MCQ
Integrating factor of the differential equation $\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$ is:
- A$\cos\text{x}$
- B$\tan\text{x}$
- ✓$\sec\text{x}$
- D$\sin\text{x}$
✓
Answer
Correct option: C.
$\sec\text{x}$
We have,
$\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$
Dividing both sides by, we get
$\frac{\text{dy}}{\text{dx}}+\frac{\sin\text{x}}{\cos\text{x}}\text{y}=\frac{1}{\cos\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+(\tan\text{x})\text{y}=\frac{1}{\cos\text{x}}$
Comparing with we get,
$\text{P}=\tan\text{x}, \text{Q}=\frac{2}{\cos\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\tan\text{x}\text{dx}}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec{\text{x}}$
$\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$
Dividing both sides by, we get
$\frac{\text{dy}}{\text{dx}}+\frac{\sin\text{x}}{\cos\text{x}}\text{y}=\frac{1}{\cos\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+(\tan\text{x})\text{y}=\frac{1}{\cos\text{x}}$
Comparing with we get,
$\text{P}=\tan\text{x}, \text{Q}=\frac{2}{\cos\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\tan\text{x}\text{dx}}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec{\text{x}}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $\angle A = {90^o}$ in the triangle ABC, then ${\tan ^{ - 1}}\left( {\frac{c}{{a + b}}} \right) + {\tan ^{ - 1}}\left( {\frac{b}{{a + c}}} \right) = $
→The differential equation of all circles passing through the origin and having their centres on the $x-$axis is
→$\int_{}^{} {\frac{1}{{x{{(\log x)}^2}}}} \;dx = $
→Let $f(x)=x \mid \sin x |, x \in R$. Then,
→The feasible region of an LPP is given in the following figure
Then, the constraints of the LPP are $x \geq 0, y \geq 0$ and
→Then, the constraints of the LPP are $x \geq 0, y \geq 0$ and
If $\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{bx}}\Big)^\text{x}$ then $\text{x}^3\text{y}_2=$
→Solve : $\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\ \text{dx}}=$
→Given that $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ and $A^2=31,$ then
→A particle is moving in a straight line according as $s = 45\,t + 11{t^2} - {t^3}$ then the time when it will come to rest, is ......... $\sec$.
→If $3{\sin ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}$ then $x$ =
→