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.
Solution of the differential equation $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$is:
  1. $\tan\text{x}+\tan\text{y}=\text{k}$
  2. $\tan\text{x}-\tan\text{y}=\text{k}$
  3. $\frac{\tan\text{x}}{\tan\text{y}}=\text{k}$
  4. $\tan\text{x}.\tan\text{y}=\text{k}$
✓

Answer

  1. $\tan\text{x}.\tan\text{y}=\text{k}$

Solution:

We have, $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$

$\Rightarrow\tan\text{y}\sec^2\text{xdx} =- \tan\text{x }\sec^2\text{ydy}$

$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$

$\Rightarrow\int\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\int\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$

$\Rightarrow\log\tan\text{x}=-\log\tan\text{y}+\log\text{k}$

$\Rightarrow\log\tan\text{x}+\log\tan\text{y}=\log\text{k}$

$\Rightarrow\log(\tan\text{x}\tan\text{y})=\log\text{k}$

$\Rightarrow\tan\text{x}\tan\text{y}=\text{k}$

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

Let $a, b, c \in R$ be all non-zero and satisfy $a^{3}+b^{3}+c^{3}=2 .$ If the matrix $A=\left(\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right)$ satisfies $\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I},$ then a value of $abc$ can be
→
A point from a vector starts is called and where it ends is called its:
  1. Terminal point, endpoint.
  2. Initial point, terminal point
  3. Origin, endpoint
  4. Initial point, endpoint
→
Derivative of ${x^6} + {6^x}$ with respect to $x $ is
→
Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ be three zero vectors such that $\vec{\text{c}}$ is a unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$ If the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6},$ then $\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$ is equal to:
  1.  $0$
  2. $1$
  3. $\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
  4. $\frac{3}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
→
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio $8: 15$ is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is $100$ , the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are

$(A)$ $24$ $(B)$ $32$ $(C)$ $45$ $(D)$ $60$

→
Let $f: R \rightarrow R$ be a function defined $f(x)=\frac{2 e^{2 x}}{e^{2 x}+\varepsilon}$. Then $f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+\ldots .+f\left(\frac{99}{100}\right)$ is equal to
→
$\left( \vec{a}\times \vec{b} \right)\times \left[ \left( \vec{b}\times \vec{c} \right)\times \left( \vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a} \right) \right]$ is
→
For real $x$ with $-10 \leq x \leq 10$ define $f(x)=\int_{-10}^x 2^{[t]} d t$ where for a real number $r$ we denote by $[r]$ the largest integer less than or equal to $r$. The number of points of discontinuity of $f$ in the interval $(-10,10)$ is
→
Let $f(x)=\left|(x-1)\left(x^{2}-2 x-3\right)\right|+x-3, x \in R$. If $m$ and $M$ are respectively the number of points of local minimum and local maximum of $f$ in the interval $(0,4)$, then $m + M$ is equal to
→
If AB = A and BA = B, where A and B are square matrices, then:
  1. B2 = B and A2 = A
  2. B2 ≠ B and A2 = A
  3. A2 ≠ A, B2 = B
  4. A2 ≠ A, B2 ≠ B
→