Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
Question
Choose the correct answer from the given four option.
The integrating factor of the differential equation $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\frac{1+\text{y}}{\text{x}}$ is:
  1. $\frac{\text{x}}{\text{e}^{\text{x}}}$
  2. $\frac{\text{e}^{\text{x}}}{\text{x}}$
  3. ${\text{x}}\text{e}^{\text{x}}$
  4. $\text{e}^{\text{x}}$
✓

Answer

  1. $\frac{\text{e}^{\text{x}}}{\text{x}}$

Solution:

We have, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\frac{1+\text{y}}{\text{x}}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}}{\text{x}}-\text{y}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}-\text{xy}}{\text{x}}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1}{\text{x}}+\frac{\text{y}(1-\text{x})}{\text{x}}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}-\Big(\frac{1-\text{x}}{\text{x}}\Big)$

$\Rightarrow\text{y}=\frac{1}{\text{x}}$

Here, $\text{P}=\frac{-(1-\text{x})}{\text{x}},\text{Q}=\frac{1}{\text{x}}$

$\text{I.F.}=\text{e}^{\int\text{Pdx}}$

$=\text{e}^{-\int\frac{1+\text{x}}{\text{x}}\text{dx}}=\text{e}^{\frac{\text{x}-1}{\text{x}}}$

$=\text{e}^{{\int\Big(1-\frac{1}{\text{x}}\Big)}\text{dx}}=\text{e}^{\int\text{x}-\log\text{x}}$

$=\text{e}^{\text{x}}.\text{e}^{\log\Big(\frac{1}{\text{x}}\Big)}=\text{e}^{\text{x}}.\frac{1}{\text{x}}$

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

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right| = $
→
If $\int {\frac{{2{x^2} + 3}}{{({x^2} - 1)\,\,({x^2} + 4)}}dx = a\log \left( {\frac{{x - 1}}{{x + 1}}} \right) + } \,b{\tan ^{ - 1}}\frac{x}{2} + c$, then values of $a$ and $b$ are
→
If $\left[ {a \times b\;\;b \times c\;\;c \times a} \right] = \alpha \;{\left[ {a\;\;b\;\;c} \right]^2}$ then $\lambda$ is equal to 
→
The matrix $\left( {\begin{array}{*{20}{c}}1&a&2\\1&2&5\\2&1&1\end{array}} \right)$ is not invertible, if  $‘a’ $ has the value
→
If $A$ is nilpotent matrix of index $2$, then $A(I_2+A)^{51}$ is equal to (where $I_2$ is the identity matrix of order $2$)
→
Let $\alpha$ be a non-zero real number. Suppose $f: \mathrm{R} \rightarrow$ $\mathrm{R}$ is a differentiable function such that $f(0)=2$ and $\lim _{\mathrm{x} \rightarrow-\infty} \mathrm{f}(\mathrm{x})=1$. If $f^{\prime}(\mathrm{x})=\alpha f(x)+3$, for all $\mathrm{x} \in \mathrm{R}$, then $f\left(-\log _e 2\right)$ is equal to . . . . . . . . . 
→
The area of a triangle is $5$ and two of its vertices are $A(2, 1), B(3, -2)$. The third  vertex which lies on line $y = x + 3$ is-
→
If $A$ is a square matrix of order 3 such that the value of $|\operatorname{adj} A|=8$, then the value of $\left|A^{\top}\right|$ is
→
If S is the samle space and $\text{P(A)}=\frac{1}{3}, \text{P(B)}$ and $\text{S}=\text{A}\cup\text{B,}$ where A and B are tow mutually exclusive events, then P(A) =
  1. $\frac{1}{4}$
  2. $\frac{1}{2}$
  3. $\frac{3}{4}$
  4. $\frac{3}{8}$
→
For $\mathrm{a}, \mathrm{b}>0$, let $f(x)=\left\{\begin{array}{l}\frac{\tan ((a+1) x)+b \tan x}{x}, x<0 \\ \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, x>0\end{array}\right.$ be a continous function at $x=0$. Then $\frac{b}{a}$ is equal to
→