M.C.Q (1 Marks)

MCQ 511 Mark

If $\log _e\left(1+\frac{d^2 y}{d x^2}\right)=x$, then find the sum of order and degree of given differential equation.

A
2
B
4
✓
3
D
5

Answer

Correct option: C.

(c) : Given differential equation is $\log _e\left(1+\frac{d^2 y}{d x^2}\right)=x$
$\Rightarrow 1+\frac{d^2 y}{d x^2}=e^x$
Here, highest order derivative is $\frac{d^2 y}{d x^2}$, whose power is 1
$\therefore \quad$ Its order is 2 and degree is 1 .
$\therefore \quad$ Required sum $=2+1=3$.

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MCQ 521 Mark

If $\frac{d y}{d x}=y \sin 2 x, y(0)=1$, then solution is

✓
$y=e^{\sin ^2 x}$
B
$y=\sin ^2 x$
C
$y=\cos ^2 x$
D
$y=e^{\cos ^2 x}$

Answer

Correct option: A.

$y=e^{\sin ^2 x}$

(a) : We have, $\frac{d y}{d x}=y \sin 2 x$
$
\Rightarrow \quad \frac{d y}{y}=\sin 2 x d x \Rightarrow \log y=-\frac{\cos 2 x}{2}+C
$
Since $y(0)=1 \Rightarrow x=0, y=1 \Rightarrow C=1 / 2$
$
\therefore \quad \log y=\frac{1}{2}(1-\cos 2 x) \Rightarrow \log y=\sin ^2 x \Rightarrow y=e^{\sin ^2 x}
$

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MCQ 531 Mark

Integrating factor of the differential equation $\left(1-x^2\right) \frac{d y}{d x}-x y=1$ is

A
$-x$
B
$\frac{x}{1+x^2}$
✓
$\sqrt{1-x^2}$
D
$\frac{1}{2} \log \left(1-x^2\right)$

Answer

Correct option: C.

$\sqrt{1-x^2}$

(c): $\left(1-x^2\right) \frac{d y}{d x}-x y=1 \Rightarrow \frac{d y}{d x}-\frac{x}{1-x^2} \cdot y=\frac{1}{1-x^2}$
$\therefore \quad$ I.F. $=e^{-\int \frac{x}{1-x^2} d x}=e^{\frac{1}{2} \int \frac{-2 x}{1-x^2} d x}$
$=e^{\frac{1}{2} \log \left(1-x^2\right)}=e^{\log \left(1-x^2\right)^{\frac{1}{2}}}=\sqrt{1-x^2}$

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MCQ 541 Mark

The degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^2=x \sin \frac{d y}{d x}$ is

A
1
B
2
C
3
✓
not defined

Answer

Correct option: D.

not defined

(d) : Since, the given differential equation is not a poly-nomial in $\frac{d y}{d x}$. Therefore, its degree is not defined.

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MCQ 551 Mark

The solution of the differential equation $\frac{d y}{d x}=\frac{x^2+1}{2 x y}$ satisfying $y(1)=1$, is

✓
$y^2=\frac{x^2}{2}+\ln |x|+\frac{1}{2}$
B
$y^2=\frac{x^2}{2}-\ln |x|+\frac{1}{2}$
C
$y^2=x(1+x)-10$
D
$(x-2)^2+(y-3)^2=5 x y$

Answer

Correct option: A.

$y^2=\frac{x^2}{2}+\ln |x|+\frac{1}{2}$

(a) : Given, $\frac{d y}{d x}=\frac{x^2+1}{2 x y}$
$
\Rightarrow 2 y d y=\frac{\left(x^2+1\right)}{x} d x \Rightarrow 2 y d y=\frac{(x+1)}{x} d x
$
Integrating on both sides, we get
$
\begin{aligned}
& y^2=\frac{x^2}{2}+\ln |x|+C \\
\Rightarrow \quad & y^2=\frac{x^2}{2}+C
\end{aligned}
$
When $x=1, y=1$
$
\therefore \quad C=\frac{1}{2} \quad \therefore \quad y^2=\frac{x^2}{2}+\ln |x|+\frac{1}{2}
$

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MCQ 561 Mark

The integrating factor for solving the differential equation $x \frac{d y}{d x}-y=2 x^2$ is

A
$e^{-y}$
B
$e^{-e}$
C
$x$
✓
$\frac{1}{x}$

Answer

Correct option: D.

$\frac{1}{x}$

(d) : We have, $x \frac{d y}{d x}-y=2 x^2$
i.e., $\frac{d y}{d x}-\frac{y}{x}=2 x \quad \therefore \quad$ I.F. $=e^{\int \frac{-1}{x} d x}=e^{-\ln x}=e^{\ln x^{-1}}=\frac{1}{x}$
$\therefore$ Integrating factor is $\frac{1}{x}$.

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MCQ 571 Mark

The differential equation having solution as $y=17 e^x+a e^{-x}$ is

A
$y^{\prime \prime}-x=0$
✓
$y^{\prime \prime}-y=0$
C
$y^{\prime}-y=0$
D
$y^{\prime}-x=0$

Answer

Correct option: B.

$y^{\prime \prime}-y=0$

(b): We have, $y=17 e^x+a e^{-x} \Rightarrow y^{\prime}=17 e^x-a e^{-x}$
$
\Rightarrow y^{\prime \prime}=17 e^x+a e^{-x} \Rightarrow y^{\prime \prime}=y \Rightarrow y^{\prime \prime}-y=0
$

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MCQ 581 Mark

Which of the following is the general solution of the differential equation $\frac{d y}{d x}=\frac{y}{x}$ ?

A
$y=\frac{k}{x}$
B
$y=k \log x$
✓
$y=k x$
D
$\log y=k x$

Answer

Correct option: C.

$y=k x$

(c) : $\frac{d y}{d x}=\frac{y}{x} \Rightarrow \frac{d x}{x}=\frac{d y}{y}$
$
\Rightarrow \log x=\log y+\log C \Rightarrow x=y C \Rightarrow y=\frac{1}{C} x \Rightarrow y=k x \text {. }
$

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MCQ 591 Mark

The integrating factor of the differential equation $\left(x+3 y^2\right) \frac{d y}{d x}=y$ is

A
$y$
B
$-y$
✓
$\frac{1}{y}$
D
$-\frac{1}{y}$

Answer

Correct option: C.

$\frac{1}{y}$

(c) : We have, $\left(x+3 y^2\right) \frac{d y}{d x}=y$
$
\Rightarrow \frac{x+3 y^2}{y}=\frac{d x}{d y} \Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y
$
This is a linear differential equation.
$
\therefore \quad \text { I.F. }=e^{-\int \frac{d y}{y}}=e^{-\log y}=e^{\log y^{-1}}=\frac{1}{y}
$

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Maths M.C.Q (1 Marks)