Choose the correct answer from the given four options. The genera… - Vidyadip

Question

Choose the correct answer from the given four options.

The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\ \text{e}^{\frac{\text{x}^2}{2}}+\text{xy}$ is:

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

✓

Answer

$\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$

Solution:

Given that, $\frac{\text{dy}}{\text{dx}}\ \text{e}^{\frac{\text{x}^2}{2}}+\text{xy}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}-\text{xy}=\text{e}^{\frac{\text{x}^2}{2}}$

It is a linear differential equation.

Here, $\text{P}=-\text{x},\text{ Q}=\text{e}^{\frac{\text{x}^2}{2}}$

$\therefore\text{I.F.}=\text{e}^{\int-\text{xdx}}=\text{e}^{\frac{-\text{x}^2}{2}}$

The general solution is

$\text{y}\text{e}^{\frac{-\text{x}^2}{2}}=\int\text{e}^{\frac{-\text{x}^2}{2}}.\text{e}^{\frac{-\text{x}^2}{2}}\text{dx}+\text{c}$

$\Rightarrow\text{y}{\text{e}}^{\frac{-\text{x}^2}{2}}=\int1\text{dx}+\text{c}$

$\Rightarrow\text{y}\text{e}^{\frac{-\text{x}^2}{2}}=\text{x}+\text{c}$

$\Rightarrow\text{y}=\text{x}\text{e}^{\frac{\text{x}^2}{2}}+\text{c}\text{e}^{\frac{\text{x}^2}{2}}$

$\Rightarrow\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from DIFFERENTIAL EQUATIONS→DIFFERENTIAL EQUATIONS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int_{}^{} {x{{\tan }^{ - 1}}} xdx = $

Let $f(x)=x^3+a x^2+b x+c$ where $a, b, c$ are real numbers. If $f(x)$ has a local minimum at $x=1$ and a local maximum at $x=-\frac{1}{3}$ and $f(2)=0$, then $\int_{-1}^1 f(x) d x$ equals

The left hand derivative of $f(x) = \{x\} \, sin(\pi x)$ at $x = k$ ( $k$ is an integer), is $(\{ \}$ denotes the fraction part function $)$

Three integers are chosen at random from the first 20 integers. The probability that their product is even is,

The area of a trapezium is defined by function $f$ and given by $f(x)=(10+x) \sqrt{100-x^2}$, then the area when it is maximised is

If the function f(x) = x³ - 9kx² + 27x + 30 is increasing on R, then:

$-1\leq\text{k}\leq1$
k < -1 or k > 1
0 < k < 1
-1 < k < 0

$A (2,6,2), B (-4,0, \lambda), C (2,3,-1)$ and $D (4,5,0)$, $|\lambda| \leq 5$ are the vertices of a quadrilateral $A B C D$. If its area is $18$ square units, then $5-6 \lambda$ is equal to $.........$.

Let $S=\left\{\left(\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right) ; a, b \in\{1,2,3, \ldots 100\}\right\}$ and let $T_{n}=\left\{A \in S: A^{n(n+1)}=I\right\}$. Then the number of elements in $\bigcap \limits_{n=1}^{100} T_{n}$ is

The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for x = 2. in order to make f(x) continuous at x = 2, here f(2) should be defined as:

0
1
2
3

Let $S=\left\{E, E_{2} \ldots . E_{8}\right\}$ be a sample space of random experiment such that $P\left(E_{n}\right)=\frac{n}{36}$ for every $n =1,2 \ldots .$. Then the number of elements in the set $\left\{ A \subset S : P ( A ) \geq \frac{4}{5}\right\}$ is