Download App

DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS1 Mark
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:
  1. $\text{y}=\text{c}\text{e}^{\frac{-\text{x}^2}{2}}$
  2. $\text{y}=\text{c}\text{e}^{\frac{\text{x}^2}{2}}$
  3. $\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
  4. $\text{y}=(\text{c}-\text{x})\text{e}^{\frac{\text{x}^2}{2}}$
✓

Answer

  1. $\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
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→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

The derivative of function $\cos (\sin x)$ is :
→
$\int\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}=$
  1. $-\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
  2. $-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
  3. $-\frac{1}{2}\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
  4. $-\frac{1}{2}\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
→
Let $\text{A}=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ be such that $A^{-1} = kA,$ then $k$ equals:
→
A rea bounded by the circle $x^2 + y^2 = 1$ and the curve $| x | + | y | = 1$ is$:$
→
The vector equation of a line which passes through the point $(2,-4,5)$ and is parallel to the line $\frac{x+3}{3}=\frac{4-y}{2}=\frac{z+8}{6}$ is :
→
The distance of the point (-1, -5, -10) from the point of intersection of the line $\vec{\text{r}}.=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$ and the plane $\vec{\text{r}}.=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$ is:
  1. 9
  2. 13
  3. 17
  4. None of these
→
If $\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}-1=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix},$ then:
  1. $\text{a}=1,\text{b}=1$
  2. $\text{a}=\cos2\theta,\text{b}=\sin2\theta$
  3. $\text{a}=\sin2\theta,\text{b}=\cos2\theta$
  4. None of these.
→
The function $f(x)=\left\{\begin{array}{cc}|x-3|, & x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1\end{array}\right.$ is
→
Choose the correct answer from the given four options.
Find the value of $\lambda$ such that the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ are orthogonal:
  1. $0$
  2. $1$
  3. $\frac{3}{2}$
  4. $-\frac{5}{2}$
→
The matrix $\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$ is:
  1. A skew-symmetric matrix.
  2. A symmetric matrix.
  3. A diagonal matrix.
  4. An uppertriangular matrix.
→