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 options.

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ is:

  1. $\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$

  2. $\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$

  3. $\text{y}\log(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$

  4. $\text{y}(1+\text{x}^2)=\text{C}+\sin^{-1}\text{x}$

✓

Answer

  1. $\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$

Solution:

Given is, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$

Here, $\text{P}=\frac{2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\frac{1}{(1+\text{x}^2)^2}$

This ia a linerar differential equation.

$\therefore\text{I.F.}=\text{e}^{\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$

Put $1+\text{x}^2=\text{t}\Rightarrow2\text{x}\text{ dx}=\text{dt}$

$\therefore\text{I.F.}=\text{e}^{\int\frac{\text{dt}}{\text{t}}}=\text{e}^{\log\text{t}}$

$=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2$

Thus, the general solution is

$\text{y}.(1+\text{x}^2)=\int(1+\text{x}^2)\frac{1}{(1+\text{x}^2)}+\text{C}$

$\Rightarrow\text{y}(1+\text{x}^2)=\int\frac{1}{(1+\text{x}^2)}\text{dx}+\text{C}$

$\Rightarrow\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}+\text{C}$

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

The points $A\,(a),\,B\,(b),\,C\,(c)$ will be collinear if
→
If ${D_p} = \left| {\,\begin{array}{*{20}{c}}p&{15}&8\\{{p^2}}&{35}&9\\{{p^3}}&{25}&{10}\end{array}\,} \right|$, then ${D_1} + {D_2} + {D_3} + {D_4} + {D_5} = $
→
For any positive integer $n$, let $S_n:(0, \infty) \rightarrow R$ be defined by

$S_n(x)=\sum_{k=1}^n \cot ^{-1}\left(\frac{1+k(k+1) x^2}{x}\right)$

where for any $x \in R , \cot ^{-1} x \in(0, \pi)$ and $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then which of the following

statements is (are) $TRUE$?

$(A)$ $S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^2}{10 x }\right)$, for all $x >0$

$(B)$ $\lim _{n \rightarrow \infty} \cot \left(S_n(x)\right)=x$, for all $x>0$

$(C)$ The equation $S_3(x)=\frac{\pi}{4}$ has a root in $(0, \infty)$

$(D)$ $\tan \left( S _{ n }( x )\right) \leq \frac{1}{2}$, for all $n \geq 1$ and $x >0$

→
If $h(a) = h(b),$ the value of the integral$\int_a^b {{{[f(g(h(x)))]}^{ - 1}}f'(g(h(x)))\,g'(h(x))\,h'(x)\,dx = } $
→
$A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right];\,\,I = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$${A^{ - 1}} = \frac{1}{6}[{A^2} + cA + dI]$ where $c,d \in R$, then pair of values $(c,d)$
→
$\int_{\,0}^{\,3} {|2 - x|dx} $ equals
→
If $\int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} \,dx = k\pi ,} $ then $k = $
→
The area bounded by the curve $4 y^{2}=x^{2}(4-x)(x-2)$ is equal to ...... .
→
If A and B are two events associated to a random experiment such that $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P(B)}=\frac{17}{20}$, then P(A|B) =
  1. $\frac{14}{17}$
  2. $\frac{17}{20}$
  3. $\frac{7}{8}$
  4. $\frac{1}{8}$
→
If the multiplicative group of $2 × 2$ matrices of the form $\left( {\begin{array}{*{20}{c}}a&a\\a&a\end{array}} \right)$, for $a \ne 0$ and $a \in R$, then the inverse of $\left( {\begin{array}{*{20}{c}}2&2\\2&2\end{array}} \right)$ is
→