Question
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$ is:
  1. $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
  2. $\text{x}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
  3. $\phi(\frac{\text{y}}{\text{x}})=\text{Ky}$
  4. $\text{y}\phi(\frac{\text{y}}{\text{x}})=\text{K}$ 

Answer

  1. $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$

Solution:

We have,

$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$

Let $\text{y}=\text{ux}$

$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$

$\therefore \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\text{u}+\frac{\phi(\text{u})}{\phi'(\text{u})}$

$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=\frac{\phi(\text{u})}{\phi'(\text{u})}$

$\Rightarrow \frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\frac{1}{\text{x}}\text{dx}$

Integrating both sides, we get

$ \int\frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow \log|\phi(\text{v})|=\log|\text{x}|+\log|\text{K}|$

$\Rightarrow \log|\phi(\frac{\text{y}}{2})|-\log|\text{x}|=\log\text{K}$

$\Rightarrow \log|\phi(\frac{\text{y}}{2})|=\log\text{K}$

$\Rightarrow\phi(\frac{\text{y}}{2})|=\text{Kx}$

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

$f: N \rightarrow N$, is defined by $f(x)=x^6$ then, ___________ .
If $ x=-1 $ and $ x=2 $ are extreme point of $f\left( x \right) = \alpha \log \left| x \right| + \beta {x^2} + x$ then $\left( {\alpha ,\beta } \right)$ 
Let $\text{f(x)}=\text{x}+\text{b}|\text{x}|+\text{c}|\text{x}|^4,$ where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if:
  1. a = 0
  2. b = 0
  3. c = 0
  4. None of these.
For which value of $x$, are the determinants $\left|\begin{array}{ll}2 x & -3 \\ 5 & x\end{array}\right|$ and $\left|\begin{array}{cc}10 & 1 \\ -3 & 2\end{array}\right|$ equal?
The domain of the function

$f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} \text { is }$

$\int {{e^{{x^2}}}}  \cdot {e^x}\left( {2{x^2} + x + 1} \right)dx = {e^{{x^2}}}\left( {f\left( x \right)} \right) + c$ where $c$ is constant of integration. If the minimum value of $f(x) $ is equal to $'m'$ then find the value of $\left[ { - \frac{1}{m}} \right]$ , $[·]$ denotes $[GIF]$ functions
If $\overrightarrow {{F_1}} = i - j + k,$ $\overrightarrow {{F_2}} = - i + 2j - k,$ $\overrightarrow {{F_3}} = j - k,$ $\vec A = 4i - 3j - 2k$ and $\vec B = 6i + j - 3k,$ then the scalar product of $\overrightarrow {{F_1}} + \overrightarrow {{F_2}} + \overrightarrow {{F_3}} $and $\overrightarrow {AB} $ will be
Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x},$ where $b$ is a constant such that $0 < b < 1$. Then
Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem (using simplex), we find that.
  1. The values of decision variables obtained by rounding off are always very close to the optimal values.
  2. The value of the objective function for a maximization problem will likely be less than that for the simplex solution.
  3. The value of the objective function for a minimization problem will likely be less than that for the simplex solution.
  4. All constraints are satisfied exactly.
  5. None of the above.
The value of c in Rolle's theorem when
f(x) = 2x3 - 5x2 - 4x + 3, $\text{x}\in\Big[\frac{1}{3},3\Big]$ is:
  1. $2$
  2. $-\frac{1}{3}$
  3. $-2$
  4. $\frac{2}{3}$