MCQ

MCQ 511 Mark

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

A
$\log\text{y}=\text{kx}$
✓
$\text{y}=\text{kx}$
C
$\text{xy}=\text{k}$
D
$\text{y}=\text{k}\log\text{x}$

Answer

Correct option: B.

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

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
$\int\frac{\text{1}}{\text{y}}\text{dy}=\int\frac{\text{1}}{\text{x}}\text{dx}$
$\log\text{y}=\log\text{x}+\log\text{k}$
$\log\text{y}-\log\text{x}=\log\text{k}$
$\log\frac{\text{y}}{\text{x}}=\log\text{k}$
$\Rightarrow\frac{\text{y}}{\text{x}}=\text{k}$
$\Rightarrow\text{y}=\text{k}{\text{x}}$

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MCQ 521 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.

✓
$\text{y}=\frac{1}{\text{x}^{2}}$
B
$\text{x}=\frac{1}{\text{y}^{2}}$
C
$\text{x}=\frac{1}{\text{y}}$
D
$\text{y}=\frac{1}{\text{x}}$

Answer

Correct option: A.

$\text{y}=\frac{1}{\text{x}^{2}}$

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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MCQ 531 Mark

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

A
$3$
B
$2$
C
$1$
✓
$0$

Answer

Correct option: D.

$0$

The number of arbitray constant in the particular solution of a differential equation is always zero.

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MCQ 541 Mark

The integrating factor of the differential equation$(1-\text{y}^{2})\frac{\text{dx}}{\text{dy}}+\text{yx}=\text{ay}(-1<\text{y}<1)$ is:

A
$\frac{1}{\text{y}^{2}-1}$
B
$\frac{1}{\sqrt{\text{y}^{2}+1}}$
C
$\frac{1}{1-\text{y}^{2}}$
✓
$\frac{1}{\sqrt{1-\text{y}^{3}}}$

Answer

Correct option: D.

$\frac{1}{\sqrt{1-\text{y}^{3}}}$

We have,
$(1-\text{y}^{2})\frac{\text{dx}}{\text{dy}}+\text{yx}=\text{ay}$
$\frac{\text{dx}}{\text{dy}}+\frac{\text{y}}{1-\text{y}^{2}}\text{x}=\frac{\text{ay}}{1-\text{y}^{2}}$
Comparing with we get,
$\text{P}=\frac{\text{y}}{1-\text{y}^{2}}, \text{Q}=\frac{\text{ay}}{1-\text{y}^{2}}$
Now,
$\text{I.F}=\text{e}^{\int\frac{\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\int\frac{-2\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\log|1-\text{y}^{2}|}$
$=\text{e}^{\log\Big|\frac{1}{\sqrt{1-\text{y}^{2}}}\Big|}$
$=\frac{1}{\sqrt{1-\text{y}^{2}}}$

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Maths MCQ