Fill In The Blanks[1 Marks ]

Question 11 Mark

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

Answer

The degree of the differential equation $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{e}^{\frac{\text{dy}}{\text{dx}}}=0$ is not defined. Solution:
Given differential equation is $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{e}^{\frac{\text{dy}}{\text{dx}}}=0$
Degree of this equation is not defined.

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Question 21 Mark

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The solution of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2$ is ________.

Answer

The solution of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2$ is $\text{y}=\frac{\text{x}^4}{4}+\text{C}\text{x}^{-2}.$Solution:
We have, $\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=\text{x}$
This equation the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{2}{\text{x}}\text{dx}}$
$=\text{e}^{2\log\text{x}}=\text{x}^2$
The general solution is
$\text{y}\text{x}^2=\int\text{x}.\text{x}^2\text{dx}+\text{C}$
$\Rightarrow\text{y}^{\text{x}^2}=\frac{\text{x}^4}{4}+\text{C}$
$\Rightarrow\text{y}=\frac{\text{x}^4}{4}+\text{C}\text{x}^{-2}$

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Question 31 Mark

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The degree of the differential equation $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\text{x}$ is _________.

Answer

The degree of the differential equation $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\text{x}$ is two.
Solution:
Given differential equation is $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\text{x}$
So, degree of this equation is two.

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Question 41 Mark

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General solution of the differential equation of the type $\frac{\text{dy}}{\text{dx}}+\text{P}_1\text{x}=\text{Q}_1$ is given by _________.

Answer

General solution of the differential equation of the type $\frac{\text{dy}}{\text{dx}}+\text{P}_1\text{x}=\text{Q}_1$ is given by $\text{x}\text{e}^{\int{\text{P}_1\text{dy}}}=\int\text{Q}_1\text{e}^{\int\text{P}_1\text{dy}}\text{dy}+\text{C}.$Solution:
The general solution is $\text{x}\text{ I.F.}=\int\text{Q}(\text{I.F.})\text{dy}+\text{C}\ ...(\text{i.e.})$ $\text{xe}\int^{\text{Pdy}}=\int\text{Q}\left\{\text{e}^{\int\text{Pdy}}\right\}\text{dy}+\text{C}$ $\text{x}\text{e}^{\int{\text{P}_1\text{dy}}}=\int\text{Q}_1\text{e}^{\int\text{P}_1\text{dy}}\text{dy}+\text{C}$

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Question 51 Mark

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The solution of $(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+2\text{xy}-4\text{x}^2=0$ is _________.

Answer

Given differential equation is $(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+2\text{xy}-4\text{x}^2=0$
Dividing both sides by $(1 + x^2),$ we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2\text{x}}{1+\text{x}^2}-\frac{4\text{x}^2}{1+\text{x}^2}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2\text{x}}{1+\text{x}^2}\text{y}=\frac{4\text{x}^2}{1+\text{x}^2}$
$\therefore\text{I.F.}=\text{e}^{\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
Put $1+\text{x}^2=\text{t}\Rightarrow2\text{xdx}=\text{dt}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{\text{dt}}{\text{t}}}=\text{e}^{\log\text{t}}$
$\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2$
The general solution is
$\text{y}.(1+\text{x}^2)=\int(1+\text{x}^2)\frac{4\text{x}^2}{(1+\text{x}^2)}\text{dx}+\text{C}$
$\Rightarrow(1+\text{x}^2)\text{y}=\int4\text{x}^2\text{dx}+\text{C}$
$\Rightarrow(1+\text{x}^2)\text{y}=4\frac{\text{x}^3}{3}+\text{C}$
$\Rightarrow\text{y}=\frac{4\text{x}^3}{3(1+\text{x}^2)}+\text{C}(1+\text{x}^2)^{-1}$

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Question 61 Mark

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The number of arbitrary constants in the general solution of a differential equation of order three is ________.

Answer

The number of arbitrary constants in the general solution of a differential equation of order three is three.
Solution:
The number of arbitrary constants in a solution of a differential equation of order n is equal to its order.
Now, the given differential equation is of order three.
Therefore the number of arbitrary constants in the general solution of given differential equation is three.

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Question 71 Mark

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General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ is _________.

Answer

General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ is $\text{y}=\frac{1}{2}(\sin\text{x}-\cos\text{x})+\text{C}\text{e}^{-\text{x}}.$Solution:
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ Which is of the from $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{I.F.}=\int\text{e}^{1\text{dx}}=\text{e}^{\text{x}}$ So, the general solution is $\text{y}.\text{e}^{\text{x}}=\int\text{e}^{\text{x}}\sin\text{x dx}+\text{C}$ $\Rightarrow\text{y}.\text{e}^{\text{x}}=\frac{1}{2}\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})+\text{C}$ $\Rightarrow\text{y}=\frac{1}{2}(\sin\text{x}-\cos\text{x})+\text{C}\text{e}^{-\text{x}}$

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Question 81 Mark

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The solution of differential equation $\cot\text{y dx}=\text{x dy} $ is _________.

Answer

The solution of differential equation $\cot\text{y dx}=\text{x dy} $ is $\text{x}=\text{C}\sec\text{ y}.$Solution:
Given differential equation is $\cot\text{y dx}=\text{x dy} $ $\Rightarrow\frac{1}{\text{x}}\text{dx}=\tan\text{y dy}$ On integrating both sides, we get $\Rightarrow\int\frac{1}{\text{x}}\text{dx}=\int\tan\text{y dy}$ $\Rightarrow\log(\text{x})=\log(\sec\text{y})+\log\text{C}$ $\Rightarrow\log\Big(\frac{\text{x}}{\sec\text{y}}\Big)=\log\text{C}$ $\Rightarrow\frac{\text{x}}{\sec\text{y}}=\text{C}$ $\Rightarrow\text{x}=\text{C}\sec\text{y}$

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Question 91 Mark

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The integrating factor of $\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{1+\text{y}}{\text{x}}$ is ________.

Answer

The integrating factor of $\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{1+\text{y}}{\text{x}}$ is $\text{e}^{\text{x}}.\text{e}^{-\log\text{x}}=\frac{\text{e}^{\text{x}}}{\text{x}}.$Solution:
$\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{1+\text{y}}{\text{x}}$ $\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{1}{\text{x}}+\frac{\text{y}}{\text{x}}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}+\text{y}\Big(1-\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}$ $\therefore\text{I.F.}=\text{e}^{\Big(1-\frac{1}{\text{x}}\Big)\text{dx}}$ $=\text{e}^{\text{x}-\log\text{x}}$ $=\text{e}^{\text{x}}.\text{e}^{-\log\text{x}}=\frac{\text{e}^{\text{x}}}{\text{x}}$

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Question 101 Mark

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The solution of the differential equation $\text{ydx}+(\text{x}+\text{xy})\text{dy}=0$ is ________.

Answer

The solution of the differential equation $\text{ydx}+(\text{x}+\text{xy})\text{dy}=0$ is $\text{xy}=\text{C}\text{e}^{-\text{y}}.$
Solution:
We have, $\text{ydx}+(\text{x}+\text{xy})\text{dy}=0$
$\Rightarrow\text{ydx}+\text{x}(1+\text{y})\text{dy}=0$
$\Rightarrow\frac{\text{dx}}{-\text{x}}=\Big(\frac{1+\text{y}}{\text{y}}\Big)\text{dy}$
$\Rightarrow\int\frac{1}{\text{x}}\text{dx}=-\int\Big(\frac{1}{\text{y}}+1\Big)\text{dy}$
$\Rightarrow\log\text{x}=-\log\text{y}-\text{y}+\log\text{C}$
$\Rightarrow\log\text{x}+\log\text{y}-\log\text{C}=-\text{y}$
$\Rightarrow\log\frac{\text{xy}}{\text{C}}=-\text{y}$
$\Rightarrow\frac{\text{xy}}{\text{C}}=\text{e}^{-\text{y}}$
$\Rightarrow\text{xy}=\text{C}\text{e}^{-\text{y}}$

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Question 111 Mark

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$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ is an equation of the type _________.

Answer

$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ is an equation of the type $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}.$
Solution:
The equation is of the type $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}.$

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Question 121 Mark

Degree of differential equation $\frac{d^2 y}{d x^2}+\sin \left(\frac{d y}{d x}\right)=0$ is ________

Answer

Not defined

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