2 Marks Questions

Question 12 Marks

Find the general solution of the differential equation $\left(1+x^2\right) d y=\left(1+y^2\right) d x$.

Answer

The given differential equation can be written in the following form:
$
\frac{d y}{d x}=\frac{1+y^2}{1+x^2}
$
$\because 1+y^2, 1+x^2 \neq 0$, Therefore separating the variables, the given differential equation can be written in the following form:
$
\frac{d y}{1+y^2}=\frac{d x}{1+x^2}
$
Integrating
$
\begin{aligned}
\int \frac{d y}{1+y^2} & =\int \frac{d x}{1+x^2} \\
\text { or } \quad \tan ^{-1} y & =\tan ^{-1} x+C
\end{aligned}
$
This is the general solution of the differential equation.

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Question 22 Marks

Find the solution of the following differential equation :
$
x \sqrt{1+y^2} d x+y \sqrt{1+x^2} d y=0
$

Answer

$
x \sqrt{1+y^2} d x+y \sqrt{1+x^2} d y=0
$
or$\qquad$$
x \sqrt{1+y^2} d x=-y \sqrt{1+x^2} d y
$
or$\qquad$$
\frac{x d x}{\sqrt{1+x^2}}=\frac{-y d y}{\sqrt{1+y^2}}
$
On integrating $\int \frac{x d x}{\sqrt{1+x^2}}=-\int \frac{y d y}{\sqrt{1+y^2}}$
Let $1+x^2=u^2$ and $1+y^2=v^2$
$\therefore 2 x d x=2 u d u$ and $2 y d y=2 v d v$
$\Rightarrow x d x=u d u$ and $y d y=v d v$
Now
$
\int \frac{u d u}{u}=-\int \frac{v d v}{v}
$
$
\Rightarrow \quad u=-v+c_1
$
Putting the values $\sqrt{1+x^2}=-\sqrt{1+y^2}+c_1$
$
\Rightarrow \quad \sqrt{1+x^2}+\sqrt{1+y^2}=c_1
$

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Question 32 Marks

For the following differential equation find the sum of its order and degree :
$
y=x\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}
$

Answer

The highest order derivative present in the differential equation is $\frac{d^2 y}{d x^2}$ and therefore its order is 2 . This differential equation is a polynomial equation in $\frac{d^2 y}{d x^2}$ and $\frac{d y}{d x}$ and the highest exponent of $\frac{d^2 y}{d x^2}$ is 1 , therefore the degree of this differential equation is 1 . Hence the sum of order and degree $=2+1=3$.

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MATHS 2 Marks Questions

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