MCQ
If $\frac{d y}{d x}=y \sin 2 x, y(0)=1$, then solution is
  • $y=e^{\sin ^2 x}$
  • B
    $y=\sin ^2 x$
  • C
    $y=\cos ^2 x$
  • D
    $y=e^{\cos ^2 x}$

Answer

Correct option: A.
$y=e^{\sin ^2 x}$
(a) : We have, $\frac{d y}{d x}=y \sin 2 x$
$
\Rightarrow \quad \frac{d y}{y}=\sin 2 x d x \Rightarrow \log y=-\frac{\cos 2 x}{2}+C
$
Since $y(0)=1 \Rightarrow x=0, y=1 \Rightarrow C=1 / 2$
$
\therefore \quad \log y=\frac{1}{2}(1-\cos 2 x) \Rightarrow \log y=\sin ^2 x \Rightarrow y=e^{\sin ^2 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

Similar questions

The value of $\int_0^1 {{x^2}{e^x}dx} $ is equal to
Let $D$ be the domain of the function $f(x)=\sin ^{-1}$ $\left(\log _{3 x}\left(\frac{6+2 \log _3 x}{-5 x}\right)\right)$. If the range of the function $g: D \rightarrow R$ defined by $g( x )= x -[ x ],([ x ]$ is the greatest integer function), is ( $\alpha, \beta)$, then $\alpha^2+\frac{5}{\beta}$ is equal to
If A and B are two events such that $\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{4}.$ and $\text{P}(\text{A}|\text{B})\times\text{P}(\overline{\text{A}}\cap\text{B})$ is equals to.
Let $f : R -\{0,1\} \rightarrow R$ be a function such that $f(x)+f\left(\frac{1}{1-x}\right)=1+x$. Then $f(2)$ is equal to :
If $a \times r = b + \lambda a$ and $a\,\,.\,\,r = 3,$ where $a = 2i + j - k$ and $b = - i - 2j + k,$ then $r$ and $\lambda$ are equal to
Let f : R - {n} → R be a function defined by $\text{f(x)}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}},$ where $\text{m}\neq\text{n.}$ Then,
Let $Q$ and $R$ be two points on the line $\frac{ x +1}{2}=\frac{ y +2}{3}=\frac{ z -1}{2}$ at a distance $\sqrt{26}$ from the point $P (4,2,7)$. Then the square of the area of the triangle $PQR$ is $....$
If $\theta$ is an acute angle and the vector $(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}$ is perpendicular to the vector $\hat{\text{i}}-\sqrt{3}\hat{\text{j}},$
If $y = 2x + {\cot ^{ - 1}}\,x + \log \left( {\sqrt {1 + {x^2}}  - x} \right),$ then $y$
The integrating factor of the differential equation$(1-\text{y}^{2})\frac{\text{dx}}{\text{dy}}+\text{yx}=\text{ay}(-1<\text{y}<1)$ is: