\tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$

Solution:
We have,
$(1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}+1+\text{y}^{2}=0$
$\Rightarrow (1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^{2})$
$\Rightarrow \frac{1}{(1+\text{y}^{2})}\text{dy}=-\frac{1}{(1+\text{x}^{2})}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{(1+\text{y}^{2})}\text{dy}=-\int\frac{1}{(1+\text{x}^{2})}\text{dx}$
$\Rightarrow \tan^{-1}\text{y}=-\tan^{-1}\text{x}+\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$

View full question & answer→

Question 411 Mark

The general solution of the differntial equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{coses}\ \text{x}$ is:

$\text{x}+\text{y}\sin\text{x}=\text{C}$
$\text{x}+\text{y}\cos\text{x}=\text{C}$
$\text{y}+\text{x}(\sin\text{x}+\cos\text{x})=\text{C}$
$\text{y}\sin\text{x}=\text{x}+\text{C}$

Answer

$\text{y}\sin\text{x}=\text{x}+\text{C}$

Solution:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{coses}\ \text{x}$
Comparting with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ we get
$\text{P}=\cot\text{x}$
$\text{Q}=\text{coses} \ \text{x}$
Now,
$\text{I.F}=\text{e}^{\int\cot\text{x}\text{dx}}$
$=\text{e}^{\log(\sin\text{x})}$
$=\sin\text{x}$
So, the solution is given by
$\Rightarrow \text{y}\sin\text{x}=\int\sin\text{x}\times\text{cosec}\ \text{x}\text{dx}+\text{C}$
$\Rightarrow \text{y}\sin\text{x}=\text{x}+\text{C}$

View full question & answer→

Question 421 Mark

The degree of the differential equation $2\text{x}^{2}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+3\frac{\text{dy}}{\text{dx}}+\text{y}=0$ is:

2
1
0
Not defined.

Answer

2

Solution:
We have,
$2\text{x}^{2}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+3\frac{\text{dy}}{\text{dx}}+\text{y}=0$
Here, the highest order is $\frac{\text{d}^{2}\text{y}}{\text{d}^{2}\text{x}}.$
Hence, the order is 2.

View full question & answer→

MCQ 431 Mark

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax}+\text{g}}{\text{by}+\text{f}}$ represents a circle when,

A
$\text{a}=\text{b}$
✓
$\text{a}=-\text{b}$
C
$\text{a}=-2\text{b}$
D
$\text{a}=2\text{b}$

Answer

Correct option: B.

$\text{a}=-\text{b}$

We have,$\frac{\text{dy}}{\text{dx}}=\frac{\text{ax}+\text{g}}{\text{by}+\text{f}}$
$\Rightarrow (\text{by}+\text{f})\text{dy}=(\text{ax}+\text{g})\text{dx}$
Intergrating both sides, we get
$\Rightarrow \int(\text{by}+\text{f})\text{dy}=\int(\text{ax}+\text{g})\text{dx}$
$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}=\text{a}\frac{\text{x}^{2}}{2}+\text{gx}+\text{C}$
$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}-\text{a}\frac{\text{x}^{2}}{2}-\text{gx}=\text{C}$
$\Rightarrow \text{b}\text{y}^{2}+2\text{fy}-\text{a}\text{x}^{2}-2\text{gx}-2\text{C}=0$
The above equation resprasents a circle.
Therefore, the coffrcients of $x^2$ and $y^2$ must be equal.
$-\text{a}=\text{b}$
$\Rightarrow \text{a}=-\text{b}$

View full question & answer→

Question 441 Mark

Solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$ is:

$\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$
$\text{x}(\text{y}-\cos\text{x})=\sin\text{x}+\text{C}$
$\text{x}(\text{y}+\cos\text{x})=\cos\text{x}+\text{C}$
None of these.

Answer

$\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}=\sin\text{x}\ ...(\text{i})$
Comparing with we get,
$\text{P}=\frac{1}{\text{x}}$
$\text{Q}=\sin\text{x}$
Now,
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log|\text{x}|}$
$=\text{x}$
Therefore, intergration of (i) is given by,
$\text{y}\times\text{I.F}=\int\text{x}^{2}\times\text{I.F.}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}=\int\text{x}\ \sin\text{x}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}=\text{x}\int\sin\text{x}\ \text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x}\ \text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{yx}=-\text{x}\cos\text{x}+\int\cos\text{x}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}+\text{x}\cos\text{x}=\sin\text{x}+\text{C}$
$\Rightarrow\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$

View full question & answer→

Question 451 Mark

The equation of the curve aatisfying the differential $\text{y}(\text{x}+\text{y}^{3})\text{dx}=\text{x}(\text{y}^{3}-\text{x})$ dy and passing through the point (1, 1) is:

$\text{y}^{3}-2\text{x}+3\text{x}^{2}\text{y}=0$
$\text{y}^{3}+2\text{x}+3\text{x}^{2}\text{y}=0$
$\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$
None of these.

Answer

$\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$

Solution:
We have,
$\text{y}(\text{x}+\text{y}^{3})\text{dx}=\text{x}(\text{y}^{3}-\text{x})\text{dy}$
$\Rightarrow (\text{xy}+\text{y}^{4})\text{dx}=(\text{xy}^{3}-\text{x}^{2})\text{dy}=0$
$\Rightarrow\text{xy}\ \text{dx}+\text{y}^{4}\text{dx}-\text{xy}^{3}\text{dy}+\text{x}^{2}\text{dy}=0$
$\Rightarrow \text{x}(\text{y}\text{dx}+\text{x}\text{dy})+\text{y}^{3}(\text{y}\text{dx}-\text{x}\text{dy})=0$
$\Rightarrow \text{xd}(\text{xy})+\text{x}^{2}\text{y}^{3}\frac{(\text{y}\text{dx}-\text{x}\text{dy})}{\text{x}^{2}}=0$
$\Rightarrow \frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}-\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)=0$
$\Rightarrow \frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}-\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)$
Integrating both sides we get,
$\Rightarrow \int\frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}=\int\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow-\frac{1}{\text{xy}}=\frac{\Big(\frac{\text{y}}{\text{x}}\Big)^{2}}{2}-\text{C}$
$\Rightarrow-\frac{1}{\text{xy}}-\frac{1}{2}\Big(\frac{\text{y}^{2}}{\text{x}^{2}}\Big)+\text{C}=0$
$\Rightarrow \text{y}^{3}+2\text{x}+2\text{Cx}^{2}\text{y}=0$
It is given that the curve passes through (1, 1).
Hence,
$\text{y}^{3}+2\text{x}+2\text{Cx}^{2}\text{y}=0$
$(1)^{3}+2(1)+2\text{C}(1)(1)=0$
$1+2+2\text{C}=0$
$2\text{C}=-3$
$\text{C}=-\frac{3}{2}$
The required curve is,
$\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$

View full question & answer→

MCQ 461 Mark

The solution of $\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$ is :

A
$x^2 + y^2 = 12x + C$
B
$x^2 + y^2 = 3x + C$
C
$x^3 + y^3 = 3x + C$
✓
$x^3 + y^3 = 12x + C$

Answer

Correct option: D.

$x^3 + y^3 = 12x + C$

We have,
$\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$
$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4-\text{x}^{2}$
$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=(4-\text{x}^{2})\text{dx}$
Integrating both sides, we get
$\int\text{y}^{2}\frac{\text{dy}}{\text{dx}}=\int(4-\text{x}^{2})\text{dx}$
$\Rightarrow \frac{\text{y}^{3}}{3}=4\text{x}-\frac{\text{x}^{3}}{3}+\text{D} $
$\Rightarrow \text{y}^{3}=12\text{x}-\text{x}^{3}+3\text{D}$
$\Rightarrow \text{x}^{3}+\text{y}^{3}=12\text{x}+\text{C}$

View full question & answer→

Question 471 Mark

Which of the following transformation reduce the differential quation into the form $\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$ into the from $\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}$

$\text{u}=\log\text{x}$
$\text{u}=\text{e}^{\text{z}}$
$\text{u}=(\log\text{z})^{-1}$
$\text{u}=(\log\text{z})^{2}$

Answer

$\text{u}=(\log\text{z})^{-1}$

Solution:
We have,
$\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}\ ...(\text{i})$
Let $\text{u}=(\log\text{z})^{-1}$
$\frac{\text{du}}{\text{dx}}=-\frac{1}{(\log)^{2}}\times\frac{1}{\text{z}}\times\frac{\text{dz}}{\text{dx}}$
$\frac{\text{du}}{\text{dx}}=-\text{z}(\log\text{z})^{2}\frac{\text{du}}{\text{dx}}$
Substituting the value of the equation (i),
$-\text{z}(\log\text{z})^{2}\frac{\text{du}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z}^{2})$
$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}\frac{1}{\log\text{z}}=-\frac{1}{\text{x}^{2}}$
$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}({\log\text{z}})^{-1}=-\frac{1}{\text{x}^{2}}$
$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}(\text{u})=-\frac{1}{\text{x}^{2}}$
It can be written as,
$\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$
Where, $\text{p}(\text{x})=-\frac{1}{\text{x}}$
$\text{q}(\text{x})=-\frac{1}{\text{x}^{2}}$

View full question & answer→

MCQ 481 Mark

The general solution of differention eqution $\frac{\text{y}\ \text{dx}-\text{x}\ \text{dy}}{\text{y}}=0$ is:

A
$xy = C$
B
$x = Cy^2$
✓
$y = Cx$
D
$y = Cx^2$

Answer

Correct option: C.

$y = Cx$

We have,$\frac{\text{y}\ \text{dx}-\text{x}\ \text{dy}}{\text{y}}=0$
$\Rightarrow \text{y}\ \text{dx}=\text{x}\ \text{dy}$
$\Rightarrow \frac{1}{\text{y}}\ \text{dy}=\frac{1}{\text{x}}\ \text{dx}$
Integrating both sides, we get,
$\int\frac{1}{\text{y}}\ \text{dy}=\int\frac{1}{\text{x}}\ \text{dx}$
$\Rightarrow \log\text{y}=\log\text{x}+\text{D}$
$\Rightarrow \log\text{y}-\log\text{x}=\text{C}$
$\Rightarrow \log\big(\frac{\text{y}}{2}\big)=\log\text{C}$
$\Rightarrow \frac{\text{y}}{2}=\text{C}$
$\Rightarrow \text{y}=\text{Cx}$

View full question & answer→

Question 491 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:

3
2
1
Not defined.

Answer

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.

View full question & answer→

Question 501 Mark

The general solution of the differntial equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$ is:

$\log\text{y}=\text{kx}$
$\text{y}=\text{kx}$
$\text{xy}=\text{k}$
$\text{y}=\text{k}\log\text{x}$

Answer

$\text{y}=\text{kx}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{1}}{\text{y}}\text{dy}=\frac{\text{1}}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{1}}{\text{y}}\text{dy}=\int\frac{\text{1}}{\text{x}}\text{dx}$
$\log\text{y}=\log\text{x}+\log\text{k}$
$\log\text{y}-\log\text{x}=\log\text{k}$
$\log\frac{\text{y}}{\text{x}}=\log\text{k}$
$\Rightarrow\frac{\text{y}}{\text{x}}=\text{k}$
$\Rightarrow\text{y}=\text{k}{\text{x}}$

View full question & answer→

Generate Paper Free All Differential Equations topics

M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip