MCQ
If $\text{xy}-\log_\text{e}\text{y}=1$ satisfies the equation $\text{x}(\text{yy}_2+\text{y}_1^2)-\text{y}_2+\lambda\text{yy}_1=0,$ then $\lambda=$
  • A
    $-3$
  • B
    $1$
  • $3$
  • D
    None of these

Answer

Correct option: C.
$3$
$\text{xy}-\log_\text{e}\text{y}=1$
$\Rightarrow\text{xy}_1+\text{y}-\frac{\text{y}_1}{\text{y}}=0$
$\Rightarrow\text{xyy}_1+\text{y}^2-\text{y}_1=0$
$\Rightarrow\text{yy}_1+\text{xy}_1\text{y}_1+\text{xyy}_2+2\text{yy}_1-\text{y}_2=0$
$\Rightarrow\text{x}(\text{y}_1^2+\text{yy}_2)-\text{y}_2+3\text{yy}_2=0$
$\therefore\lambda=3$

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 differention equation $\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}},\text{n}>2$ can be reduced to linear from by substituting:
The value of the determinant $\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{nx}&\cos(\text{n}+1)\text{x}&\cos(\text{n}+2)\text{x}\\\sin\text{nx}&\sin(\text{n}+1)\text{x}&\sin(\text{n}+2)\text{x}\end{vmatrix}$ is independent of:
The area of figure bounded by $y = {e^x},\,y = {e^{ - x}}$ and the straight line $x = 1$ is
A line in the $3-$ dimensional space makes an angle $\theta \left( {0 < \theta  \le \frac{\pi }{2}} \right)$ with both the $x$ and $y$ axes. Then the set ofall values of $\theta $ is the interval
The minimum distance between a point on the curve $y=e^x$ and a point on the curve $y=\log _e x$ is
The value of the integral $\int_0^{\pi / 2} \frac{3 \sqrt{\cos \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$ equals
Let $\vec u = a\hat i + b\hat j + c\hat k$  , $\vec v = b\hat i + c\hat j + a\hat k\,\,$  $\vec w = c\hat i + a\hat j + b\hat k = \lambda \vec x + \mu \vec y$ where $\left[ {\vec u\,\,\vec v\,\,\vec w} \right] = 0\,\  \,\,\left( {a + b + c} \right),\,\,\lambda ,\mu  \ne 0$ then the vectors $\vec x,\vec y,\vec u,\vec v,\vec w$ are
If A and B are square matrices of order 2, then det (A + B) = 0 is possible only when:
The area bounded by the curve $y=x^4-2 x^3+x^2+3$ with $x-$ axis and ordinates corresponding to the minima of $y$ is :
Let the function $f$ be defined by $f(x) = \frac{{2x + 1}}{{1 - 3x}}$, then ${f^{ - 1}}(x)$ is