MCQ
The solution of the differential equation $\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{\phi \,\left( {\frac{y}{x}} \right)}}{{\phi '\,\left( {\frac{y}{x}} \right)}}$ is
  • $\phi \,\left( {\frac{y}{x}} \right) = kx$
  • B
    $x\,\phi \,\left( {\frac{y}{x}} \right) = k$
  • C
    $\phi \,\left( {\frac{y}{x}} \right) = ky$
  • D
    $y\,\phi \left( {\frac{y}{x}} \right) = k$

Answer

Correct option: A.
$\phi \,\left( {\frac{y}{x}} \right) = kx$
a
(a) $\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{\phi \,\left( {\frac{y}{x}} \right)}}{{\phi '\,\left( {\frac{y}{x}} \right)}}$. Put $y = vx$ ==> $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

 The given differential equation becomes

$v + x\frac{{dv}}{{dx}} = v + \frac{{\phi \,(v)}}{{\phi '\,(v)}}$ ==> $\frac{{\phi '(v)}}{{\phi (v)}}dv = \frac{{dx}}{x}$

==> $\log \phi (v) = \log x + \log k$ ==> $\phi (v) = kx$ ==> $\phi \,\left( {\frac{y}{x}} \right) = kx$.

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

If a, b, c are position vectors of the vertices of a $\Delta\text{ABC}$ then $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=$
If $R = \{(6, 6), (9, 9), (6, 12), (12, 12), (12,6)\}$ is a relation on set $A = \{3, 6, 9, 12\}$ , then relation $R$ is
Function $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ then Domain of $f (x)$ is
Choose the correct answer from given four options in each of the Exercise : If $A, B$ and $C$ are angles of a triangle, then the determinant $\begin{vmatrix}-1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&-1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&-1\end{vmatrix}$ is equal to :
Let ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx,\,{I_2} = \int_a^{\pi - a} {\,\,f(\sin x)dx} } $, then ${I_2}$ is equal to
If in a right angled triangle $ABC$, the hypotenuse $AB = p,$ then $\overrightarrow {AB} \,\,.\,\,\overrightarrow {AC} + \overrightarrow {BC} \,.\,\,\overrightarrow {BA} + \overrightarrow {CA} \,\,.\,\,\overrightarrow {CB} $ is equal to
The point$(s),$ at which the function $f$ given by $f(x)=\left\{\begin{array}{l}\frac{x}{|x|}, x<0 \\ -1, x \geq 0\end{array}\right.$ is continuous, is$/$are
The vector$(s)$ which is/are coplanar with vectors $\hat{i}+\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}+\hat{k}$, and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$ is/are

$(A)$ $\hat{j}-\hat{k}$ $(B)$ $-\hat{i}+\hat{j}$ $(C)$ $\hat{i}-\hat{j}$ $(D)$ $-\hat{j}+\hat{k}$

If $\int_0^{2a} {f(x)\,dx = 2\int_0^a {f(x)\,dx,} } $ then
The solution of the equation $\frac{d y}{d x}+2 y=4 x$ is :